Re: [PHP] Select box won't display in Netscape 4.xx
Have you double and triple checked to be sure your form elements are nested properly? [EMAIL PROTECTED] wrote: >Hi, > >I'm developing a website for multiple browsers. In Netscape 4.xx (both >Win98 and Linux versions), the php-scripts display the select boxes in >my forms only as plain text. I cannot make any selections. In other >browsers (NS 6 and IE5), it works fine. > >Anyone seen this behavior before? > >Regards, >Edwin > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] Using a HTML button
JAVASCRIPT VERSION
mytitle
function multiply(x){
var myresult = 2 * x;
document.write(myresult);
}
PHP VERSION
mytitle
if($mybutton){
$myresult = 2 * $mybutton;
print $myresult;
}else{
print "Click the button to see what 2 mulitplied by 5 is!";
}
You can do it either way, don't take my code at face "value" (sorry,
couldn't resist) I haven't tested it, but it should be enough to get you
going. I would suggest that you use the PHP version, that way you don't
have to worry about cross browser JavaScript compatibility.
>> > Hi,
>> >
>> > I am rather new to PHP so I hope somebody can help me.
>> >
>> > I have a normal button (HTML code). Then I have made a function
>> >
>> > function multiply($X)
>> > {
>> > return 2*$X;
>> > }
>> >
>> > The button has an OnClick, which calls my function.
>> >
>> >
>> > I then writes $X to the screen.
>> >
>> >
>> >
>> > The problem is I dont even have to press the button. When the page
>> loads it
>> > already has calculated 2*$X. Why is that. I would like to only call my
>> > function when I press my button.
>>
>> What value is it returning 5, or 10?
>>
>> Nick Wilson
>>
>> Tel:+45 3325 0688
>> Fax:+45 3325 0677
>> Web:www.explodingnet.com
>
>
>
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[PHP] mysql_fetch_row win32 to Linux
Hi everybody, I have some scripts running on win32 Apache mySQL and they
all run great. I have copied the entire application to my Linux box
which I think I have running pretty smoothly now. The database is there
and I can access it just fine etc etc.
Here is the hitch (there's always a hitch) I can't get PHP to retrieve
my data after a SELECT query. I know the query works simply by setting
and displaying flags. It seems to choke on mysql_fetch_row. Here is an
example.
Here is a snippet I have written
$myquery = "SELECT * FROM xpackage";
print "myquery = $myquery";
$myresult = mysql_query($myquery);
while($myrowz = mysql_fetch_row($myresult)){
print "" . $myrowz[0] . "";
}
On the win32 box (identical snippet and identical db on both machines)
I get this back
myquery = SELECT * FROM xpackage
62
61
60
59
58
57
56
55
54
53
115
114
127
126
111
125
On the Linux box I get this
myquery = SELECT * FROM xpackage
Warning: Supplied argument is not a valid MySQL result resource in /path
to page with error/default.php on line 5
Line 5 is this part
while($myrowz = mysql_fetch_row($myresult)){
I am at a total loss, I have not seen anywhere stating that
mysql_fetch_row is depreciated, does anybody have anything to add to
this to help me out?
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Re: [PHP] help with include
I think you can use a URL when you specify your value in the include statement, can you not? [EMAIL PROTECTED] wrote: >Hi, I need some help with the include function. >I have a phpBB script and want to put the config.inc.php file in a password >protected directory, Easy peasy you might say, and I know it is, but the >problem comes because the password protected directory isn't on the site the >script is on, mainly because the host says I've got to upgrade my account to >use password protected dir's. >I've got one site with a password protected dir available and would like to >put it on there. > >Any idea's how to do it would be appreciated. > >Many thanks > >Brian > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] mysql_fetch_row win32 to Linux
Like this
$myquery = "SELECT * FROM xpackage";
print "myquery = $myquery";
$myresult = mysql_query($myquery);
while($myrowz = mysql_fetch_row($myresult)){
print "" . $myrowz[0] . "";
}
[EMAIL PROTECTED] wrote:
>on 1/12/02 11:45 AM, sundogcurt at [EMAIL PROTECTED] wrote:
>
>>myquery = SELECT * FROM xpackage
>>Warning: Supplied argument is not a valid MySQL result resource in /path
>>to page with error/default.php on line 5
>>
>>Line 5 is this part
>>while($myrowz = mysql_fetch_row($myresult)){
>>
>
>this tells me that $myresult is not a valid mysql result resource
>identifier. how do you obtain $myresult?
>
>
> -- mike cullerton
>
>
>
[PHP] SELECT DAYOFYEAR(NOW())
Hi! I'm storing a date in this format :
2002-01-13 11:08:40
-mm-dd hh:mm:ss
This is great because it gives me the date and the time, which I need.
I want to compare this date from the database to the current DAYOFYEAR()
but when I do this
select DAYOFYEAR('2002-01-13 11:08:40')
I get this
Error - Perl execution failed
Modification of non-creatable array value attempted, subscript -1 at ./mysql-lib.pl
line 187, line 2.
It works great when I do select DAYOFYEAR(NOW()) though, I'm totally
confused, is this a known bug or is there a work around that anybody
knows of that you can let me in on?
Ulitmately I want to compare todays DAYOFYEAR to the DAYOFYEAR from
fields in the database to establish entries that were created "today".
I'm looking at what I have written here and I think I have confused
myself, if this is not clear, please feel free to flame me.
I'm looking forward to any light that can be shed on this problem.
