MySQL query returns non-object?

Question

I'm writing code for a user to update his password on a user account system.

This is the portion of code I am referring to:

$checker = mysqli_query($db, "SELECT userid FROM tbl_user WHERE userpassword = '".md5($current)."'");
//echo "SELECT userid FROM tbl_user WHERE userpassword = '".md5($current)."'";
if ($checker == $_SESSION['exp_user']['userid']) {$check = true;} else {$check = false;}

For some reason, $checker is returning an object array rather than an object itself.

Doing a print_r($checker) produces this:

mysqli_result Object ( [current_field] => 0 [field_count] => 1 [lengths] => [num_rows] => 1 [type] => 0 )

Any ideas on what I should be doing?

Progman · Accepted Answer · 2010-11-20 10:09:47Z

1

It doesn't return an "object array", just an object. The class of the object is mysqli_result (as expected) and it got some public fields like num_rows as seen on the manual of the MySQLi_Result class. If you want to read data from the result set you still have to use fetch methods like mysqli_result::fetch_assoc(), even if you result set contains only one row and one column.

answered Nov 20, 2010 at 10:09

Progman

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