php 报错如下:Notice: Trying to get property of non-object
2024-08-30 20:03:02
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
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