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$sql = "SELECT * FROM `$table` WHERE `$column` = '$var'";

 

this will work as long as there is nothing but that value in the column. If there is more text you will get back nothing. You would have to use LIKE and wildcards to get more results

 

$sql = "SELECT * FROM `$table` WHERE `$column` LIKE '%".$var."%'";

 

Ray

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