Question

我有一个webservice返回，有时，状态401。它带有一个JSON主体，如：

{"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json正文，但是getResponseMessage只是给了我一个通用的“Unauthorized”...那么如何检索那个JSON？

Answer 1

如果是非200响应，您可以致电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); }：

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URL连接：如何以状态返回正文！= 200？

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