试图获取非对象的属性 [英] Trying to get property of non-object in
本文介绍了试图获取非对象的属性的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
在控制页面上:
<?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);
?>
在查看页面上:
<?php foreach ($sidemenus as $sidemenu): ?>
<?php echo $sidemenu->mname."<br />";?>
<?php endforeach; ?>
错误是:
注意:尝试在第22行的C:\ wamp \ www \ phone \ pages \ init.php中获取非对象的属性
Notice: Trying to get property of non-object in C:\wamp\www\phone\pages\init.php on line 22
您可以修复它吗?我不知道发生了什么.
Can you fix it? I don't have any idea what happened.
推荐答案
检查手册中的 mysql_fetch_object()
.它返回一个对象,而不是对象数组.
Check the manual for mysql_fetch_object()
. It returns an object, not an array of objects.
我猜你想要这样的东西
$results = mysql_query("SELECT * FROM sidemenu WHERE `menu_id`='".$menu."' ORDER BY `id` ASC LIMIT 1", $con);
$sidemenus = array();
while ($sidemenu = mysql_fetch_object($results)) {
$sidemenus[] = $sidemenu;
}
可能我建议您看看PDO. PDOStatement::fetchAll(PDO::FETCH_OBJ)
会执行您假设的mysql_fetch_object()
做的事
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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