Question

我正在尝试使用用户名和密码登录。如果用户名是正确的，则转到另一个Activity，否则保留在当前Activity。如果我第一次登录它成功，但第二次出错了。

错误.....

    04-21 07:41:13.915: E/AndroidRuntime(2125): java.lang.IllegalStateException: Cannot execute task: the task has already been executed (a task can be executed only once)
    04-21 07:41:13.915: E/AndroidRuntime(2125):     at android.os.AsyncTask.executeOnExecutor(AsyncTask.java:578)
    04-21 07:41:13.915: E/AndroidRuntime(2125):     at android.os.AsyncTask.execute(AsyncTask.java:534)
    04-21 07:41:13.915: E/AndroidRuntime(2125):     at com.bis.storyanimationapp.LoginActivity$1.onClick(LoginActivity.java:35)

代码： LoginActivity

//enter code here

package com.bis.storyanimationapp;

import android.app.Activity;
import android.os.Bundle;
import android.view.Menu;
import android.view.View;
import android.view.View.OnClickListener;

import com.bis.storyanimationapp.asynchtask.AsynchTask1;
import com.bis.storyanimationapp.view.LoginView;
import com.example.storyanimationapp.R;



public class LoginActivity extends Activity {




    LoginView loginView;

    AsynchTask1 objLoginAsycTask;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        loginView=new LoginView(this);
        super.onCreate(savedInstanceState);
        setContentView(loginView.getLayoutmain());
        objLoginAsycTask=new AsynchTask1(LoginActivity.this,loginView);
        loginView.getBtnLogin().setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                // TODO Auto-generated method stub
                objLoginAsycTask.execute("");
                //objLoginAsycTask.cancel(true);


            }
        });
    }


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

}

AsynchTask1

package com.bis.storyanimationapp.asynchtask;

import android.R.integer;
import android.app.Activity;
import android.app.ProgressDialog;
import android.content.Context;
import android.content.Intent;
import android.os.AsyncTask;
import android.widget.Toast;

import com.bis.storyanimationapp.LoginActivity;
import com.bis.storyanimationapp.MainActivity;
import com.bis.storyanimationapp.view.LoginView;

public class AsynchTask1 extends AsyncTask<String,integer, String>  {
    Activity activity;
    Context context;
    private ProgressDialog signUpProcess;
    LoginView view;

//  private AsyncInterface asyInter;
    public AsynchTask1(Activity activity,LoginView view){
        this.activity=activity;

        this.view=view;
        context=activity.getApplicationContext();

    }
    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        signUpProcess = ProgressDialog.show(this.activity, "Please Wait !!!",
                "Submitting Data For Registration!!!", true, false);
        Toast.makeText(context, "inpree ex.",10).show();

    }
    @Override
    protected void onPostExecute(String result) 
    {
        // TODO Auto-generated method stub
        super.onPostExecute(result);
        signUpProcess.dismiss();
        Toast.makeText(context,"post",5).show();
        myMethod(result);

    }
    private void myMethod(String result) {
        // TODO Auto-generated method stub
        if(view.getEdUserName().getText().toString().equals("abc")){

            Intent i=new Intent(context,MainActivity.class);
            activity.startActivity(i);
            //this.activity.finish();
    }else{
        Toast.makeText(context,"invalid user name",5).show();
//      Intent i=new Intent(context,LoginActivity.class);
//      activity.startActivity(i);
//      

    }

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

        return view.getEdUserName().getText().toString();
        //return null;
    }
}

Answer 1

正如@Raghunandan建议的那样，每次都创建一个asyntask的新实例：

public void onClick(View v) {
            new AsynchTask1().execute("");
        }
    });

Answer 2

你试试这个代码：

 loginView.getBtnLogin().setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub
   objLoginAsycTask=new AsynchTask1(LoginActivity.this,loginView);
            objLoginAsycTask.execute((Void) null);


        }
    });
}

Answer 3

AsyncTask个实例只能使用一次。

相反，只需将您的任务称为new AsynchTask1().execute("");

即可

来自AsyncTask API文档

Answer 4

每次登录时尝试创建AsyncTask的新实例。

@Override
   public void onClick(View v) {
            AsyncTask1 at = new AsyncTask1()
            at.execute()
   }

无法使用异步任务第二次登录

4 个答案: