Question

我是android的新手我知道这个问题多次问过，但我找不到合适的解决方案。我想将password和username发送到服务器并进行检查。它返回JSON对象，如果它是零或一，我检查对象的值。问题是在catch中获取消息

请求失败：android.os.networkonmainthreadException

这是我的代码

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_loginpage);
   editTextUserName = (EditText) findViewById(R.id.user);
    editTextPassword = (EditText) findViewById(R.id.password);
    message=(TextView) findViewById(R.id.mess);
    String username = editTextUserName.getText().toString();
    String password = editTextPassword.getText().toString();
    Button send =(Button)findViewById(R.id.send);
    send.setOnClickListener(
            new View.OnClickListener(){
                @Override
                public void onClick(View view) {

                  // invokeLogin();
                    clickbuttonRecieve();

                }
            }

    );
}


public void clickbuttonRecieve() {
    try {
        JSONObject json = new JSONObject();
        String username = editTextUserName.getText().toString();
        String password = editTextPassword.getText().toString();
        json.put("userName",username);
        json.put("password", password);
        int timeconnection=3000;
        HttpParams httpParams = new BasicHttpParams();
        HttpConnectionParams.setConnectionTimeout(httpParams,
                timeconnection);
        HttpConnectionParams.setSoTimeout(httpParams, timeconnection);
        HttpClient client = new DefaultHttpClient(httpParams);
        //
        //String url = "http://10.0.2.2:8080/sample1/webservice2.php?" +
        //             "json={\"UserName\":1,\"FullName\":2}";
        String url = "http://phone.tmsline.com/checkuser";

        HttpPost request = new HttpPost(url);
        request.setEntity(new ByteArrayEntity(json.toString().getBytes(
                "UTF8")));
        request.setHeader("json", json.toString());
        HttpResponse response = client.execute(request);
        HttpEntity entity = response.getEntity();
        // If the response does not enclose an entity, there is no need
        if (entity != null) {
            JSONObject jsonget = new JSONObject();
            String login = jsonget.getString("msg");
            if (login.toString().equalsIgnoreCase("1")){
                Toast.makeText(this, "Request success: " + login,
                        Toast.LENGTH_LONG).show();

            }
            else{
                Toast.makeText(this, "Request failed: " + login,
                        Toast.LENGTH_LONG).show();
            }

        }
    } catch (Throwable t) {
        Toast.makeText(this, "Request failed: " + t.toString(),
                Toast.LENGTH_LONG).show();
    }
}

php代码（注意：我没有编写php代码，另一个负责或者那个）

public function check_user(Request $request){
    $username = $request->username;
    $password = $request->password;
     if (Auth::attempt(['username' => $username, 'password' => $password])) {
    // return view('test',['user'=>$username]);
        return response()->json(['msg','1']);
}

return response()->json(['msg','0']);
}

Answer 1

您应该使用不同的线程发送Epoch 1/1 23/718 [..............................] - ETA: 522s - loss: 8.4146请求而不是主（UI）线程。您只需使用AsyncTask，Here就是一个很好的例子。或者甚至更好地使用volley或okhttp，因为它们默认情况下异步处理请求。

Answer 2

requested failed:android.os.networkonmainthreadException

您无法在主线程中运行网络请求。你应该创建新的线程。获取它的最简单方法是使用异步任务https://stackoverflow.com/a/24399320/2717821

但在生产中不建议使用。非常流行的方法是使用RxJava https://github.com/ReactiveX/RxJava

Answer 3

我已经改变了你的方法（clickbuttonRecieve）并编写了Convertor Method for Convert InputStream to String（ConvertInputStreamToString）：

public void clickbuttonRecieve() {
try {
    JSONObject json = new JSONObject();
    String username = editTextUserName.getText().toString();
    String password = editTextPassword.getText().toString();
    json.put("userName",username);
    json.put("password", password);
    int timeconnection=3000;
    HttpParams httpParams = new BasicHttpParams();
    HttpConnectionParams.setConnectionTimeout(httpParams,
            timeconnection);
    HttpConnectionParams.setSoTimeout(httpParams, timeconnection);
    HttpClient client = new DefaultHttpClient(httpParams);
    //
    //String url = "http://10.0.2.2:8080/sample1/webservice2.php?" +
    //             "json={\"UserName\":1,\"FullName\":2}";
    String url = "http://phone.tmsline.com/checkuser";

    HttpPost request = new HttpPost(url);
    request.setEntity(new ByteArrayEntity(json.toString().getBytes(
            "UTF8")));
    request.setHeader("json", json.toString());
    HttpResponse response = client.execute(request);
    HttpEntity entity = response.getEntity();


    InputStream inputStream = response.getEntity().getContent();
    String result = ConvertInputStreamToString(inputStream);
    // If the response does not enclose an entity, there is no need
    if (result!= null) {
        JSONObject jsonget = new JSONObject(result);

       if(jsonget.has("msg")){
        String login = jsonget.getString("msg");
        if (login.toString().equalsIgnoreCase("1")){
            Toast.makeText(this, "Request success: " + login,
                    Toast.LENGTH_LONG).show();

        }
        else{
            Toast.makeText(this, "Request failed: " + login,
                    Toast.LENGTH_LONG).show();
        }
      }
    }
  } catch (Throwable t) {
    Toast.makeText(this, "Request failed: " + t.toString(),
            Toast.LENGTH_LONG).show();
  }
}

private static String ConvertInputStreamToString(InputStream inputStream)
        throws NaabException {
    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) {
        throw new NaabException(e);
    } catch (Exception e) {
        throw new NaabException(e);
    }

}

使用JSON检查用户名和密码

3 个答案: