Question

我正在尝试在我的项目中实现一个线程示例，我希望将用户输入的用户名和密码传入线程，这样就可以检测到帐户是否存在，但我是难以正确实施它。任何帮助将不胜感激。

以下是我收到的错误之一。

类'checkPassword2'必须在'AsyncTask'中声明为abstract或实现抽象方法＆gt;'doInBackground（Params ...）

package com.example.liam.ca3;

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


public class LoginScreen extends Activity {

    private final static String TAG = "Sleep";

    private static final String defName = "admin";
    private static final String defPassword = "admin1";
    private ProgressBar mProgressBar;



    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.loginscreen);

        mProgressBar = (ProgressBar) findViewById(R.id.progressBar);


        final EditText uname = (EditText) findViewById(R.id.username_edittext);
        final EditText passwd = (EditText) findViewById(R.id.password_edittext);

        final Button loginButton = (Button) findViewById(R.id.login_button);
        loginButton.setOnClickListener(new View.OnClickListener() {

            public void onClick(View v) {
                new checkPassword2().execute(uname, passwd);


                if (checkPassword(uname.getText(), passwd.getText())) {
                    Log.i("8", "accepted password");
                    // Create an explicit Intent for starting the HelloAndroid Activity
                    Intent startApp = new Intent(LoginScreen.this,
                            MainActivity.class);

                    // Use the Intent to start the HelloAndroid Activity
                    startActivity(startApp);

                } else {
                    Log.i("8", "failed password");
                    uname.setText("");
                    passwd.setText("");
                }
            }
        });
    }

    private boolean checkPassword(Editable uname, Editable passwd) {

        Log.i("2", uname.toString());
        Log.i("2", passwd.toString());

        if(uname.toString().equalsIgnoreCase(defName) && passwd.toString().equalsIgnoreCase(defPassword))
        {
            Log.i("8", "this should work");
            return true;
        }
        Log.i("8", "this shouldn't work");
        return false;
    }

    private class checkPassword2 extends AsyncTask<Integer, Integer, String> {
        protected int doInBackground(String... values) {
            int result = 0;

            String nameTry = values[0];
            String passTry = values[1];

            if(nameTry.equalsIgnoreCase(defName) && passTry.equalsIgnoreCase(defPassword))
            {
                Log.i("8", "this should work");
                result = 1;
            }
            else {
                Log.i("8", "this shouldn't work");
                result = 2;
            }

            for (int i = 1; i < 11; i++) {
                sleep();
                publishProgress(i * 10);
            }
            return result;
        }

        protected void onProgressUpdate(Integer... progress) {
            mProgressBar.setProgress(progress[0]);
        }

        protected void onPostExecute(Long result) {
            mProgressBar.setVisibility(ProgressBar.INVISIBLE);
        }

        private void sleep() {
            try {
                Thread.sleep(500);
            } catch (InterruptedException e) {
                Log.e(TAG, e.toString());
            }
        }
    }

}

Answer 1

查看文档：{{3}}

扩展AsyncTask时，需要指定三个泛型：

Params是传递给doInBackground（）
Progress是传递给onProgressUpdate（）和publishProgress（）的参数类型
Result是传递给onCancelled（）和onPostExecute（）的参数的类型，是doInBackground（）的返回类型

在您的代码中，声明的泛型类型与参数类型之间存在不匹配。你的泛型是＆＃34;整数，整数，字符串＆＃34;：

Params = Integer
进度=整数
结果=字符串

现在让我们看看你的代码......

doInBackground(String... values)

Misatch：它需要Integer参数，指定String参数。但是你确实返回了一个整数。

onPostExecute(Long result)

不匹配：它需要Integer参数，指定Long参数。

确保方法中的类型与generics子句中指定的类型相符，一切都应该没问题。

Android：将字符串传递给asyncTask

1 个答案: