Question

我有一个Android应用程序，用于响应回复JSON响应的Web查询。当查询格式正确时，一切正常。但是，当查询错误时，代码崩溃，我在堆栈跟踪中得到以下内容：

java.io.FileNotFoundException: https://******

我已将*****放在我放置隐私的查询位置。

如果我在浏览器中输入相同的查询，我会收到以下回复：

{"statusCode":401,"error":"Unauthorized","message":"Missing authentication"}

我的问题是 - 为什么我的代码会抛出错误而不是将其视为有效的JSON响应然后我可以解析？

这是我的代码：

@Override
protected String doInBackground(String... params) {

try {



URL u = new URL(params[0]);
    HttpURLConnection conn = (HttpURLConnection) u.openConnection();
    conn.setRequestMethod("GET");
    conn.connect();
    InputStream is = conn.getInputStream();

    // Read the stream              

    byte[] b = new byte[1024];
    ByteArrayOutputStream baos = new ByteArrayOutputStream();

    while ( is.read(b) != -1) {
            baos.write(b);
    }

    String JSONResp = new String(baos.toByteArray());

        return JSONResp;
    }
    catch(Throwable t) {
        t.printStackTrace();
    }
    return null;
}

Answer 1

您可以使用以下代码检查httpurlconnection对象的状态代码：

 if(conn.getResponseCode()==HttpURLConnection.HTTP_ACCEPTED)
            {
                //handle the accepted response
            }
            else
            {
                //handle the failed response
            }

Answer 2

您应该使用getErrorStream()来读取后端调用的错误流

如果出现错误，则从服务器返回输入流因为在远程服务器上找不到请求的文件。这个 stream可用于读取服务器将发回的数据。

InputStream is = null;
if (responseCode == 200) {
  is = urlConnection.getInputStream();
} else {
  is = urlConnection.getErrorStream();
}

Answer 3

使用以下代码来获取json响应表单服务器：

 private InputStream is = null;
   private String jsonString = "";
   private JSONObject jsonObj = null;

    public JSONObject getJSONFromUrl(final String url) { //url is your url

            try {
                DefaultHttpClient httpClient = new DefaultHttpClient();
                HttpPost httpPost = new HttpPost(url);

                HttpResponse httpResponse = httpClient.execute(httpPost);
                HttpEntity httpEntity = httpResponse.getEntity();

                is = httpEntity.getContent();

            } catch (UnsupportedEncodingException e) {
                e.printStackTrace();
            } catch (ClientProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }

            try {

                BufferedReader reader = new BufferedReader(new InputStreamReader(
                        is, "iso-8859-1"), 8);
                StringBuilder sb = new StringBuilder();
                String line;
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                is.close();
                if (sb.length() != 0)
                    jsonString = sb.toString();
            } catch (Exception e) {
                Log.e("_TAG", "Buffer Error: " + e.getMessage());
            }
            try {
                jsonObj = new JSONObject(jsonString);
            } catch (JSONException e) {
                Log.e("_TAG", "Parsing Error: " + e.getMessage());
            }
            return jsonObj;
        }

现在通过上面的方法从返回的jsonObj获取 statusCode 。

捕获404 JSON响应而不是崩溃

3 个答案: