I keep getting the same mysql error code but I dont know how to correct it.
Error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /websites/123reg/LinuxPackage22/we/ez/y_/weezy.co.uk/public_html/search.php on line 29
I have marked out line 29 on the code below.
What does this mean?
Thanks
<?php
// Change the fields below as per the requirements
$db_host="";
$db_username="";
$db_password="";
$db_name="";
$db_tb_name="data";
$db_tb_atr_name="name";
$db_tb_atr_name="email";
$db_tb_atr_name="location";
//Now we are going to write a script that will do search task
// leave the below fields as it is except while loop, which will display results on screen
mysql_connect("$db_host","$db_username","$db_password");
mysql_select_db("$db_name");
$query=mysql_real_escape_string($_GET['query']);
$query_for_result=mysql_query("
SELECT name, email, location
FROM data WHERE
$db_tb_atr_name like '%".$query."%'");
echo "<h2>Search Results</h2><ol>";
while ($row = mysql_fetch_array($result)) <<<<<<<<<<<<<<< LINE 29
{
echo "<li>";
echo substr($row["name"], $row["email"], $row["location"]);
echo "</li><hr/>";
}
echo "</ol>";
mysql_close();
?>
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