method returns Resource id#14 PHP/MySQL

Question

New to classes.

public function get_feed($member_id) {



    $sql = "SELECT mc. * , cp. * 
            FROM `my_connections` mc
            LEFT JOIN `thoughts` cp ON cp.`member_id` = mc.`connection_id` 
            WHERE mc.`member_id` = '".$member_id."'
            AND mc.`approved` = 1";

    $result = @mysql_query($sql,$db); check_sql(mysql_error(), $sql, 0);

    //echo '<pre>'.$sql.'</pre>';

            return $result; //this returns Resource id#14 or NULL


    /*while($list = mysql_fetch_array($result)) {

        $feed_item[] = $list;
    }

    return $feed_item;*/ 

}// EO get_feed()

when I call it

$feed = new my_feed;

echo $feed->get_feed($uid);

it returns Null.

But when I use the same sql outside the method it returns expected result.

what should I be returning in a query like above.

$db looks like your connection string and you are using inside the function so you may need to use as global $db ; inside the function before using it. — Abhik Chakraborty, Jan 12 '14 at 19:44
Perhaps if you didn't suppress errors on your `mysql_query` call you would see why it's failing. — jszobody, Jan 12 '14 at 19:46
You are both correct, I missed that- Still getting the same result though when returning $result, i.e. Resource id #14 — Angad Dubey, Jan 12 '14 at 19:47
@maximl337 `mysql_query` returns a resource. That's how it is supposed to work. What are you expecting? See http://us2.php.net/mysql_query — jszobody, Jan 12 '14 at 19:47
@AbhikChakraborty `global` shouldn't be used in OOP. Better to use `$this->db` or similar. — user555, Jan 12 '14 at 19:48
@jszobody You are correct, what I want to know is, is it proper to just return $result then loop through it or to loop through it with a while loop and put the resources in an array and return that (like in the commented section of the code) — Angad Dubey, Jan 12 '14 at 19:49
Yes if your class has the $db access or if its a part of its parent class — Abhik Chakraborty, Jan 12 '14 at 19:50
@maximl337 That's entirely up to you. That's a design choice. — jszobody, Jan 12 '14 at 19:50
Also see http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php?rq=1 — MECU, Jan 12 '14 at 19:52
And while you're at it stop using `mysql_*` and use `mysqli` or `PDO`. See http://www.phptherightway.com/#databases. — user555, Jan 12 '14 at 19:52
I have inherited a legacy application, I am slowly modernizing it, will make the changes soon.thank you — Angad Dubey, Jan 12 '14 at 19:53
@jszobody anyway I can mark your answer as complete through comments. sry fairly new to SO aswell. — Angad Dubey, Jan 12 '14 at 19:54

jszobody · Accepted Answer · 2014-01-12T20:06:15.433

Several things:

You are suppressing errors, leaving yourself in the dark:

$result = @mysql_query($sql,$db); // Remove the @ symbol
You should then find that $db isn't a valid database connection object, due to scope. Make sure you have it setup as a class variable, then reference $this->db instead.
Your Resource id#14 result is expected, that's what mysql_query returns. You can use the various mysql_fetch_* methods to parse it here, or simply return it and let the caller handle parsing.
Stop using mysql_* methods! Switch to mysqli or PDO as soon as you can (I understand that can take time with legacy applications).

score 1 · Answer 2 · edited May 23 '17 at 12:29

$result always return Resource id#14 is the way php handle that, to get the data you have to fetch it using the functions php have for that, like mysql_fetch_array or mysql_fetch_assoc

$result = @mysql_query($sql,$db); check_sql(mysql_error(), $sql, 0);

while($list = mysql_fetch_array($result)) {

    $feed_item[] = $list;
}

return $feed_item;

$feed _item will have the query result into an array

NOTE: Why shouldn't I use mysql_* functions in PHP?

Try to avoid suppressing errors like the @ in @mysqli_query you will need get the error to debug

method returns Resource id#14 PHP/MySQL

2 Answers2