Re: [PHP] SELECT DAYOFYEAR(NOW())
I am a dork!
And you answered that almost before I sent it, you're like Mighty Mouse
on Ginseng!
[EMAIL PROTECTED] wrote:
>This doesn't appear to be a PHP question. Are you sure you are on the
>right list?
>
>On Sun, 13 Jan 2002, sundogcurt wrote:
>
>>Hi! I'm storing a date in this format :
>>
>>2002-01-13 11:08:40
>>-mm-dd hh:mm:ss
>>
>>This is great because it gives me the date and the time, which I need.
>>I want to compare this date from the database to the current DAYOFYEAR()
>>
>>but when I do this
>>select DAYOFYEAR('2002-01-13 11:08:40')
>>
>>I get this
>>
>>
>> Error - Perl execution failed
>>
>>Modification of non-creatable array value attempted, subscript -1 at ./mysql-lib.pl
>line 187, line 2.
>>
>>
>>
>>It works great when I do select DAYOFYEAR(NOW()) though, I'm totally
>>confused, is this a known bug or is there a work around that anybody
>>knows of that you can let me in on?
>>
>>Ulitmately I want to compare todays DAYOFYEAR to the DAYOFYEAR from
>>fields in the database to establish entries that were created "today".
>>
>>I'm looking at what I have written here and I think I have confused
>>myself, if this is not clear, please feel free to flame me.
>>
>>I'm looking forward to any light that can be shed on this problem.
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>
Re: [PHP] PHP as a cron job
I know this is a total hack job but could you say, set the location of the script as your default page in a browser, then open the browser via cron and it would run it as the page opens? I am interested in this too, I am VERY new to Linux and have only just started to look at cron, any links to SIMPLE tutorials or something similar would be appreciated. [EMAIL PROTECTED] wrote: >on 1/16/02 11:26 AM, Tiago Luchini at [EMAIL PROTECTED] wrote: > >>>Unable to open /home/admin/webmin-0.85/ in Unknown on line 0 >>> >this is possibly a path or env issue. have you tried using absolute >filenames? > Yes. Same thing happens with absolute or relative paths. >>>what's the line in your code where this error occurs? >>> >>it doesn't run any line at all. don't matter what I put on my code, it just >>says "unable to open" >> >> > >what's the first line of the file being called by cron? is it > > #!/usr/local/bin/php -q > > > > -- mike cullerton michaelc at cullerton dot com > > >
Re: [PHP] RTFM
I've been (sorta) following this thread for the last little while and I
have to say that I'm on the side of "one list for all!". The problem
with splitting the list is the different levels of "newbie-ism" on
different subjects, There are things that advanced users can learn from
new comers to the language too.
For instance. I don't really consider myself a "total newb" when it
comes to PHP anymore, but there are areas of the language that I still
haven't touched upon, which would make me a newbie. I agree though that
some questions are asked a little too often, and that could be due to
unclear documentation. If it is bothersome to you, write a nice simple
tutorial and post it!
For example, quite often I see people ask how to INSERT multiple values
from one selectbox. I have yet to see anybody post any help save for
telling people to use checkboxes instead. Using a selectbox for this
isn't all that difficult at all!
//FIRST YOU START WITH A SELECTBOX
//NOTE THE "[]" AT THE END OF THE NAME FOR THE SELECTBOX, ONCE SUBMITTED
THIS WILL BE TREATED AS AN ARRAY CONTAINING THE SELECTED VALUES.
//THE "MULTIPLE" STATEMENT ALLOWS THE USER TO SELECT MULTIPLE OPTIONS
INSTEAD OF JUST ONE.
0
1
2
/// AFTER THE FORM IS SUBMITTED ///
//GET A COUNT ON THE NUMBER OF ENTRIES SELECTED
$mycount = count($selectpackage);
//USE THE RESULT FOR A COUNTER
while($mycount){
$mycount--;
//RUN YOUR INSERT COMMAND UNTIL THE COUNTER HAS EXPIRED
mysql_query("INSERT INTO mytable (myvalue1,myvalue2) VALUES
('$selectpackage[$mycount]', '$mycount')");
}
Maybe some of use should get together and start putting things like this
in a repository with an easy flexible search. I know from experience
that some of the documentation that is available to new users is a bit
on the stuffy side. Sometimes a better approach is to give up some
"starter code" and let them go from there.
Just my two cents worth and a mini tutorial that can be used to get
things started.
Now if you'll excuse me, I have to continue banging my forehead on my
keyboard figuring out this "%$#%$#@" file upload. (C:
We all benefit from "new and used" users alike.
~Curt~
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Re: [PHP] RTFM
I would be more than happy to put some mini-tutorials together to send
you, I'm not sure of the relevance of them though. Maybe adding a
"suggest a tut" area where people that are having trouble can request a
tutorial and anybody interested can fire one in?
Just a thought
~Curt~
[EMAIL PROTECTED] wrote:
>I've actually been working on just that. I've been reading this list for a
>long time and noticed the repetition. I have 65+ faqs on my site
>http://www.php-faq.com/ that are there basically from my own experiences.
>I haven't posted it yet, because I don't think it's "full" enough. It even
>has a suggest-a-faq page, that'd I'd need to activate. I do like your idea
>of mini-tutorials. I have a lot of programming experience (12+ years),
>mostly in business applications, and know what it's like to struggle. The
>mailing list is awesome; I've never really posted a question because the
>answers are in the archive.
