I rarely use LIKE so just wanted to verify this is correct, it seems to work..
$searchsql is the field name, choice of several
$searchforthis is what they entered into the input.
The wildcard % needs to be on both sides to truly be wild right.
$query="SELECT * FROM recipes WHERE $searchsql LIKE '%$searchforthis%' AND rec_approved = 1 ORDER BY $searchsql";