URL 连接：如何返回状态为 != 200 的正文？

Question

我有一个 web 服务有时会返回状态 401。它带有一个 JSON 主体，类似于：

{"status": { "message" : "Access Denied", "status_code":"401"}}

现在，这是我用来发出服务器请求的代码：

HttpURLConnection conn = null;
try{
   URL url = new URL(/* url */);
   conn = (HttpURLConnection)url.openConnection(); //this can give 401
   JsonReader reader = new JsonReader(new InputStreamReader(conn.getInputStream()));

   JsonObject response = gson.fromJson(reader, JsonObject.class);
   //response handling
}catch(IOException ex){
       System.out.println(conn.getResponseMessage()); //not working
}

当请求失败时，我想阅读那个 json 正文，但是 getResponseMessage 只给了我一个通用的 "Unauthorized"...那么如何检索那个 JSON？

Answer 1

在非200响应的情况下可以调用conn.getErrorStream()：

HttpURLConnection conn = null;
try {
    URL url = new URL(/* url */);
    conn = (HttpURLConnection)url.openConnection(); //this can give 401
    JsonReader reader = new JsonReader(new InputStreamReader(conn.getInputStream()));

    JsonObject response = gson.fromJson(reader, JsonObject.class);
} catch(IOException ex) {
    JsonReader reader = new JsonReader(new InputStreamReader(conn.getErrorStream()));
    JsonObject response = gson.fromJson(reader, JsonObject.class);
}

通过 Stack Overflow 数据库进行粗略搜索会带您到 this article，其中提到了这个解决方案。

URL 连接：如何返回状态为 != 200 的正文？

URL connection: how to get body returned with status != 200?

java

httpurlconnection

gson