6 replies to this topic

[Hacker-X]

I just uploaded Gio's Affiliate script, and I got this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/name/public_html/display.php on line 5

This is the display.php file:

<?
include "connect.php";

$result = mysql_query("SELECT * FROM affilliates ORDER by RAND() LIMIT 0,3");
while($row = mysql_fetch_array($result)){

echo "<p><a href=\"www.domain.com/counter.php?outid=" . $row['id'] . "\" target=\"_blank\">
<img src=\"" . $row['ws_button'] . "\" width=88 height=31 alt=\"". $row['ws_name'] ."\" border=\"0\"></a><br>in: " . $row['countin'] . " | out: " . $row['countout'] . "</p>";
}
?>

The same error occurs when I attempt to open Affiliate.php the code can be found here:

<?
include "connect.php";

$result = mysql_query(" SELECT * FROM affilliates ORDER by RAND();");
while($row = mysql_fetch_array($result)){

echo "<p><a href=\"www.domain.com/counter.php?outid=" . $row['id'] . "\" target=\"_blank\">
<img src=\"" . $row['ws_button'] . "\" width=88 height=31 alt=\"". $row['ws_name'] ."\" border=\"0\"></a><br>in: " . $row['countin'] . " | out: " . $row['countout'] . "</p>";
}
?>

Edited by Hacker-X, 25 November 2005 - 06:36 AM.

[Mr. Matt]

RAND(); that is causing your problem, its not correct MySQL syntax...

[Chaos King]

Actually, it is correct synthax. That is how you select random rows from the selected query.

Are you sure that there are actually affiliates in your database? That is usually the error, also, you should add or die(mysql_error()); to the end of your queries to ensure that it is accurate also.

$result = mysql_query("MYSQL STUFF GOES HERE") or die(mysql_error());

Edited by Chaos King, 25 November 2005 - 12:51 PM.

[liveman]

try making it

$result = mysql_query("SELECT * FROM `affilliates` ORDER BY RAND() LIMIT 0,3");
while($row = mysql_fetch_array($result,.MYSQL_NUM))

$result = mysql_query(" SELECT * FROM affilliates ORDER by RAND();");
while($row = mysql_fetch_array($result)){

Get rid of that ; after the RAND() and the bottom should work to see if its actaully your sql query ... try running
SELECT * FROM `affilliates` ORDER BY RAND() in phpmyadmin and if it works, then its somethin in your php coding

Edited by liveman, 25 November 2005 - 12:31 PM.

[rc69]

1. RAND() is correct, i use it myself.
2. MYSQL_NUM, won't fix there error.
3. That thing with the semi-colon may fix the bottom query.

To figure out what WILL fix the error. Try using this

$result = mysql_query("SELECT * FROM affilliates ORDER BY RAND() LIMIT 0,3") or die(mysql_error());
while($row = mysql_fetch_array($result)){

Also note, i'm assuming the actual query i selected for the example is to select random affiliates. If that's the case, you'll only randomly select the first 3 affiliates, no matter what.
To select any 3 affiliates, remove the "0," from the query.

[Ruben K]

Note that 'order by RAND() limit 3' counts as 3 queries instead of just one

Edited by Cliff, 28 November 2005 - 11:33 AM.

[HaloprO]

<?php
include 'connect.php';
$q = mysql_query('SELECT * FROM `affiliates` ORDER BY RAND() LIMIT 3') or die(mysql_error());
$r = mysql_fetch_assoc($q);
do {
	echo "<p><a href=\"www.domain.com/counter.php?outid=" . $r['id'] . "\" target=\"_blank\"> 
	<img src=\"" . $r['ws_button'] . "\" width=88 height=31 alt=\"". $r['ws_name'] ."\"
	border=\"0\"></a><br>in: " . $r['countin'] . " | out: " . $r['countout'] . "</p>";
} while($r = mysql_fetch_assoc($q));
?>

This should work with what you are doing, if you get errors it has something to do with the database or table..