I currently made a script, but does not run properly. The line that is wrong is($number is a variable that holds a variable from 1 - 9):
[php]
$result = mysql_query(“SELECT * from links where prog = ‘$name’ and name LIKE ‘%$numberx%’ ORDER BY name;”) or die(mysql_error());
[/php]
It runs fine with no $number(So it works if it is [php]SELECT * from links where prog = '$name’and name LIKE ‘%x%’ ORDER BY name;[/php]). I need it to be able to look for any name that has $number (e.g. 1) with the a x next to it. If $number was 5 Then it would be name LIKE ‘%5x%’. I do not know how put the variable in so it works.
I desperately need it as my website cannot function properly without it.
Thanks,