Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

[simon2x1]

how can i fix these error and what does it mean
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in C:\wamp\www\test_php\log\search\result.php on line 8

<?php
 
@mysql_connect('localhost','user2','point')or die('could not connect');
@mysql_select_db('files') or die('could not select database');
 
$term = $_GET['term'];
$sql = mysql_query("SELECT * FROM user WHERE Fname LIKE %term%");
while ($row = mysql_fetch_array($sql))
{
echo 'Fname : '.$row ['Fname'];
echo '<br/>Lname :'.$row ['Lname'];
echo '<br/>Address : '.$row ['Address'];
echo '<br/>phone no: '.$row ['phone no'];
echo '<br/>Email: '.$row ['Email'];
echo '<br/><br/>';
}
?>
 

[dumm]

When you have a mysql error, you can output the error by echoing mysql_error() after the query has been sent.

example:

$sql = mysql_query("SELECT * FROM user WHERE Fname LIKE '%term%'");
if(!$sql)
echo mysql_error();
 

In your query you must put %term% between quotes.

$sql = mysql_query("SELECT * FROM user WHERE Fname LIKE '%term%' ");
 

[Dormilich]

actually, what is %term% (or did you mean $term or %$term%)?

[ak1dnar]

Hi Simon,
I changed the original thread title (how can i fix this error) with the existing one.
while posting threads there are some rules/guidelines that everybody should read and follow.Please be good enough to have a look here. good luck.
Posting Guidelines

MODERATOR

The meaning of that error is, there is no "result set" after executing that query. may be your SQL statement is in correct.

[Markus]

You get this because your query is returning no results. You should check using mysql_num_rows(), and, as mentioned before, it is very good practice to use the mysql_error() function on any query you do.

http://bytes.com/topic/php/answers/6...gging-messages