试图获取非对象的属性 [英] Trying to get property of non-object in

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:第 22 行试图获取 C:\wamp\www\phone\pages\init.php 中非对象的属性

Notice: Trying to get property of non-object in C:\wamp\www\phone\pages\init.php on line 22

你能修好吗?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

查看<的手册代码>mysql_fetch_object().它返回一个对象，而不是一个对象数组.

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;
}

<小时>

我建议你看看 PDO.PDOStatement::fetchAll(PDO::FETCH_OBJ) 做你认为 mysql_fetch_object() 要做的事情

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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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