java - URL 连接 : how to get body returned with status ! = 200？

Question

我的网络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，我想读取该 json 正文，但是 getResponseMessage 只给我一个通用的“未经授权”...那么如何检索该 JSON？

Answer 1

如果响应不是 200，您可以调用 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，您会找到 this article其中提到了这个解决方案。