Re: [PHP] session lost problem
bug zhu hat am 24. April 2012 um 08:28 geschrieben: > thank you for your explanation, > when i write to $_SESSION after session_commit(),$_SESSION is just a > regular array Yes. Actually session_commit does not "terminate" the session as mentioned earlier but is closes it for writing. You cann still read session values. The benefit of using session_commit is that the server saved associated session file is no longer locked, so that parallel requests can both access the values. The approach ist as follows: Call session_commit() as early in you code (after session_open) as possible to avoid locking. So first do all the writing to the $_SESSION array, then do write close (or commit). After that you can still read all session relevant information. If you want to write afterwards to your $_SESSIOn array you simply have to call session_start to re-open the write context. Afterwards you can commit it again to remove the lock. But be careful! session_start and session_commit perform write operations on your harddisk or whatever storage you use. Many calls to start and commit will result in losing performance. Regards, Marco -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] session lost problem
2012/4/24 ma...@behnke.biz > > > bug zhu hat am 24. April 2012 um 08:28 geschrieben: > > > thank you for your explanation, > > when i write to $_SESSION after session_commit(),$_SESSION is just a > > regular array > > Yes. Actually session_commit does not "terminate" the session as mentioned > earlier but is closes it for writing. You cann still read session values. > > The benefit of using session_commit is that the server saved associated > session file is no longer locked, so that parallel requests can both access > the values. > > The approach ist as follows: > Call session_commit() as early in you code (after session_open) as possible > to avoid locking. So first do all the writing to the $_SESSION array, then > do write close (or commit). After that you can still read all session > relevant information. > > If you want to write afterwards to your $_SESSIOn array you simply have to > call session_start to re-open the write context. Afterwards you can commit > it again to remove the lock. > > But be careful! session_start and session_commit perform write operations > on your harddisk or whatever storage you use. Many calls to start and > commit will result in losing performance. > > Regards, > Marco > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > got it, very appreciate you explanation:-) -- thanks, bugzhu
Re: [PHP] Re: No error reporting on
Sounds like a very good reason to me? It's a development tool. Add it now, remove it later. "Dotan Cohen" wrote in message news:cakdxfkn+6mn9chuqdwevx5ap0km-rkls0q+carbf5hktvuo...@mail.gmail.com... On Mon, Apr 23, 2012 at 16:53, Jim Lucas wrote: >> Possibly, thanks. I actually don't have access to that! >> > > That line should be placed in your script. not the php.ini file > Yes, I'm working on a functions file that is include()ed by the main script. I'm not supposed to touch the main script without a very good reason. -- Dotan Cohen http://gibberish.co.il http://what-is-what.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Hmm.. this is strange..
Hello. I have a upload form in a html file and a corresponding PHP file that shall take care of the information. But I am doing something newbie error here.. What am I doing wrong? (The text is norwegian, but you still see and understand the PHP code) bildegalleri.html - Opplasting til Fotogalleri Opplasting til Fotogalleri Velg bilde for opplasting Filbane bildegalleri.php - Bildegalleri Bildegalleri When I run this script, I always get "Ok". Even when the input file field is empty.. Can someone tell me what I am doing wroing in this? Thanks for you time and effort to help me out. Karl -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Hmm.. this is strange..
Hi,
Instead of just checking if the variable is not set, additionally
check if it is empty
Kind regards/met vriendelijke groet,
Serge Fonville
http://www.sergefonville.nl
Convince Google!!
They need to add GAL support on Android (star to agree)
http://code.google.com/p/android/issues/detail?id=4602
2012/4/24 Karl-Arne Gjersøyen :
> Hello.
> I have a upload form in a html file and a corresponding PHP file that
> shall take care of the information.
> But I am doing something newbie error here..
>
> What am I doing wrong? (The text is norwegian, but you still see and
> understand the PHP code)
>
> bildegalleri.html
> -
>
>
>
>
> Opplasting til Fotogalleri
>
>
>
> Opplasting til Fotogalleri
>
>
> Velg bilde for opplasting
> Filbane
>
>
>
>
>
>
>
> bildegalleri.php
> -
> if(!isset($_POST['last_opp_fil'])){
> header('Location: bildegalleri.html');
> }
> elseif(empty($_FILES['filbane'])){
> header('Location: bildegalleri.html');
> } else {
> ?>
>
>
>
>
> Bildegalleri
>
>
> Bildegalleri
> echo "OK";
> }
> ?>
>
>
>
> When I run this script, I always get "Ok". Even when the input file
> field is empty.. Can someone tell me what I am doing wroing in this?