>
>- Original Message -
>From: "sundogcurt" <[EMAIL PROTECTED]>
>To: "GENERAL PHP LIST" <[EMAIL PROTECTED]>
>Sent: Saturday, January 19, 2002 12:56 PM
>Subject: Re: [PHP] RTFM
>
>
>>I've been (sorta) following this thread for the last little while and I
>>have to say that I'm on the side of "one list for all!". The problem
>>with splitting the list is the different levels of "newbie-ism" on
>>different subjects, There are things that advanced users can learn from
>>new comers to the language too.
>>
>>For instance. I don't really consider myself a "total newb" when it
>>comes to PHP anymore, but there are areas of the language that I still
>>haven't touched upon, which would make me a newbie. I agree though that
>>some questions are asked a little too often, and that could be due to
>>unclear documentation. If it is bothersome to you, write a nice simple
>>tutorial and post it!
>>
>>For example, quite often I see people ask how to INSERT multiple values
>>from one selectbox. I have yet to see anybody post any help save for
>>telling people to use checkboxes instead. Using a selectbox for this
>>isn't all that difficult at all!
>>
>>
>>file://FIRST YOU START WITH A SELECTBOX
>>
>>
>>file://NOTE THE "[]" AT THE END OF THE NAME FOR THE SELECTBOX, ONCE
>>
>SUBMITTED
>
>>THIS WILL BE TREATED AS AN ARRAY CONTAINING THE SELECTED VALUES.
>>
>>file://THE "MULTIPLE" STATEMENT ALLOWS THE USER TO SELECT MULTIPLE OPTIONS
>>INSTEAD OF JUST ONE.
>>
>>0
>>1
>>2
>>
>>
>>/// AFTER THE FORM IS SUBMITTED ///
>>
>>file://GET A COUNT ON THE NUMBER OF ENTRIES SELECTED
>>$mycount = count($selectpackage);
>>
>>file://USE THE RESULT FOR A COUNTER
>>while($mycount){
>>$mycount--;
>>
>>file://RUN YOUR INSERT COMMAND UNTIL THE COUNTER HAS EXPIRED
>>mysql_query("INSERT INTO mytable (myvalue1,myvalue2) VALUES
>>('$selectpackage[$mycount]', '$mycount')");
>>}
>>
>>
>>
>>Maybe some of use should get together and start putting things like this
>>in a repository with an easy flexible search. I know from experience
>>that some of the documentation that is available to new users is a bit
>>on the stuffy side. Sometimes a better approach is to give up some
>>"starter code" and let them go from there.
>>
>>Just my two cents worth and a mini tutorial that can be used to get
>>things started.
>>
>>Now if you'll excuse me, I have to continue banging my forehead on my
>>keyboard figuring out this "%$#%$#@" file upload. (C:
>>
>>We all benefit from "new and used" users alike.
>>~Curt~
>>
>>
>>--
>>PHP General Mailing List (http://www.php.net/)
>>To unsubscribe, e-mail: [EMAIL PROTECTED]
>>For additional commands, e-mail: [EMAIL PROTECTED]
>>To contact the list administrators, e-mail: [EMAIL PROTECTED]
>>
>>
>
[PHP] ftp functions
Before I get flamed for not reading the manual or trying on my own, I
would just like to say that I have been working on this for a few days
and have read the manual and looked at quite a few examples, all of
which seem to elude me.
I want to upload an image, and I want to use the ftp function to do so.
I want to use the ftp functions for two reasons, (1) I have not used
them yet, and (2) I will be loading an array with a list of files to be
uploaded and I think the ftp functions are better suited for this (as
opposed to the copy function).
On to the trouble
I can connect just fine.
";
echo "Attempted to connect to $ftp_server for user
$ftp_user_name";
die;
} else {
echo "Connected to $ftp_server, for user $ftp_user_name";
}
//The file, even if I write the path out on my own, does not upload
eventually I will have the path put in from a form.
// upload the file
$destination_file = "/public_html/valeriemacwilliam/pics/$SID_.jpg";
//(I have tried this without the beginning slash as well)
$source_file = "C:\\temp\\prestigecars\\bk_jag.jpg";
$upload = ftp_put($conn_id, $destination_file, $source_file, FTP_BINARY);
// check upload status
if (!$upload) {
echo "Ftp upload has failed!";
} else {
echo "Uploaded $source_file to $ftp_server as
$destination_file";
}
//Even the ftp_close(function fails and causes an error);
// close the FTP stream
ftp_close($conn_id);
?>
Are the ftp functions depreciated? I have taken this example directly
from the manual and editted the values to reflect my own situation but
none of the function calls have been editted. Usually when this is the
case I have just over looked something VERY simple and easy, please take
a look and see what you can come up with.
I would even be happy with an FAQ, I have been through the one on
PHP.NET with no joy \C:
PHP is the lastest release running on Mandrake 8.1 (w / plenty of
horsepower)
Thank you for your time.
~Curt~
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Re: [PHP] ftp functions
Would you suggest that I use the copy function instead?
I have tried the copy function in a prelim test and I had trouble with
the function "finding" the file on the local machine. From what I have
seen in the FAQs on PHPBUILDER and this is not an uncommon problem.
One possible solution was that the path to the file to be uploaded must
be relative to the location of the script. This doesn't make much sense
to me though seeing as the file to be uploaded is on the local machine
and the script on the remote machine.
I'm drowning here, anybody have a rope?
[EMAIL PROTECTED] wrote:
>one problem, the ftp functions that you are trying to use are on the server.
