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.
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[] = $sidemenu;
}
Might I suggest you have a look at PDO. PDOStatement::fetchAll(PDO::FETCH_OBJ) does what you assumed mysql_fetch_object() to do
<?php foreach ($sidemenus->mname as $sidemenu): ?>
<?php echo $sidemenu ."<br />";?>
or
$sidemenus = mysql_fetch_array($results);
then
<?php echo $sidemenu['mname']."<br />";?>
Notice: Trying to get property of non-object in C:\wamp\www\phone\pages\init.php on line 22
@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.
$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.
Source: Stackoverflow.com