>
> Thanks for you time and effort to help me out.
>
> Karl
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
--
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Re: [PHP] Hmm.. this is strange..
Here I try to check if the input file is empty or not. But still it
did not work. I receive OK even when the input file field is empty..
Bildegalleri
Bildegalleri
Den 16:48 24. april 2012 skrev Serge Fonville
følgende:
> Hi,
>
> Instead of just checking if the variable is not set, additionally
> check if it is empty
>
> Kind regards/met vriendelijke groet,
>
> Serge Fonville
>
> http://www.sergefonville.nl
>
> Convince Google!!
> They need to add GAL support on Android (star to agree)
> http://code.google.com/p/android/issues/detail?id=4602
>
>
> 2012/4/24 Karl-Arne Gjersøyen :
>> Hello.
>> I have a upload form in a html file and a corresponding PHP file that
>> shall take care of the information.
>> But I am doing something newbie error here..
>>
>> What am I doing wrong? (The text is norwegian, but you still see and
>> understand the PHP code)
>>
>> bildegalleri.html
>> -
>>
>>
>>
>>
>> Opplasting til Fotogalleri
>>
>>
>>
>> Opplasting til Fotogalleri
>>
>>
>> Velg bilde for opplasting
>> Filbane
>>
>>
>>
>>
>>
>>
>>
>> bildegalleri.php
>> -
>> > if(!isset($_POST['last_opp_fil'])){
>> header('Location: bildegalleri.html');
>> }
>> elseif(empty($_FILES['filbane'])){
>> header('Location: bildegalleri.html');
>> } else {
>> ?>
>>
>>
>>
>>
>> Bildegalleri
>>
>>
>> Bildegalleri
>> > echo "OK";
>> }
>> ?>
>>
>>
>>
>> When I run this script, I always get "Ok". Even when the input file
>> field is empty.. Can someone tell me what I am doing wroing in this?
>>
>> Thanks for you time and effort to help me out.
>>
>> Karl
>>
>> --
>> PHP General Mailing List (http://www.php.net/)
>> To unsubscribe, visit: http://www.php.net/unsub.php
>>
--
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Re: [PHP] Hmm.. this is strange..
Thank you very much Tolga!
if(empty($_FILES['filbane']['name']))
24.04.2012 17:54 tarihinde, Karl-Arne Gjersøyen yazdı:
> Here I try to check if the input file is empty or not. But still it
> did not work. I receive OK even when the input file field is empty..
>
> if(!isset($_POST['last_opp_fil'])){
> header('Location: bildegalleri.html');
> } else {
> $bildefil = $_FILES['filbane'];
> if(empty($bildefil)){
> header('Location: bildegalleri.html');
> } else {
> ?>
>
>
>
>
> Bildegalleri
>
>
> Bildegalleri
> echo "OK";
> }
> }
> ?>
>
>
>
> Den 16:48 24. april 2012 skrev Serge Fonville
> følgende:
>>
>> Hi,
>>
>> Instead of just checking if the variable is not set, additionally
>> check if it is empty
>>
>> Kind regards/met vriendelijke groet,
>>
>> Serge Fonville
>>
>> http://www.sergefonville.nl
>>
>> Convince Google!!
>> They need to add GAL support on Android (star to agree)
>> http://code.google.com/p/android/issues/detail?id=4602
>>
>>
>> 2012/4/24 Karl-Arne Gjersøyen:
>>>
>>> Hello.
>>> I have a upload form in a html file and a corresponding PHP file that
>>> shall take care of the information.
>>> But I am doing something newbie error here..
>>>
>>> What am I doing wrong? (The text is norwegian, but you still see and
>>> understand the PHP code)
>>>
>>> bildegalleri.html
>>> -
>>>
>>>
>>>
>>>
>>> Opplasting til Fotogalleri
>>>
>>>
>>>
>>> Opplasting til Fotogalleri
>>>
>>>
>>> Velg bilde for opplasting
>>> Filbane
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>> bildegalleri.php
>>> -
>>> >> if(!isset($_POST['last_opp_fil'])){
>>> header('Location: bildegalleri.html');
>>> }
>>> elseif(empty($_FILES['filbane'])){
>>> header('Location: bildegalleri.html');
>>> } else {
>>> ?>
>>>
>>>
>>>
>>>
>>> Bildegalleri
>>>
>>>
>>> Bildegalleri
>>> >> echo "OK";
>>> }
>>> ?>
>>>
>>>
>>>
>>> When I run this script, I always get "Ok". Even when the input file
>>> field is empty.. Can someone tell me what I am doing wroing in this?