>therefor you are trying to connect one "server" to another "server". you
>are not connecting from your workstation to the remote server. for what you
>are attempting to do, ftp will not work.
>
>Jim Lucas
>- Original Message -
>From: "sundogcurt" <[EMAIL PROTECTED]>
>To: "GENERAL PHP LIST" <[EMAIL PROTECTED]>
>Sent: Monday, January 21, 2002 10:47 AM
>Subject: [PHP] ftp functions
>
>
>>
>>Before I get flamed for not reading the manual or trying on my own, I
>>would just like to say that I have been working on this for a few days
>>and have read the manual and looked at quite a few examples, all of
>>which seem to elude me.
>>
>>
>>I want to upload an image, and I want to use the ftp function to do so.
>>I want to use the ftp functions for two reasons, (1) I have not used
>>them yet, and (2) I will be loading an array with a list of files to be
>>uploaded and I think the ftp functions are better suited for this (as
>>opposed to the copy function).
>>
>>On to the trouble
>>
>>I can connect just fine.
>>
>>>$ftp_server = "xx.xx.xxx.xx";
>>$ftp_user_name = "username";
>>$ftp_user_pass = "password";
>>
>>// set up basic connection
>>$conn_id = ftp_connect($ftp_server);
>>
>>// login with username and password
>>$login_result = ftp_login($conn_id, $ftp_user_name, $ftp_user_pass);
>>
>>// check connection
>>if ((!$conn_id) || (!$login_result)) {
>>echo "Ftp connection has failed!";
>>echo "Attempted to connect to $ftp_server for user
>>$ftp_user_name";
>>die;
>>} else {
>>echo "Connected to $ftp_server, for user $ftp_user_name";
>>}
>>
>>
>>//The file, even if I write the path out on my own, does not upload
>>eventually I will have the path put in from a form.
>>
>>
>>// upload the file
>>$destination_file = "/public_html/valeriemacwilliam/pics/$SID_.jpg";
>>//(I have tried this without the beginning slash as well)
>>
>>$source_file = "C:\\temp\\prestigecars\\bk_jag.jpg";
>>
>>$upload = ftp_put($conn_id, $destination_file, $source_file, FTP_BINARY);
>>
>>// check upload status
>>if (!$upload) {
>>echo "Ftp upload has failed!";
>>} else {
>>echo "Uploaded $source_file to $ftp_server as
>>$destination_file";
>>}
>>
>>//Even the ftp_close(function fails and causes an error);
>>// close the FTP stream
>>ftp_close($conn_id);
>>?>
>>
>>Are the ftp functions depreciated? I have taken this example directly
>>from the manual and editted the values to reflect my own situation but
>>none of the function calls have been editted. Usually when this is the
>>case I have just over looked something VERY simple and easy, please take
>>a look and see what you can come up with.
>>
>>I would even be happy with an FAQ, I have been through the one on
>>PHP.NET with no joy \C:
>>
>>PHP is the lastest release running on Mandrake 8.1 (w / plenty of
>>horsepower)
>>
>>Thank you for your time.
>>~Curt~
>>
>>
>>--
>>PHP General Mailing List (http://www.php.net/)
>>To unsubscribe, e-mail: [EMAIL PROTECTED]
>>For additional commands, e-mail: [EMAIL PROTECTED]
>>To contact the list administrators, e-mail: [EMAIL PROTECTED]
>>
>>
>
>
Re: [PHP]Upload file to system
http://www.php.net/manual/en/features.file-upload.php http://www.faqts.com/knowledge_base/view.phtml/aid/988 : ) [EMAIL PROTECTED] wrote: >I want to make something like when you are attaching something to an >email in hotmail. I want to make something that searches your own >computer for files on your harddisk that you want to upload to the >server. > >But I don't know if this is possible. If this is possible.Can somebody >tell me which function I should use? Or if someone have an example... >please... > >Duky > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP]Upload file to system
You could try contacting your server admin and seeing if he / she will increase the limit... That would be where I would start. If that doesn't work, my hosting fees are quite reasonable ahah [EMAIL PROTECTED] wrote: >Thnx. it's working now. But.. Howcome I can only upload files no bigger >then 100 kb? I have changed the MAX_FILE_SIZE to 10. Also, I can't >change the php.ini, coz the webserver isn't mine.. what to do now?? > >Sundogcurt wrote: > >>http://www.php.net/manual/en/features.file-upload.php >>http://www.faqts.com/knowledge_base/view.phtml/aid/988 >> >>: ) >> >>[EMAIL PROTECTED] wrote: >> >>>I want to make something like when you are attaching something to an >>>email in hotmail. I want to make something that searches your own >>>computer for files on your harddisk that you want to upload to the >>>server. >>> >>>But I don't know if this is possible. If this is possible.Can somebody >>>tell me which function I should use? Or if someone have an example... >>>please... >>> >>>Duky >>> >>> > >
Re: [PHP]Upload file to system
