为什么在Android Studio中进行JSON解析时无法访问URL?

时间:2019-08-20 12:00:44

标签: android json retrofit2 httphandler

我在做一个android项目。我为此项目使用MySQL,PHP,Android Studio。我创建一个服务器和数据库。我想在任何地方访问服务器。因此,我使用一个静态IP并添加一个端口。当我检查此端口时,我看到它是打开的,并且在任何地方都可以访问该IP。我尝试。

然后,我为JSON值创建一个表和php页面。

我的PHP页面返回以下内容:

{
  "News": [
    {
      "NewsID": "1",
      "NewsTitle": "A",
      "NewsDesc": "A",
      "NewsImage": "A.jpg",
      "NewsDate": "2019-08-20"
    }
  ]
}

现在,我想在android中解析此JSON值。我尝试使用基本的HTTP Request和Retrofit 2,但是我无法访问URL。我尝试其他网址,它们正在工作。但是,当我写我的IP地址时;我取空值。我不能解决这个问题。我该怎么办?

这是我在Java中的HttpHandler类:

public class HttpHandler {

    private static final String TAG = HttpHandler.class.getSimpleName();

    public HttpHandler() {
    }

    public String makeServiceCall(String reqUrl) {
        String response = null;
        try {
            URL url = new URL(reqUrl);
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            conn.setRequestMethod("GET");
            // read the response
            InputStream in = new BufferedInputStream(conn.getInputStream());
            response = convertStreamToString(in);
        } catch (MalformedURLException e) {
            Log.e(TAG, "MalformedURLException: " + e.getMessage());
        } catch (ProtocolException e) {
            Log.e(TAG, "ProtocolException: " + e.getMessage());
        } catch (IOException e) {
            Log.e(TAG, "IOException: " + e.getMessage());
        } catch (Exception e) {
            Log.e(TAG, "Exception: " + e.getMessage());
        }
        return response;
    }

    private String convertStreamToString(InputStream is) {
        BufferedReader reader = new BufferedReader(new InputStreamReader(is));
        StringBuilder sb = new StringBuilder();

        String line;
        try {
            while ((line = reader.readLine()) != null) {
                sb.append(line).append('\n');
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                is.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return sb.toString();
    }
}

然后在MainActivity类上对此进行解析:

   HttpHandler hh = new HttpHandler();
   String JSON = hh.makeServiceCall(url);

0 个答案:

没有答案