java - Android 应用程序在模拟器上运行正常，但在真实设备上崩溃

Question

我正在开发一个使用 MySQL 数据库和 json 的 android 应用程序，它可以在模拟器上正常运行，但是当我在我的设备 (Galaxy S5) 上启动它时，我面临着强制关闭这是我对服务器的请求并获取 json 对象:

    btnFaalSazi.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View arg0) {

            String validatePhoneNumber = phoneNumber.getText().toString();
            if (validatePhoneNumber.matches("^(?:0\\d{10}|9\\d{9})$")) {

                txtError.setText("");
                structUsers.register_number = phoneNumber.getText().toString();
                String phone = structUsers.register_number;

                Log.i("LOG", phone);
                Log.i("LOG", "HELOOOOOO");

                final ArrayList<NameValuePair> params = new ArrayList<NameValuePair>();
                params.add(new BasicNameValuePair("register_number", phone));

                String result = Webservice.readUrl("http://192.168.10.110:2233/api/register", params);

                if (result != null) {
                    try {
                        G.users.clear();
                        JSONObject object = new JSONObject(result);
                        String status = object.optString("status");
                        String code = object.optString("code");
                        String message = object.optString("message");

                        Log.i("LOG", "status hast " + status);
                        Log.i("LOG", "code hast  " + code);
                        Log.i("LOG", "mesage hast  " + message);

                        if (status != null && code != null) {
                            if (Integer.parseInt(status) == -1) {
                                Intent intent = new Intent(ActivityRegisterNumber.this, ActivityRegisterCode.class);
                                intent.putExtra("REGISTERNUMBER", structUsers.register_number);
                                ActivityRegisterNumber.this.startActivity(intent);
                            }
                        }

                        if (status != null && message != null) {
                            if (Integer.parseInt(status) == 100) {
                                Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();
                                Log.i("LOG", "after error 100");
                            } else if (Integer.parseInt(status) == 101) {
                                Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();
                                Log.i("LOG", "after error 101");
                            } else if (Integer.parseInt(status) == 102) {
                                Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();
                                Log.i("LOG", "after error 102");
                            } else if (Integer.parseInt(status) == 103) {
                                Toast.makeText(getApplicationContext(), message, Toast.LENGTH_LONG).show();
                                Log.i("LOG", "after error 103");
                            }
                        }
                    }
                    catch (JSONException e) {
                        e.printStackTrace();
                    }
                }
            } else {
                txtError.setText("Wrong phone number");
            }

        }
    });

我认为应用程序在执行这一行时会崩溃:

                    String result = Webservice.readUrl("http://192.168.10.110:2233/api/register", params);

这是我的网络服务模块:

public class Webservice {

public static String readUrl(String url, ArrayList<NameValuePair> params) {

    try {

        HttpClient client = new DefaultHttpClient();
        HttpPost method = new HttpPost(url);

        if (params != null) {
            method.setEntity(new UrlEncodedFormEntity(params));
        }

        HttpResponse response = client.execute(method);

        InputStream inputStream = response.getEntity().getContent();
        String result = convertInputStreamToString(inputStream);

        return result;
    }
    catch (ClientProtocolException e) {
        e.printStackTrace();
    }
    catch (IOException e) {
        e.printStackTrace();
    }

    return null;
}


public static String convertInputStreamToString(InputStream inputStream) {
    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream));
        StringBuilder builder = new StringBuilder();

        String line = "";

        while ((line = reader.readLine()) != null) {
            builder.append(line);
        }

        return builder.toString();
    }
    catch (IOException e) {
        e.printStackTrace();
    }

    return null;
}
}

Answer 1

您的网络服务的 URL 在您的机器上是本地的。模拟器工作正常，因为它可能在同一网络上。

但是这个网址:http://192.168.10.110:2233/您的设备无法访问。这就是它出现超时错误并且您的应用程序崩溃的原因。

如果你想在你的设备上测试这个，也许你需要使用公共(public)网络或使用 wifi 和一些代理工具，如 Charles

希望你现在清楚了这个问题。