获取一个数字的PHP对象属性 [英] Get a PHP object property that is a number

问题描述

我正在尝试从解码为PHP对象的JSON数据中获取一个属性.这只是一个YouTube数据API请求，它返回一个视频对象，该视频对象具有类似的内容对象；

I'm trying to get a property from JSON data decoded into a PHP object. It's just a YouTube data API request that returns a video object that has a content object liks so;

[content] => stdClass Object
                (
                    [5] => https://www.youtube.com/v/r4ihwfQipfo?version=3&f=videos&app=youtube_gdata
                    [1] => rtsp://v4.cache7.c.youtube.com/CiILENy73wIaGQn6pSL0waGIrxMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp
                    [6] => rtsp://v6.cache3.c.youtube.com/CiILENy73wIaGQn6pSL0waGIrxMYESARFEgGUgZ2aWRlb3MM/0/0/0/video.3gp
                )

做

$object->content->5

抛出意外的T_DNUMBER"-完全有道理.但是，如何获取数字属性的值呢?

Throws "unexpected T_DNUMBER" - which makes perfect sense. But how do I get the value of a property that is a number?

我确定我应该知道这一点.预先感谢.

I'm sure I should know this. Thanks in advance.

推荐答案

这应该有效:

$object->content->{'5'}

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