Java 返回 2xx 以外的代码时，如何使用 HttpURLConnection 获取响应正文？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/25011927/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to get response body using HttpURLConnection, when code other than 2xx is returned?
提问by kiedysktos
I have problem with retrieving Json response in case when server returns error. See details below.
我在检索 Json 响应时遇到问题，以防服务器返回错误。请参阅下面的详细信息。
How I perform the request
我如何执行请求
I use java.net.HttpURLConnection. I setup request properties, then I do:
我用java.net.HttpURLConnection. 我设置请求属性，然后我做：
conn = (HttpURLConnection) url.openConnection();
After that, when request is successful, I get response Json:
之后，当请求成功时，我得到响应 Json：
br = new BufferedReader(new InputStreamReader((conn.getInputStream())));
sb = new StringBuilder();
String output;
while ((output = br.readLine()) != null) {
sb.append(output);
}
return sb.toString();
... and the problem is:
......问题是：
I can't retrieve Json received when the server returns some error like 50x or 40x,. Following line throws IOException:
当服务器返回诸如 50x 或 40x 之类的错误时，我无法检索收到的 Json。以下行抛出 IOException：
br = new BufferedReader(new InputStreamReader((conn.getInputStream())));
// throws java.io.IOException: Server returned HTTP response code: 401 for URL: www.example.com
The server sends body for sure, I see it in external tool Burp Suite:
服务器肯定会发送正文，我在外部工具 Burp Suite 中看到它：
HTTP/1.1 401 Unauthorized
{"type":"AuthApiException","message":"AuthApiException","errors":[{"field":"email","message":"Invalid username and/or password."}]}
I can get response message (i.e. "Internal Server Error") and code (i.e. "500") using following methods:
我可以使用以下方法获取响应消息（即“内部服务器错误”）和代码（即“500”）：
conn.getResponseMessage();
conn.getResponseCode();
But I can't retrieve request body... maybe there is some method I didn't notice in the library?
但是我无法检索请求正文......也许图书馆中有一些我没有注意到的方法？
采纳答案by user207421
If the response code isn't 200 or 2xx, use getErrorStream()instead of getInputStream().
如果响应代码不是 200 或 2xx，请使用getErrorStream()代替getInputStream().
回答by kiedysktos
To make things crystal clear, here is my working code:
为了让事情变得清晰，这是我的工作代码：
if (200 <= conn.getResponseCode() && conn.getResponseCode() <= 299) {
br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
} else {
br = new BufferedReader(new InputStreamReader(conn.getErrorStream()));
}
回答by Sharhabeel Hamdan
This is an easy way to get a successful response from the server like PHP echo otherwise an error message.
这是一种从服务器获得成功响应的简单方法，例如 PHP echo，否则会出现错误消息。
BufferedReader br = null;
if (conn.getResponseCode() == 200) {
br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String strCurrentLine;
while ((strCurrentLine = br.readLine()) != null) {
System.out.println(strCurrentLine);
}
} else {
br = new BufferedReader(new InputStreamReader(conn.getErrorStream()));
String strCurrentLine;
while ((strCurrentLine = br.readLine()) != null) {
System.out.println(strCurrentLine);
}
}
