注意:尝试获取非对象的属性 [英] Notice: Trying to get property of non-object in

问题描述

我正在学习一些PHP.它应该打印0，但是出现错误:

I'm learning a bit of PHP. It's supposed to print 0, however I get an error:

注意:尝试获取...中非对象的属性

Notice: Trying to get property of non-object in...

<?php 
$json = '[{"assetref":"","qty":0,"raw":0}]';

$obj = json_decode($json);
print $obj->{'qty'}; // Result 0 

?>

推荐答案

JSON字符串外面的括号使它成为数组内部的对象.

The brackets on the outside of your JSON string are causing it to be an object inside of an array.

您可以通过$obj[0]->{'qty'};

或更改您的json字符串，使其直接实例化为对象.

OR change your json string so it instantiates into an object directly.

$json = '{"assetref":"","qty":0,"raw":0}';

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