[英]PHP Notice: Trying to get property of non-object
When I get file content from javascript, decode it and get the value of object. 当我从javascript获取文件内容时,对其进行解码并获取对象的值。 It display trying to get property of non-project.
显示试图获取非项目的属性。 However, the object variable is successfully pass to another php class.
但是,对象变量已成功传递给另一个php类。 May I know why it cause this problem?
我可以知道为什么会导致这个问题吗? I had try all the solution from stackoverflow but it not work and it also display the same error.
我已经尝试了所有来自stackoverflow的解决方案,但是它不起作用,并且还会显示相同的错误。 The following is my code:
以下是我的代码:
$chat_info = file_get_contents("php://input");
$chat_request = json_decode($chat_info);
$username = $chat_request->username;
$messageContent = $chat_request->messageContent;
$dateTime = $chat_request->date_time;
$channel = $chat_request->channel;
$event = $chat_request->event;
You can also do this: 您也可以这样做:
$chat_info = file_get_contents("php://input");
$chat_request = json_decode($chat_info,true);
$username = $chat_request['username'];
$messageContent = $chat_request['messageContent'];
$dateTime = $chat_request['date_time'];
$channel = $chat_request['channel'];
$event = $chat_request['event'];
And after it if it show any error like undefined index
then the $chat_info
variable does not contain the value of a key that you want. 然后,如果它显示任何错误(
$chat_info
undefined index
则$chat_info
变量将不包含所需键的值。
Probably one of the object which you use get from time to time NULL value. 您使用的对象之一可能会不时获得NULL值。 Check this out, i hope that is correct answer.
检查一下,我希望这是正确的答案。
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