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trying to get property of non object in


trying to get property of non object in  using -'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.'

asked Oct 6, 2015 by mtabakade
edited Oct 6, 2015 by rajesh
0 votes
20 views



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3 Answers

0 votes

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

answered Oct 6, 2015 by pradip.bhoge
0 votes

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 mname."
";?>

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

That is the reason for your error.

answered Oct 6, 2015 by suyesh.lokhande
0 votes
mname as $sidemenu): ?>
";?>

or

$sidemenus = mysql_fetch_array($results);

then

";?>
answered Oct 6, 2015 by manju bhargava

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