Trying to get property of non-object in

Question

on Control page:

<?php
  include 'pages/db.php'; 
  $results = mysql_query("SELECT * FROM sidemenu WHERE `menu_id`='".$menu."' ORDER BY `id` ASC LIMIT 1", $con);
  $sidemenus = mysql_fetch_object($results);
?>

on View Page:

<?php foreach ($sidemenus as $sidemenu): ?>
  <?php echo $sidemenu->mname."<br />";?>
<?php endforeach; ?>

Error is:

Notice: Trying to get property of non-object in C:\wamp\www\phone\pages\init.php on line 22

Can you fix it? I don't have any idea what happened.

Answer 1

Check the manual for mysql_fetch_object(). It returns an object, not an array of objects.

I'm guessing you want something like this

$results = mysql_query("SELECT * FROM sidemenu WHERE `menu_id`='".$menu."' ORDER BY `id` ASC LIMIT 1", $con);

$sidemenus = array();
while ($sidemenu = mysql_fetch_object($results)) {
    $sidemenus[] = $sidemenu;
}

---

Might I suggest you have a look at PDO. PDOStatement::fetchAll(PDO::FETCH_OBJ) does what you assumed mysql_fetch_object() to do

Answer 2

Your error

Notice: Trying to get property of non-object in C:\wamp\www\phone\pages\init.php on line 22

Your comment

@22 is <?php echo $sidemenu->mname."<br />";?>

$sidemenu is not an object, and you are trying to access one of its properties.

That is the reason for your error.

Answer 3

$sidemenu is not an object, so you can't call methods on it. It is probably not being sent to your view, or $sidemenus is empty.

Answer 4

<?php foreach ($sidemenus->mname as $sidemenu): ?>
<?php echo $sidemenu ."<br />";?>

or

$sidemenus = mysql_fetch_array($results);

then

<?php echo $sidemenu['mname']."<br />";?>