Is that a bad word? \ : [EMAIL PROTECTED] wrote: >Did someone say hosting fees? > >- Original Message - >From: "sundogcurt" <[EMAIL PROTECTED]> >To: "GENERAL PHP LIST" <[EMAIL PROTECTED]> >Sent: Monday, January 28, 2002 3:32 PM >Subject: Re: [PHP]Upload file to system > > >>You could try contacting your server admin and seeing if he / she will >>increase the limit... >> >>That would be where I would start. >>If that doesn't work, my hosting fees are quite reasonable ahah >> >> >> >>[EMAIL PROTECTED] wrote: >> >>>Thnx. it's working now. But.. Howcome I can only upload files no bigger >>>then 100 kb? I have changed the MAX_FILE_SIZE to 10. Also, I >>> >can't > >>>change the php.ini, coz the webserver isn't mine.. what to do now?? >>> >>>Sundogcurt wrote: >>> >>>>http://www.php.net/manual/en/features.file-upload.php >>>>http://www.faqts.com/knowledge_base/view.phtml/aid/988 >>>> >>>>: ) >>>> >>>>[EMAIL PROTECTED] wrote: >>>> >>>>>I want to make something like when you are attaching something to an >>>>>email in hotmail. I want to make something that searches your own >>>>>computer for files on your harddisk that you want to upload to the >>>>>server. >>>>> >>>>>But I don't know if this is possible. If this is possible.Can somebody >>>>>tell me which function I should use? Or if someone have an example... >>>>>please... >>>>> >>>>>Duky >>>>> >>>>> >>> >> > > >
Re: [PHP] Date function
Is this date stored in mysql by anychance? [EMAIL PROTECTED] wrote: >Hi All, > >Is there any function in PHP4.X or in Oracle8i to calculate number of >Business days ( excluding Saturday and Sunday) between two given dates? > >Thanks in advance. >Sridhar. > >-Original Message- >From: qartis [mailto:[EMAIL PROTECTED]] >Sent: Monday, January 28, 2002 1:38 PM >To: [EMAIL PROTECTED]; [EMAIL PROTECTED] >Subject: [PHP-DB] Re: [PHP] File uploading like hotmail > > >I think you might be looking for HTML form element. > > >"Duky Yuen" <[EMAIL PROTECTED]> wrote in message >[EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > >>I want to make something like when you are attaching something to an >>email in hotmail. I want to make something that searches your own >>computer for files on your harddisk that you want to upload to the >>server. >> >>But I don't know if this is possible. If this is possible.Can somebody >>tell me which function I should use? Or if someone have an example... >>please... >> >>Duky >> > > > >-- >PHP Database Mailing List (http://www.php.net/) >To unsubscribe, e-mail: [EMAIL PROTECTED] >For additional commands, e-mail: [EMAIL PROTECTED] >To contact the list administrators, e-mail: [EMAIL PROTECTED] > >
[PHP] upload permission denied!
I'm not sure if this is a problem I can overcome with PHP or if I have to go directly to the Linux box to fix (I hope not). I have moved my development from a win32/apache/mysql box to a Mandrake/apache/mysqly box and have only one problem (that I know of) to overcome. When I try to upload an image (with scripts that worked just fine on the win32 system) I get a permission denied error. The dir that I am trying to write to has permissions for everybody to do anything! (777 or 0777 how ever you want to format it). The scripts are located in my user dir under public_html etc etc. I can create directories in there but I can't seem to convince the darn thing to copy an image. It seems to upload (no errors) but that is where it ends. I don't know what other information you need from me or if this is even the proper list to be submitting to. Again I say I am able to create directories but not copy an image. ??? Thanks and I'm sorry that my question has anything to do with uploading stuff eheheh ~Curtis~ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] upload permission denied!
Like so?
error message :
Warning: Unable to create 'sdf.jpg': Permission denied in
/home/gruesome/public_html/valeriemacwilliam/includes/content/addpic.php
on line 21
code :
done";
}else{
print "not done";
}
}
?>
Add Picture
new file name
[EMAIL PROTECTED] wrote:
>
>At least give us the error you get.
>
>And it sounds like I'd need to see your code too.
>
>bvr.
>
>On Mon, 28 Jan 2002 15:23:50 -0800, sundogcurt wrote:
>
>>I'm not sure if this is a problem I can overcome with PHP or if I have
>>to go directly to the Linux box to fix (I hope not).
>>
>>I have moved my development from a win32/apache/mysql box to a
>>Mandrake/apache/mysqly box and have only one problem (that I know of) to
>>overcome.
>>
>>When I try to upload an image (with scripts that worked just fine on the
>>win32 system) I get a permission denied error. The dir that I am trying
>>to write to has permissions for everybody to do anything! (777 or 0777
>>how ever you want to format it). The scripts are located in my user dir
>>under public_html etc etc. I can create directories in there but I can't
>>seem to convince the darn thing to copy an image. It seems to upload (no
>>errors) but that is where it ends. I don't know what other information
>>you need from me or if this is even the proper list to be submitting to.
>>
>>Again I say I am able to create directories but not copy an image. ???
>>
>>Thanks and I'm sorry that my question has anything to do with uploading
>>stuff eheheh
>>
>
>
>
Re: [PHP] upload permission denied!
the directory permissions are set to 777 / 0777 any user, any group,
etc. at least that is how I understand it. if i'm wrong please correct
me, I do appreciate the help thank you.
[EMAIL PROTECTED] wrote:
>On Tuesday 29 January 2002 11:10, sundogcurt wrote:
>
>>Like so?
>>
>>error message :
>>Warning: Unable to create 'sdf.jpg': Permission denied in
>>/home/gruesome/public_html/valeriemacwilliam/includes/content/addpic.php
>>on line 21
>>
>>code :
>>>if($myfile){
>>if(copy(stripslashes($myfile), $newname . '.jpg')){
>>print "done";
>>}else{
>>print "not done";
>>}
>>}
>>
>
>
>[snip]
>
>
>You need to change to the permissions of the directory in which you want to
>copy the file to, so that it's writable by the 'user' running the webserver.
