Database help

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will83
Forum Commoner
Posts: 53
Joined: Thu Nov 10, 2005 3:13 pm

Database help

Post by will83 »

Hi I'm getting the error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\phpdev5\www\tracking\display-contents.php on line 20

From this code:

Code: Select all

<?php  

include 'connect.php'; 

$company = $_POST['company']; 

echo "Results for ".$company."<br /><br />"; 

$query = mysql_query("SELECT * FROM link_log WHERE link_log.company_id = link_company.id;"); 

echo "<table cellspacing=\"0\" cellpadding=\"0\">\n"; 

echo "<tr>\n"; 

while($row = mysql_fetch_array($query)) { 
echo "<td>".$company."</td><td>$row[date]<\td>\n"; 
} 

echo "</tr>\n"; 

echo "</table>\n"; 

?>


Can anyone spot what is wrong with this?

Thanks for your help.

Will
User avatar
Zoxive
Forum Regular
Posts: 974
Joined: Fri Apr 01, 2005 4:37 pm
Location: Bay City, Michigan

Post by Zoxive »

Try adding a Die for the Query, see if somethings wrong with the query,

Code: Select all

$query = mysql_query("SELECT * FROM link_log WHERE link_log.company_id = link_company.id;") or die('Query failed: ' . mysql_error());
-NSF
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John Cartwright
Site Admin
Posts: 11470
Joined: Tue Dec 23, 2003 2:10 am
Location: Toronto
Contact:

Post by John Cartwright »

change

Code: Select all

$query = mysql_query("SELECT * FROM link_log WHERE link_log.company_id = link_company.id;");
to

Code: Select all

$query = mysql_query("SELECT * FROM link_log WHERE link_log.company_id = link_company.id;") or die(mysql_error());
to see why your query is failing.
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