Using LIKE to find something inside a string - php

Question

I don't understand why this doesn't work. I have the following contents in sowner : " 6 4 7 " without the quotes, but with all the spaces including in the beginning and the end. That's the sowner value of a row in the DB.
I have row id which I transform into a string of this form (id is 4) " 4 " then search for it inside sowner to see if it's there. No results returned.
Here is my code:
$sql = 'SELECT * FROM `services` WHERE `sowner` LIKE ?';
$stmt = $conn->prepare($sql);
if($stmt === false) {
trigger_error('Wrong SQL: ' . $sql . ' Error: ' . $conn->errno . ' ' . $conn->error, E_USER_ERROR);
}
$spacedid = " " . strval($row['id']) . " ";
$stmt->bind_param('s', $spacedid);
$stmt->execute();
$res3 = $stmt->get_result();
The purpose is to have an undefined number of people's ids in a string separated by spaces and only showing stuff to people if their id is inside the services sowner field.
I am positive it's the LIKE because if I remove the WHERE, all rows show up.
Also I'm using the same implementation in a on-type search suggestion form like this
SELECT * FROM `users` WHERE username LIKE ?

Answer 1

you need to add the wildcard operators to the like
so the statement becomes more like
SELECT * FROM `users` WHERE username LIKE '% 6 %'
change the line
$spacedid = " " . strval($row['id']) . " ";
to
$spacedid = "% " . strval($row['id']) . " %";