>
>
Re: [PHP] upload permission denied!
That is where the script is that uploads the file, the file should be going to : /home/gruesome/public_html/valeriemacwilliam/includes/content/ /home/gruesome/public_html/valeriemacwilliam/pics/ or /tmp/ or where ever I have tried a ton of different folders [EMAIL PROTECTED] wrote: >On Tuesday 29 January 2002 11:48, sundogcurt wrote: > >>the directory permissions are set to 777 / 0777 any user, any group, >>etc. at least that is how I understand it. if i'm wrong please correct >>me, I do appreciate the help thank you. >> > >If they're set to 777 then anybody can do anything to it which means it >should be OK. The only thing is whether you have set the *correct* directory? >According to your code, I believe the file will be copied to: > > /home/gruesome/public_html/valeriemacwilliam/includes/content/ > >Is this correct? > >
[PHP] funny unlink problem
unlink(); help the file I would like to delete (actual path) /home/gruesome/public_html/nuved/clients/prestigemotors/images/55jpg path I am feeding to unlink function /home/gruesome/public_html/nuved/clients/prestigemotors/images/55jpg error I get when I try to delete the file Warning: Unlink failed (No such file or directory) in /home/gruesome/public_html/nuved/includes/content/addvehicle.php on line 315 relative path to file (from script containing the unlink function) ../../clients/prestigemotors/images/55jpg I am copying the file with the same script ($folder = 'prestigemotors' && $tempVID = '55') $destination = '/home/gruesome/public_html/nuved/clients/' . $folder . '/images/' . $tempVID . '.jpg'; if(copy(stripslashes($ipath),$destination)) ... Mandrake / Apache / Frustration help? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] funny unlink problem
AARHHH!!! thank you! (man I hate that...) [EMAIL PROTECTED] wrote: >I see 55jpg NOT 55.jpg > >--Tony > >On Wed, 2002-01-30 at 12:40, sundogcurt wrote: > >>unlink(); help >> >>the file I would like to delete (actual path) >>/home/gruesome/public_html/nuved/clients/prestigemotors/images/55jpg >> >>path I am feeding to unlink function >>/home/gruesome/public_html/nuved/clients/prestigemotors/images/55jpg >> >>error I get when I try to delete the file >>Warning: Unlink failed (No such file or directory) in >>/home/gruesome/public_html/nuved/includes/content/addvehicle.php on line >>315 >> >>relative path to file (from script containing the unlink function) >>../../clients/prestigemotors/images/55jpg >> >>I am copying the file with the same script ($folder = 'prestigemotors' >>&& $tempVID = '55') >>$destination = '/home/gruesome/public_html/nuved/clients/' . $folder . >>'/images/' . $tempVID . '.jpg'; >>if(copy(stripslashes($ipath),$destination)) ... >> >>Mandrake / Apache / Frustration >> >>help? >> >> >>-- >>PHP General Mailing List (http://www.php.net/) >>To unsubscribe, e-mail: [EMAIL PROTECTED] >>For additional commands, e-mail: [EMAIL PROTECTED] >>To contact the list administrators, e-mail: [EMAIL PROTECTED] >> -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] PHP and listbox multiple selections
Notice the "[ ]" at the end of the select box name, this prompts php to
treat the values as an array.
$myrow[1]";
}
}
?>
When the form is submitted I fire this code
$mycount = count($options); // get the number of elements in the
array (selected in the drop down)
while($mycount){// cycle through the array decrementling your
counter var ($mycount) untill you have inserted all of the values.
$mycount--;
mysql_query("INSERT INTO table (value1,value2) VALUES
($value1,$options[$mycount])");
}
I hope this helps, it works perfectly for me and uses less code than
other examples I have seen.
> On Monday, January 28, 2002, at 10:41 AM, Simos Varelakis wrote:
>
>> A php form contains a listbox (named foolist) with n Items which
>> can be
>> multiple selected and form posted to
>> results.php how can results.php determine which items were selected
>> in order
>> to add all the items in a Mysql table ???
>
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[PHP] loading Arrays
Do I have to fully load up an array when I declare it or can I assign
more values to it later?
What I want to do is something like this :
// create empty array
$samplearray = array();
// add value to array
if($flag == "true"){
$samplearray => $samplevalue;
}
This doesn't work, can anybody tell me if I'm on the right track here or
should I scrap it and start over?
Thanks
Curt
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[PHP] passed val compare to array val
Hi guys, like every other problem I have this one is SURE to be
painfully simple to fix and make me look g0ofy again.
I have an edit user form with checkboxes like the one below :
>
When the form loads I grab the users relevant information and display
it, I am having trouble checking checkboxes that should be checked
though. Here is how I am attempting to do it.
Prior to the form loading I load an array with the required checkbox
VALUES, not names, I did it this way because I reference the checkbox
values as an array to write to a database.
As the form loads I call a function "docheck()" and pass $inputvalue
(the value of the checkbox) compare the values pulled from the database
and if there is a match, I write "CHECKED" to indicate that the user had
previously chosen this.
The function is below :
function docheck($inputvalue){
$flag = 0;
$tempcnt = 0;
$temptotal = count($editActivity);
while($tempcnt != $temptotal){
if($inputvalue == $editActivity[$tempcnt]){
$flag = 1;
}
$tempcnt++;
}
if($flag == 1){
print "CHECKED";
}
}
The problem seems to be that a "match" never occurs, I tried passing
"Boating" and re-wrote the function as such...