>>>
>>> Thanks for you time and effort to help me out.
>>>
>>> Karl
>>>
>>> --
>>> PHP General Mailing List (http://www.php.net/)
>>> To unsubscribe, visit: http://www.php.net/unsub.php
>>>
--
Hjemmeside: http://www.karl-arne.name/
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[PHP] NULL Problem
Here's my code (using MSSQL):
$conn = new COM ("ADODB.Connection")or die("Cannot start ADO");
$conn->open($connStr);
$query = "SELECT * FROM TABLE WHERE id = ".$id;
$rs = $conn->execute($query);
This code works fine, and I retrieve the values like this:
$tmp1 = $rs->fields("column1");
$tmp2 = $rs->fields("column2");
Etc...
Here's the problem - I'm trying to get a date column that I know is
NULL, but I can't seem to get my code right:
$tmp = $rs->fields("followup_on");
if(is_null($tmp)){
$followup = "";
}else{
$followup = $rs->fields("followup_on");
}
//this results in: Catchable fatal error: Object of class variant could
not be converted to string
//When I try to ECHO the $followup results (and I know the database
value is NULL)
So confused - any advice?
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Re: [PHP] NULL Problem
On Tue, Apr 24, 2012 at 7:29 PM, David Stoltz wrote:
> Here's my code (using MSSQL):
>
> $conn = new COM ("ADODB.Connection")or die("Cannot start ADO");
> $conn->open($connStr);
> $query = "SELECT * FROM TABLE WHERE id = ".$id;
> $rs = $conn->execute($query);
>
> This code works fine, and I retrieve the values like this:
>
> $tmp1 = $rs->fields("column1");
> $tmp2 = $rs->fields("column2");
> Etc...
>
>
> Here's the problem - I'm trying to get a date column that I know is
> NULL, but I can't seem to get my code right:
>
> $tmp = $rs->fields("followup_on");
> if(is_null($tmp)){
> $followup = "";
> }else{
> $followup = $rs->fields("followup_on");
> }
>
> //this results in: Catchable fatal error: Object of class variant could
> not be converted to string
> //When I try to ECHO the $followup results (and I know the database
> value is NULL)
>
>
> So confused - any advice?
>
It's been a long time ago I worked with ADO (Thank god), but shouldn't
you echo $followup->value instead of $followup?
If that's not working, try a var_dump($followup), so you can check
exactly what it is.
- Matijn
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Re: [PHP] NULL Problem
Have you considered the PHP MSSQL driver?
http://www.microsoft.com/download/en/details.aspx?id=20098
Kind regards/met vriendelijke groet,
Serge Fonville
http://www.sergefonville.nl
Convince Google!!
They need to add GAL support on Android (star to agree)
http://code.google.com/p/android/issues/detail?id=4602
2012/4/24 Matijn Woudt :
> On Tue, Apr 24, 2012 at 7:29 PM, David Stoltz wrote:
>> Here's my code (using MSSQL):
>>
>> $conn = new COM ("ADODB.Connection")or die("Cannot start ADO");
>> $conn->open($connStr);
>> $query = "SELECT * FROM TABLE WHERE id = ".$id;
>> $rs = $conn->execute($query);
>>
>> This code works fine, and I retrieve the values like this:
>>
>> $tmp1 = $rs->fields("column1");
>> $tmp2 = $rs->fields("column2");
>> Etc...
>>
>>
>> Here's the problem - I'm trying to get a date column that I know is
>> NULL, but I can't seem to get my code right:
>>
>> $tmp = $rs->fields("followup_on");
>> if(is_null($tmp)){
>> $followup = "";
>> }else{
>> $followup = $rs->fields("followup_on");
>> }
>>
>> //this results in: Catchable fatal error: Object of class variant could
>> not be converted to string
>> //When I try to ECHO the $followup results (and I know the database
>> value is NULL)
>>
>>
>> So confused - any advice?
>>
>
> It's been a long time ago I worked with ADO (Thank god), but shouldn't
> you echo $followup->value instead of $followup?
> If that's not working, try a var_dump($followup), so you can check
> exactly what it is.
>
> - Matijn
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
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RE: [PHP] NULL Problem
Matijn - it worked! Geez...
Strange - I don't need the ->value if it actually has a value, only if it's
NULL...
