Question

I’m having a problem with running this function. When it runs, it does exactly what I want, except that within my like_requests table the request_id is not the mysql query result linked to the variable $select but Resource Id #22. I thought that resource id’s appear when you are trying to echo out a result, but I’m not using echo. What’s wrong with the code?

function update_likes($band_requested, $new_likes, $session_user_id) {

    $select = mysql_query("SELECT `primary_id` FROM `requests` WHERE 

    `user_requester_id` = '$session_user_id' AND `person_requested` = 

    '$band_requested'");

    $sql_2 = "INSERT INTO `like_requests` (user_id, request_id) VALUES 

    ('$session_user_id', '$select')";   

    mysql_query($sql_2);

}

$band_requested = 'rally done';
$new_likes = 239;
$the_session_user_id = 3;

update_likes($band_requested, $new_likes, $the_session_user_id);

UPDATE WITH CORRECTED ANSWER
Here is the code corrected with help from David.

function update_likes($band_requested, $new_likes, $session_user_id)

{

$select = mysql_query("SELECT `primary_id` FROM `requests` WHERE `user_requester_id` = 

'$session_user_id' AND `person_requested` = '$band_requested'");

$row = mysql_fetch_row($select);

$request_id = $row[0];  

$sql_2 = "INSERT INTO `like_requests` (user_id, request_id) VALUES ('$session_user_id', 

'$request_id')";    

mysql_query($sql_2);

}

Answer 1

mysql_query returns a resource (http://php.net/manual/en/function.mysql-query.php) not just a scalar value. You’d need to use a function like mysql_fetch_row() to get the, presumably, one row you want, assign that row to a variable $row, then retrieve the primary_id with array syntax like $row['primary_id']. By the way, apparently mysql_query is being eased out and we should use the MySQLi API with the mysqli_query() method.