extract() expects parameter 1 to be array, Help please

[Namtip]

I'm a newbie . I've been stuck on this error message for 2 days:

I get this error message: extract() expects parameter 1 to be array, boolean given in C:\x\xampp\htdocs\user_personal.php on line 30

 

<?php
$query = 'SELECT
        about_me, job, hobbies, contact
    FROM
        site_user u JOIN
        site_user_profile p ON u.user_id = p.user_id
    WHERE
        username = "' . mysql_real_escape_string($_SESSION['username'], $db) . '"';
$result = mysql_query($query, $db) or die(mysql_error($db));
$row = mysql_fetch_array($result);
extract($row);
mysql_free_result($result);
mysql_close($db);
?>

I think joining table is hard, I just want to create a profile that would be linked to a user name. Please be my friend and help me so I can have PHP phun time.

[Pikachu2000]

The query is failing and returning the boolean value FALSE to the extract() function. Echo the query to the screen then paste it into phpMyAdmin and see what, if any results you get.

[PFMaBiSmAd]

The error means that your query matched zero rows (the WHERE clause was FALSE).

 

You should use mysql_num_rows() to find out if your query matched anything before you attempt to fetch and access the data.

 

As to why your query did not match anything, are you sure your $_SESSION variable contains what you expect? What do you get if you echo $query?

[Namtip]

Thanks guys you're pros. Ah, there isn't any data for the database to retrieve. Could this be why the error message comes up? What should I enter into my code if I want it to display nothing without an error message?

 

This is what it said when I echo'd the query

SELECT about_me, job, hobbies, contact FROM site_user u JOIN site_user_profile p ON u.user_id = p.user_id WHERE username = "Russell"

[trq]

if ($result = mysql_query($query)) {
  if (mysql_num_rows($result)) {
    // use data
  } else {
    // no data found.
  }
} else {
  // query failed
}
?>