json_decode()返回错误“注意:尝试获取非对象的属性"; [英] json_decode() returning error "Notice: Trying to get property of non-object"
问题描述
我正在尝试编写一个脚本,该脚本使用cURL从远程位置(在本例中为twitch.tv)获取JSON文件(不要以为该部分太相关,尽管无论如何我还是最好提一下).出于示例目的,假设它返回的JSON对象存储在变量中后如下所示:
I am trying to write a script that gets a JSON file from a remote location (in this case being twitch.tv) using cURL (don't think that part is too relevant, though I better mention it anyway). For example purposes, lets say the JSON object it returns looks something like this after being stored in a variable:
$json_object = {"_links":{"self":"https://api.twitch.tv/kraken/streams/gmansoliver","channel":"https://api.twitch.tv/kraken/channels/gmansoliver"},"stream":null}
我访问"stream"属性,我已经尝试了以下代码:
I access the "stream" property, I have tried the follow code:
<?php
$json_object = {"_links":{"self":"https://api.twitch.tv/kraken/streams/gmansoliver","channel":"https://api.twitch.tv/kraken/channels/gmansoliver"},"stream":null}
$json_decoded = json_decode($json_object, true);
echo $json_decoded->stream;
?>
尝试此操作时,出现错误注意:试图在第48行的D:\ Servers \ IIS \ Sites \ mysite \ getstream.php中获取非对象的属性".
When I try this, I get the error "Notice: Trying to get property of non-object in D:\Servers\IIS\Sites\mysite\getstream.php on line 48".
我使用json_decode()错了,还是我从抽搐发送的JSON对象出了问题?
Am I using json_decode() wrong, or is there something wrong with the JSON object I am being sent from twitch?
我现在有了JSON对象:
I now have the JSON object:
{"access_token": "qwerty1235","refresh_token": "asdfghjkl=","scope": ["user_read"]}
如果尝试使用json_decode()
对其进行解码,则会出现以下错误:Object of class stdClass could not be converted to string
.有什么建议吗?
If I try to decode it using json_decode()
I get the following error: Object of class stdClass could not be converted to string
. Any advice?
在此先感谢您的帮助
推荐答案
您正在将JSON解码为数组. json_decode($json_object, true);
将返回一个数组
You're decoding the JSON into an array. json_decode($json_object, true);
Will return an array
array (size=2)
'_links' =>
array (size=2)
'self' => string 'https://api.twitch.tv/kraken/streams/gmansoliver' (length=48)
'channel' => string 'https://api.twitch.tv/kraken/channels/gmansoliver' (length=49)
'stream' => null
如果删除第二个参数并将其作为json_decode($json_object)
If you remove the second parameter and run it as json_decode($json_object)
object(stdClass)[1]
public '_links' =>
object(stdClass)[2]
public 'self' => string 'https://api.twitch.tv/kraken/streams/gmansoliver' (length=48)
public 'channel' => string 'https://api.twitch.tv/kraken/channels/gmansoliver' (length=49)
public 'stream' => null
请参阅文档,如果为TRUE,则返回的对象将转换为关联数组.
See the documentation, When TRUE, returned objects will be converted into associative arrays.
这篇关于json_decode()返回错误“注意:尝试获取非对象的属性";的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!