Question

我正在尝试为应用程序创建登录。但是我有一个问题。

这是我的代码：

package com.forgetmenot.loginregister;


import java.util.ArrayList;
import java.util.List;

import org.apache.http.NameValuePair;
import org.apache.http.message.BasicNameValuePair;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import android.app.Activity;
import android.content.Intent;
import android.os.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;

public class MainActivity extends Activity {
    EditText uname, password;
    Button submit;
    // Creating JSON Parser object
    JSONParser jParser = new JSONParser();
    private static final String TAG = "Login";

    JSONObject json;
    private static String url_login = "http://localhost:8080/ForgetMeNotApplication/Login";
   //JSONArray incoming_msg = null;
   @Override
   protected void onCreate(Bundle savedInstanceState) {
       super.onCreate(savedInstanceState);
       setContentView(R.layout.activity_main);
       findViewsById();
       submit.setOnClickListener(new View.OnClickListener() {

           @Override
           public void onClick(View arg0) {
               // execute method invokes doInBackground() where we open a Http URL connection using the given Servlet URL
               //and get output response from InputStream and return it.
                new Login().execute();

           }
       });
    }
    private void findViewsById() {

        uname = (EditText) findViewById(R.id.txtUser);
        password = (EditText) findViewById(R.id.txtPass);
        submit = (Button) findViewById(R.id.login);
    }
    private class Login extends AsyncTask<String, String, String>{

        @Override
        protected String doInBackground(String... args) {
            // Getting username and password from user input

            String username = uname.getText().toString();
            String pass = password.getText().toString();

            List<NameValuePair> params = new ArrayList<NameValuePair>();
            params.add(new BasicNameValuePair("u",username));
            params.add(new BasicNameValuePair("p",pass));
            json = jParser.makeHttpRequest(url_login, "GET", params);
            String s=null;

            try {
                s= json.getString("info");
                Log.d("Msg", json.getString("info"));
                if(s.equals("success")){
                    Intent login = new Intent(getApplicationContext(), home.class);
                    login.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
                    startActivity(login);
                    finish();
                }
            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            return null;
        }

    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        // Inflate the menu; this adds items to the action bar if it is present.
        getMenuInflater().inflate(R.menu.menu_main, menu);
        return true;
    }

    @Override
    public boolean onOptionsItemSelected(MenuItem item) {
        // Handle action bar item clicks here. The action bar will
        // automatically handle clicks on the Home/Up button, so long
        // as you specify a parent activity in AndroidManifest.xml.
        int id = item.getItemId();
        if (id == R.id.action_settings) {
            return true;
        }
        return super.onOptionsItemSelected(item);
    }
}

Android工作室说必须从结构中的UI线程调用方法getText()：uname.getText().toString();和password.getText().toString();可能的解决方案??

Answer 1

尝试通过Login AsyncTask方法将您的值传递给execute(param1, param1, ..., paramN)：

submit.setOnClickListener(new View.OnClickListener() {

           @Override
           public void onClick(View arg0) {

               String username = uname.getText().toString();
               String pass = password.getText().toString();
               new Login().execute(username, pass);

           }
       });
    }
    private void findViewsById() {

        uname = (EditText) findViewById(R.id.txtUser);
        password = (EditText) findViewById(R.id.txtPass);
        submit = (Button) findViewById(R.id.login);
    }
    private class Login extends AsyncTask<String, String, String>{

        @Override
        protected String doInBackground(String... args) {
            // Getting username and password from user input

            String username = args[0];
            String pass = args[1];

Answer 2

制作username和pass login类变量并覆盖onPreExcecute()并执行此操作：

    @Override
    protected void onPreExecute() {

        username = uname.getText().toString();
        pass = password.getText().toString();

    }

Answer 3

您正在从后台线程访问UI线程：

String username = uname.getText().toString();
String pass = password.getText().toString();

您要做的只是将用户名/密码字符串传递给后台任务构造函数，或者您可以将它们直接传递给execute方法。如果要求它们（就像你的那样），我更喜欢将它们定义到构造函数中。

定义您的LoginTask，如

String uname;
String password;
public Login(String username, String password({
    this.uname = username;
    this.password = password;
}

然后在doInBackground（）中使用成员。

List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("u",this.username));
params.add(new BasicNameValuePair("p",this.pass));
json = jParser.makeHttpRequest(url_login, "GET", params);

编辑 - 这样你的新Login（）。execute（）调用看起来就像这个

new Login(uname.getText().toString(), password.getText().toString()).execute();

Answer 4

除非我不知道有什么改变，否则这应该不是问题。 UI元素无法从后台更新，但访问其getter从未成为问题。

无论如何，您可以通过向AsyncTask添加一个构造函数来解决这个问题，这个构造函数需要两个String，然后在创建任务时发送它们。

private class Login extends AsyncTask<String, String, String>{

    // member variables of the task class
    String uName, pwd
    public Login(String userName, String password) {
        uName = userName;
        pwd = password;
    }

    @Override
    protected String doInBackground(String... args) {...}

并将其传递到onClick()

 @Override
 public void onClick(View arg0) {
     // execute method invokes doInBackground() where we open a Http URL connection using the given Servlet URL
     //and get output response from InputStream and return it.

     // pass them here
     new Login(uname.getText().toString(), password.getText().toString()).execute();
 }

Answer 5

抛出异常是因为从后台线程调用doInBackground()。我会将两个String参数添加到Login类的构造函数中。

我想建议您在Android上获得有关AsyncTasks和线程的更多知识。例如，官方文档中有一个页面：http://developer.android.com/guide/components/processes-and-threads.html。

您可能还会看一下这门课程：https://www.udacity.com/course/developing-android-apps--ud853。您可以学习Android框架的许多基础知识。

方法getText（）必须从UI线程（Android Studio）调用

5 个答案: