sorry...this is the error i get:
Warning: Supplied argument is not a valid MySQL result resource in
/www/prunet/agent_result.php3 on line 110
-----Original Message-----
From: Glyndower [mailto:[EMAIL PROTECTED]]
Sent: Saturday, August 18, 2001 1:55 AM
To: Matt Wagner; Craig Westerman
Cc: MySQL-List
Subject: parameters?
I'm a newbie to MySql / Php and heres what i'm trying to accomplish:
I have a set of hard coded links that each have a sql parameter value like
below:
<a href="http://www.foo.com/agent_result.php3?$agent_id=9805";>My Link
Text</a>
I'm trying to pass that parameter into a query on another page and have it
display the results based on the variable.
<?php $db = mysql_connect("localhost", "user","password");
mysql_select_db("db_name");
$result = mysql_query("SELECT * FROM agentdb WHERE '$agent_id' = agent_id");
php?>
Then the data gets displayed:
<img src="http://www.foo.com/agentpics/<?php
printf(mysql_result($result,0,"first_name")); ?>_<?php
printf(mysql_result($result,0,"last_name")); ?>.jpg" width="89" height="92"
vspace="5" hspace="5">
I'm still thinking in an access box, i'm sure...can somebody give me a push
in the right direction?
TIA, Chris
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