Trying to get property of non-object in [duplicate] PASAR PARAMETRO -variable

This question already has an answer here: Call to a member function on a non-object 8 answers 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. php mysql object foreach fetch share|improve this question edited Sep 18 '12 at 15:20 Nikola K. 3,81161531 asked May 5 '11 at 2:06 Gerelt Od 3173717

$sidemenu is not an object. var_dump() it and see what it is. –  alex May 5 '11 at 2:08 #22 is <?php echo $sidemenu->mname."<br />";?> –  Gerelt Od May 5 '11 at 2:11 You should check your mysql query is not erroring. $menu might be an empty string. That would cause your $sidemenu to not be an object. –  Chris Henry May 5 '11 at 2:13 Query working well and returning datas. –  Gerelt Od May 5 '11 at 2:14 1   thanks for clarifying –  Phil May 5 '11 at 2:14 add comment

marked as duplicate by tereško, Danack, andrewsi, Jim, Macduff Oct 15 at 8:42

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4 Answers

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up vote 12 down vote accepted

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 share|improve this answer answered May 5 '11 at 2:13 Phil 39.1k43555

I really appreciate your help. Now it's working well. –  Gerelt Od May 5 '11 at 2:20 add comment

up vote 3 down vote	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. share|improve this answer answered May 5 '11 at 2:13 alex 164k63346539
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up vote 2 down vote	$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. share|improve this answer answered May 5 '11 at 2:09 james 9721023
	isn't mysql_fetch_object returns set of objects? –  Gerelt Od May 5 '11 at 2:12 4   @Gerelt No, I suggest you check the manual again. –  Phil May 5 '11 at 2:15 add comment

up vote 1 down vote

<?php foreach ($sidemenus->mname as $sidemenu): ?> <?php echo $sidemenu ."<br />";?> or $sidemenus = mysql_fetch_array($results); then <?php echo $sidemenu['mname']."<br />";?>