java 擷取400的錯誤資訊_java – 如何獲得HttpResponseException背後的實際錯誤？

我正在使用Apache HttpComponents Client來POST一個傳回

JSON的伺服器.問題是,如果伺服器傳回400錯誤,我似乎無法告訴Java錯誤是什麼(到目前為止不得不求助于資料包嗅探器 – 荒謬).這是代碼：

HttpClient httpclient = new DefaultHttpClient();

params.add(new BasicNameValuePair("format", "json"));

params.add(new BasicNameValuePair("foo", bar));

HttpPost httppost = new HttpPost(uri);

// this is how you set the body of the POST request

httppost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));

String responseBody = "";

try {

// Create a response handler

ResponseHandler responseHandler = new BasicResponseHandler();

responseBody = httpclient.execute(httppost, responseHandler);

} catch(HttpResponseException e) {

String error = "unknown error";

if (e.getStatusCode() == 400) {

// TODO responseBody and e.detailMessage are null here,

// even though packet sniffing may reveal a response like

// Transfer-Encoding: chunked

// Content-Type: application/json

//

// 42

// {"error": "You do not have permissions for this operation."}

error = new JSONObject(responseBody).getString("error"); // won't work

}

// e.getMessage() is ""

}

我究竟做錯了什麼？必須有一種簡單的方法來擷取400錯誤的消息.這是基本的.