$flag = 0;
if($inputvalue == 'Boating'){
$flag = 1;
}
if($flag == 1){
print "CHECKED";
}
and it worked just fine, so I am of a mind that I am doing something
wrong while attempting to match my passed value with the values in my
array.
Please help!
--
Curtis Gordon
Groovy Web Developer
168th Street
White Rock
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[PHP] passing array from one page to itself?
Hi again guys, I'm having trouble finding documentation on passing an array. Here is what I am doing. I have one page that calls itself, I want to pass an array that is setup the first time it's called to itself the second time it's called. I'm running out of both time AND hair (both of which are a rare comodoty these days). Any hint will be a HUGE help! Thanks in advance. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] passing array from one page to itself?
You didn't by any chance go by the name 99ways2die at one time did you?
I thought of that but when the second page loads the values are dumped
into a db and I have it written already and it uses an array, I would
like to keep it that way
(Gru)
[EMAIL PROTECTED] wrote:
>You could do something like the following:
>
>$array = array("value1","value2","value3");
>for($i=0;$iecho "";
>}
>--
>
>On Fri, 12 Apr 2002 17:14:14
> sundogcurt wrote:
>
>>Hi again guys, I'm having trouble finding documentation on passing an
>>array. Here is what I am doing.
>>
>>I have one page that calls itself, I want to pass an array that is setup
>>the first time it's called to itself the second time it's called.
>>I'm running out of both time AND hair (both of which are a rare comodoty
>>these days).
>>
>>Any hint will be a HUGE help!
>>
>>Thanks in advance.
>>
>>
>>--
>>PHP General Mailing List (http://www.php.net/)
>>To unsubscribe, visit: http://www.php.net/unsub.php
>>
>>
>
>
>See Dave Matthews Band live or win a signed guitar
>http://r.lycos.com/r/bmgfly_mail_dmb/http://win.ipromotions.com/lycos_020201/splash.asp
>
>
[PHP] Date help (no not that kind)
I am trying to determine if todays date ($today) is within a week
($startcheck) of a given date ($dob). This is what I have so far.
$todaydate = (date("Y-d-m"));
$today = strtotime($todaydate);
// $today prior to strtotime = 2001-10-11
$startcheck = strtotime(2001-08-11);
$dob = strtotime(2001-15-11);
//$today is CLEARLY between the two dates I'm testing with but "no" is
always returned.
if($today <= $dob && $startcheck <= $today) {
print "yes";
}
else {
print "no";
}
Here are the values assigned to the vars, why is $today so much smaller
than $startcheck and $dob?
1002783600 = today
1005452940 = startcheck
1005451980 = dob
win32 php4
Thanks in advance
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[PHP] take date and convert to day of year
Hi guys, I know that you can take todays date and display it as the
numeric day of the year, 1 - 365 / 0 - 364 etc.
But can you take a date such as (November-30-1971) and convert that to
the numeric day of the year?
I have been trying to do this but have had no joy, I don't think my code
is even close.
$dob = getdate("Nov-30-1971");
$dobnum = $dob['yday'];
print $dobnum;
$dob 'should' be an array and 'yday' should be the numeric value for the
day of the year, right?!?
This is what I am trying, and how I understand it, using 'yday' should
give you basically the same output as date ("z");
What I would like is the ability to convert any date to the days numeric
value. Any help would be GREATLY appreciated.
(C:
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Re: [PHP] take date and convert to day of year
I have tried to implement your code Martin, and I do thank you VERY MUCH
for the help, but your code seems to have the same trouble as mine, it
doesn't matter what date I start with, I end up with 364 as the day of
the year, I am on win32 though I don't know if that matters.
Here is what I tried:
//FORMAT HAS TO BE date("d-M-Y");
//GET DAY OF YEAR FOR DOB utime = UNIX time
$utine = strtotime("30-Nov-1971");
$dob = getdate($utime);
$dobnum = $dob['yday'];
print "dob is " . $dobnum . "";
//GET DAY OF YEAR FOR TODAY
//$today = date("d-M-Y");
$utoday = strtotime(date("d-M-Y"));
$today = getdate($utoday);
$todaynum = $today['yday'];
print "today is " . $todaynum . "";
There should be a different of about 17 days here right? Not if this is
returning 364 for $dobnum, then it's 48!
[EMAIL PROTECTED] wrote:
>looking at the manual, getdate() is meant to be passed a unix time stamp,
>so, you'll need to use strtotime() first thus:
>
>$utime = strtotime("30-Nov-1971");
>$dob = getdate($utime);
>$dobnum = $dob['yday'];
>print $dobnum;
>
>Notice I changed the format of the date, when I tried using the original
>format, strtotime() complained, saying it couldn't convert it.
>
>Martin T
>
>-Original Message-
>From: sundogcurt [mailto:[EMAIL PROTECTED]]
>Sent: Wednesday, November 14, 2001 7:57 AM
>To: GENERAL PHP LIST
>Subject: [PHP] take date and convert to day of year
>
>
>Hi guys, I know that you can take todays date and display it as the
>numeric day of the year, 1 - 365 / 0 - 364 etc.
>But can you take a date such as (November-30-1971) and convert that to
>the numeric day of the year?
>
>I have been trying to do this but have had no joy, I don't think my code
>is even close.
>
>$dob = getdate("Nov-30-1971");
>$dobnum = $dob['yday'];
>print $dobnum;
>
>$dob 'should' be an array and 'yday' should be the numeric value for the
>day of the year, right?!?
