Question

i am executing this sql query to update the database

$sql_select = "UPDATE `database1`.`media` SET `media` =(http://torcache.net/torrent/'" . $upload_result . "'.torrent) WHERE 'image_type'= '4' AND 'media' LIKE '"%$fname%"' ";

and i am getting the Warning: Division by zero??

I think it has something to do with this condition LIKE '"%$fname%"' "

which uses the sql LIKE %% to match any string because i have never used this condition before in my code

where would the error be coming from?

Answer 1

The percent sign is outside the quote, making it MODULUS, or remainder from division.

$sql_select = "UPDATE `database1`.`media` SET `media` =(http://torcache.net/torrent/'" . $upload_result . "'.torrent) WHERE 'image_type'= '4' AND 'media' LIKE '%".$fname."%' ";