php 报错如下:Notice: Trying to get property of non-object


 

参考文档如下解决:

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

 

问题如下:

 

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.

 

解决如下:

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


免责声明!

本站转载的文章为个人学习借鉴使用,本站对版权不负任何法律责任。如果侵犯了您的隐私权益,请联系本站邮箱yoyou2525@163.com删除。



 
粤ICP备18138465号  © 2018-2025 CODEPRJ.COM