>
>This is what I am trying, and how I understand it, using 'yday' should
>give you basically the same output as date ("z");
>
>What I would like is the ability to convert any date to the days numeric
>value. Any help would be GREATLY appreciated.
>
>(C:
>
>
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Re: [PHP] silly question
Take care to be sure that you make the update, before the rest of the page loads, things happen in order, so it's almost like creating your page backwards (I just had that problem). I can elaborate if you like. (C: [EMAIL PROTECTED] wrote: >Hi list, > >I have PHP code to insert the result of a form into mysql. When I nedd to >made an update, I pass an "id" in the url and use it to make the update >query. The problem is after I click in the update button (input submit) my >page refresh and came back in the same state, to see the changes I need to >type in the url again with the parameter?? why?? There's a way to avoid this >and get a new blank form page after the update? > >ps: I've stored the "id" in a input type hidden, so I could click the button >and still have the id > >Thank's in advance > >Rodrigo Peres > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] take date and convert to day of year fixed!
OK GOT IT!!
$utine = strtotime("26-Nov-2001");
$bday=(int)$utine;
$utoday = strtotime("");
$btoday = (int)$utoday;
if((($bday-$btoday) <= 1209600 && ($bday-$btoday)>0)){
print "you have " . round(($bday-$btoday)/86400)." Days till
'blanks' bday";
print "you are within the two week range.";
}else{
print "you are NOT within the date range.";
}
It's messy but it works, thanks everybody that helped me out on this.
(C:
[EMAIL PROTECTED] wrote:
>date("z"); // the today's ordinal number
>
>z - day of the year; i.e. "0" to "365"
>
>if you want the ordinal number of other days different than the current one
>you can play with mktime or strftime or strtotime to make an appropiate
>timestamp as you did in your earlier codes.
>
>
>- Original Message -
>From: "Martin Towell" <[EMAIL PROTECTED]>
>To: "GENERAL PHP LIST" <[EMAIL PROTECTED]>
>Sent: Wednesday, November 14, 2001 3:19 AM
>Subject: RE: [PHP] take date and convert to day of year
>
>
>>I'm running on WinNT4 w/ PHP 4.0.6 and the code I supplied came back with
>>333
>>and for today I get 317
>>dunno why you're getting 364
>>anyone, any eye-dears ??
>>
>>-Original Message-
>>From: sundogcurt [mailto:[EMAIL PROTECTED]]
>>Sent: Wednesday, November 14, 2001 12:14 PM
>>To: GENERAL PHP LIST
>>Subject: Re: [PHP] take date and convert to day of year
>>
>>
>>I have tried to implement your code Martin, and I do thank you VERY MUCH
>>for the help, but your code seems to have the same trouble as mine, it
>>doesn't matter what date I start with, I end up with 364 as the day of
>>the year, I am on win32 though I don't know if that matters.
>>
>>Here is what I tried:
>>
>>
>>//FORMAT HAS TO BE date("d-M-Y");
>>//GET DAY OF YEAR FOR DOB utime = UNIX time
>>$utine = strtotime("30-Nov-1971");
>>$dob = getdate($utime);
>>$dobnum = $dob['yday'];
>>print "dob is " . $dobnum . "";
>>
>>//GET DAY OF YEAR FOR TODAY
>>//$today = date("d-M-Y");
>>$utoday = strtotime(date("d-M-Y"));
>>$today = getdate($utoday);
>>$todaynum = $today['yday'];
>>print "today is " . $todaynum . "";
>>
>>There should be a different of about 17 days here right? Not if this is
>>returning 364 for $dobnum, then it's 48!
>>
>>
>>[EMAIL PROTECTED] wrote:
>>
>>>looking at the manual, getdate() is meant to be passed a unix time stamp,
>>>so, you'll need to use strtotime() first thus:
>>>
>>>$utime = strtotime("30-Nov-1971");
>>>$dob = getdate($utime);
>>>$dobnum = $dob['yday'];
>>>print $dobnum;
>>>
>>>Notice I changed the format of the date, when I tried using the original
>>>format, strtotime() complained, saying it couldn't convert it.
>>>
>>>Martin T
>>>
>>>-Original Message-
>>>From: sundogcurt [mailto:[EMAIL PROTECTED]]
>>>Sent: Wednesday, November 14, 2001 7:57 AM
>>>To: GENERAL PHP LIST
>>>Subject: [PHP] take date and convert to day of year
>>>
>>>
>>>Hi guys, I know that you can take todays date and display it as the
>>>numeric day of the year, 1 - 365 / 0 - 364 etc.
>>>But can you take a date such as (November-30-1971) and convert that to
>>>the numeric day of the year?
>>>
>>>I have been trying to do this but have had no joy, I don't think my code
>>>is even close.
>>>
>>>$dob = getdate("Nov-30-1971");
>>>$dobnum = $dob['yday'];
>>>print $dobnum;
>>>
>>>$dob 'should' be an array and 'yday' should be the numeric value for the
>>>day of the year, right?!?
>>>
>>>This is what I am trying, and how I understand it, using 'yday' should
>>>give you basically the same output as date ("z");
>>>
>>>What I would like is the ability to convert any date to the days numeric
>>>value. Any help would be GREATLY appreciated.
>>>
>>>(C:
>>>
>>>
>>
>>
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>>PHP General Mailing List (http://www.php.net/)
>>To unsubscribe, e-mail: [EMAIL PROTECTED]
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>>To contact the list administrators, e-mail: [EMAIL PROTECTED]
>>
>
