[PHP] --> I N C L U D E < -- O R --> F O P E N < -- ?
Hi, I am not a PHP programmer. (Sorry Guys;-) Although, I am working on a PHP portal type site, where I need to integrate as seemlessly(?) as possible an outside specialized leased Search Engine Service done with ASP (I suppose? ".asp"). I have been looking everywhere for this info. Just can't seem to get it all together. I done some tests but the images acompanying the Search form are not following. I suppose if I ask the them to put absolute paths... ? --- 1. Is it technically possible with PHP to interact with the engine's pages and the results it pushs so that it feels as if it was part of the site? --- 2. I don't understand the differences between "include" , "fopen" and even "@fopen". What would be the best way to do the job if it is "doable" of course ;-) Best regards, Yves -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: --> I N C L U D E < -- O R --> F O P E N < -- ?
To: Philip, Ninety-Nine Ways To Die and Maxim, Thank you very much guys for all your nice explanations on the INCLUDE vs FOPEN stuff. I'll let you know how I done with this. Should be in a week or so.;-) Again... Thanks! Long live this "No Gates" community! Best regards, Yves -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Broken icon?
Hello,
My problem is that I have in the table below an URL (see $indice_url)
that points to the graphic (.gif) that has to be use by the the
"$indice" variable. The graphics ( five different ones) located in
the same folder as the script " data_out.php3"
Any Guru out here could shred me the "Blue Light" on how to make the
images to show?
Thanks in advance,
Best regards,
Yves
This is the table that feeds:
CREATE TABLE indice (
id int(11) DEFAULT '0' NOT NULL auto_increment,
date text,
indice text,
indice_url varchar(255),
PRIMARY KEY (id)
);
The file to get the data out:
");
printf("%s\n",
$row["indice_url"]);
}
?>
The output that it gives me (in source mode):
image_f03.gif
image_f05.gif
image_f01.gif
--------
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Malouin Design Graphique
http://www.malouin.qc.ca
Québec (Québec) CANADA
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[PHP] Newbie: Images refs not loading (Broken Icons) How to?
Hi,
Is there anything wrong with this code below?
Any help or fix to this code would be very much appreciated.
The file to get the data out (gif file):
");
printf("%s\n",
$row["indice_url"]);
}
?>
Here's the "tiny" table that feeds it:
CREATE TABLE indice (
id int(11) DEFAULT '0' NOT NULL auto_increment,
date text,
indice text,
indice_url varchar(255),
PRIMARY KEY (id)
);
----
Merci,
Yves
--
Malouin Design Graphique
http://www.malouin.qc.ca
Québec (Québec) CANADA
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Re: [PHP] Newbie: Images refs not loading (Broken Icons) How to?
Yes SIR! Steve!
THIS IS WORKING NOW!
MANY... MANY... MANY... THANKS FOR YOUR CODE!
As a newbie... I have been working so hard on this one...
You noticed that I have a lot of reading to do...
Best regards,
Merci encore, (Thanks again,)
Yves
At 22:41 -0500 12/02/2001, Steve Werby wrote:
>"Malouin Design Graphique" <[EMAIL PROTECTED]> wrote:
>> Is there anything wrong with this code below?
>> Any help or fix to this code would be very much appreciated.
>
>Let's take a look.
>
>> $db = mysql_connect("www.server.com", "root", "password");
>> mysql_select_db("db_name", $db);
>> $sql = "select * from table_name order by 'date'";
>> $sql = "select * from table_name";
>
>The second $sql overwrites the first so hopefully you didn't want the first.
>
>> $result = mysql_query($sql);
>> while ($row = mysql_fetch_array($result)) {
>> print("");
>> printf("%s\n",
>> $row["indice_url"]);
>
>Where is $indice_url being set? You call it in printf() but it was not
>previously set in your code. I think you just don't understand how to
>properly use printf() and since you're not really using it to format a
>string, like it's intended to be used I'm not going to bother to explain how
>to use it. I have an idea what you're trying to do, but in my opinion
>printf(%s, $var) is a horrible way to pull data from a query result so
>instead of showing you how to fix it, I'll show you a better way. Why not
>do something like:
>
>$result = mysql_query( $sql );
>while ( $row = mysql_fetch_array( $result ) )
> {
> echo "$row[indice]\n";
> }
>
>--
>Steve Werby
>COO
>24-7 Computer Services, LLC
>Tel: 804.817.2470
>http://www.247computing.com/
--
Malouin Design Graphique
http://www.malouin.qc.ca
Québec (Québec) CANADA
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[PHP] How to inverse the date sorting
Hello, Anybody here would know, how I could inverse the date sorting or ordering? This script below gives me for example: 2001-01-01 2001-01-15 2001-01-31 and so on... when I wish it would gives me: 2001-01-31 2001-01-15 2001-01-01 and so on... - $row[date]$row[indice]\n"; } ?> - Thanks again for your help. Merci beaucoup, Yves -- ---- Malouin Design Graphique http://www.malouin.qc.ca Québec (Québec) CANADA -- 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] How to inverse the date sorting [IT WORKS! - THANKS!]
Just as simple as that?!
Thank you guys: Fang; Andrew and Joe.
Much appreciated.
Hope to be able to give a hand myself someday to newbies like me ;-)
Won't have to ask this question again. Promessed!
Mes salutations,
Yves
At 12:26 -0500 15/02/2001, Fang Li wrote:
>add DESC after order by 'date', it will be
>
>$sql = "select * from table_name WHERE TO_DAYS(NOW()) - TO_DAYS(date)
><= 21 order by 'date' DESC";
>
>Fang
>
>-Original Message-
>From: Malouin Design Graphique [mailto:[EMAIL PROTECTED]]
>Sent: February 15, 2001 12:15 PM
>To: [EMAIL PROTECTED]
>Subject: [PHP] How to inverse the date sorting
>
>
>Hello,
>
>Anybody here would know, how I could inverse the date sorting or ordering?
>
>This script below gives me for example:
>2001-01-01
>2001-01-15
>2001-01-31
>and so on...
>
>when I wish it would gives me:
>2001-01-31
>2001-01-15
>2001-01-01
>and so on...
>
>-
>$db = mysql_connect("servername.com", "root", "password");
>mysql_select_db("stats", $db);
>$sql = "select * from table_name WHERE TO_DAYS(NOW()) - TO_DAYS(date)
><= 21 order by 'date'";
>$result = mysql_query( $sql );
>while ( $row = mysql_fetch_array( $result ) )
> {
> echo "$row[date]src=\"$row[indice_url]\">$row[indice]size=\"1\">\n";
> }
>?>
>-
>
>
>Thanks again for your help.
>Merci beaucoup,
>
>Yves
--
Malouin Design Graphique
http://www.malouin.qc.ca
Québec (Québec) CANADA
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