php 報錯如下:Notice: Trying to get property of non-object


 

參考文檔如下解決:

https://stackoverflow.com/questions/5891911/trying-to-get-property-of-non-object-in

 

問題如下:

 

This question already has an answer here:

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.

 

解決如下:

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


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