But it works! Thanks!
-Original Message-
From: Matijn Woudt [mailto:tijn...@gmail.com]
Sent: Tuesday, April 24, 2012 1:40 PM
To: David Stoltz
Cc: php-general@lists.php.net
Subject: Re: [PHP] NULL Problem
On Tue, Apr 24, 2012 at 7:29 PM, David Stoltz wrote:
> Here's my code (using MSSQL):
>
> $conn = new COM ("ADODB.Connection")or die("Cannot start ADO");
> $conn->open($connStr);
> $query = "SELECT * FROM TABLE WHERE id = ".$id;
> $rs = $conn->execute($query);
>
> This code works fine, and I retrieve the values like this:
>
> $tmp1 = $rs->fields("column1");
> $tmp2 = $rs->fields("column2");
> Etc...
>
>
> Here's the problem - I'm trying to get a date column that I know is
> NULL, but I can't seem to get my code right:
>
> $tmp = $rs->fields("followup_on");
> if(is_null($tmp)){
> $followup = "";
> }else{
> $followup = $rs->fields("followup_on");
> }
>
> //this results in: Catchable fatal error: Object of class variant could
> not be converted to string
> //When I try to ECHO the $followup results (and I know the database
> value is NULL)
>
>
> So confused - any advice?
>
It's been a long time ago I worked with ADO (Thank god), but shouldn't
you echo $followup->value instead of $followup?
If that's not working, try a var_dump($followup), so you can check
exactly what it is.
- Matijn
RE: [PHP] NULL Problem
Serge,
We don't use MSSQL for much, mostly use MySQL...
But I don't want to switch out all the drivers for this one issue, which is now
resolved (thanks Matijn)
-Original Message-
From: Serge Fonville [mailto:serge.fonvi...@gmail.com]
Sent: Tuesday, April 24, 2012 1:45 PM
To: Matijn Woudt
Cc: David Stoltz; php-general@lists.php.net
Subject: Re: [PHP] NULL Problem
Have you considered the PHP MSSQL driver?
http://www.microsoft.com/download/en/details.aspx?id=20098
Kind regards/met vriendelijke groet,
Serge Fonville
http://www.sergefonville.nl
Convince Google!!
They need to add GAL support on Android (star to agree)
http://code.google.com/p/android/issues/detail?id=4602
2012/4/24 Matijn Woudt :
> On Tue, Apr 24, 2012 at 7:29 PM, David Stoltz wrote:
>> Here's my code (using MSSQL):
>>
>> $conn = new COM ("ADODB.Connection")or die("Cannot start ADO");
>> $conn->open($connStr);
>> $query = "SELECT * FROM TABLE WHERE id = ".$id;
>> $rs = $conn->execute($query);
>>
>> This code works fine, and I retrieve the values like this:
>>
>> $tmp1 = $rs->fields("column1");
>> $tmp2 = $rs->fields("column2");
>> Etc...
>>
>>
>> Here's the problem - I'm trying to get a date column that I know is
>> NULL, but I can't seem to get my code right:
>>
>> $tmp = $rs->fields("followup_on");
>> if(is_null($tmp)){
>> $followup = "";
>> }else{
>> $followup = $rs->fields("followup_on");
>> }
>>
>> //this results in: Catchable fatal error: Object of class variant could
>> not be converted to string
>> //When I try to ECHO the $followup results (and I know the database
>> value is NULL)
>>
>>
>> So confused - any advice?
>>
>
> It's been a long time ago I worked with ADO (Thank god), but shouldn't
> you echo $followup->value instead of $followup?
> If that's not working, try a var_dump($followup), so you can check
> exactly what it is.
>
> - Matijn
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
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[PHP] What is wrong here?
Hello again.
I can't figure out what is wrong here.
move_uploaded_file() get error message from die() and can't copy/move
temp_file into directory bilder
I have try to chmod 0777 bilder/ but it did not help.
Also I have try to chown www-data.www-data bilder/ since Ubuntu Server
run apache as www-data user...
Here is my souce code
--
// Temfil lagres midlertidig på serveren som
// spesifisert i php.ini
$tmp_fil = $_FILES['filbane']['temp_name'];
// lagre filnavnet..
$filnavn = "bilder/" . $_FILES['filbane']['name'];
// ..og legg fila i katalogen bilder
move_uploaded_file($tmp_fil, $filnavn) or die("Feilmelding: Kunne
ikke flytte $filnavn");
Karl
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