怎样根据google chrome的请求信息判断请求是否成功?
我使用jquery的$.post方法提交数据,添加一条评论,php程序执行出错,可能自动终止了程序并返回了错误信息,$.post的回调函数也没有调用,但是在General部分的Status Code却是200,Response和Request部分也没有状态码,难道php程序出错也会返回一个200的状态码,怎样判断返回是否成功?
回复内容:
我使用jquery的$.post方法提交数据,添加一条评论,php程序执行出错,可能自动终止了程序并返回了错误信息,$.post的回调函数也没有调用,但是在General部分的Status Code却是200,Response和Request部分也没有状态码,难道php程序出错也会返回一个200的状态码,怎样判断返回是否成功?
下面调试的地方有一个response题主可以看一下
<code>200只表示请求成功 </code>
在php中有一个设置是控制是否显示错误。
如果我没有记错的话,应该是display_error
如果这个开启,那么当程序错误的时候会显示对应的错误信息,因为要显示错误信息,所以一定是200
。
之所以没有东西输出,是因为程序错误的时候你并没有输出东西到网页中去。
可能你将error_reporting
设置为0了,也就是不输出错误。
PS:display_error
是显示错误,error_reporting
是输出哪一些错误
但是如果你将这个关闭后,如果一旦php错误,那么将会返回500
。
看http状态码

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