Question

我想执行一个操作，就像我按下按钮一次然后它将移动到所选屏幕，当我按两次后退按钮然后它将显示对话框并要求退出。

我在堆栈溢出中尝试了很多例子，但没有一个能帮助我..

navaigation.java

    private int clickCount = 0;
    private long delay = 100;
    Timer timer = new Timer();

    @Override
    public void onBackPressed() {
        if (clickCount == 2) {
            super.onBackPressed();
            timer.cancel();
        } else{
            clickCount++;
            timer.schedule(new TimerTask() {
                @Override
                public void run() {
                    runOnUiThread(new Runnable() {
                        @Override
                        public void run() {
                            backButtonHandler();
                        }
                    });
                }

        }, delay);
    }

}




public void backButtonHandler() {
    AlertDialog.Builder alertDialog = new AlertDialog.Builder(
            Navigation.this);
    // Setting Dialog Title
    alertDialog.setTitle("Leave application?");
    // Setting Dialog Message
    alertDialog.setMessage("Are you sure you want to leave the application?");
    // Setting Icon to Dialog
    alertDialog.setIcon(R.drawable.m_visit);
    // Setting Positive "Yes" Button
    alertDialog.setPositiveButton("YES",
            new DialogInterface.OnClickListener() {
                public void onClick(DialogInterface dialog, int which) {
                    finish();
                }
            });
    // Setting Negative "NO" Button
    alertDialog.setNegativeButton("NO",
            new DialogInterface.OnClickListener() {
                public void onClick(DialogInterface dialog, int which) {
                    // Write your code here to invoke NO event
                    dialog.cancel();
                }
            });
    // Showing Alert Message
    alertDialog.show();
}

我试过这种方式，但它无法正常工作......

Answer 1

嘿，如果用户第一次点击后退按钮，请查看此代码我已经显示Snackbar。第二次应用程序将退出。

private boolean mIsBackAlreadyPressed;

    @Override
public void onBackPressed() {
    if (drawer.isDrawerOpen(GravityCompat.START)) {
        drawer.closeDrawer(GravityCompat.START);
    } else if (mIsBackAlreadyPressed) {
            mIsBackAlreadyPressed = false;
            super.onBackPressed();
            overridePendingTransition(R.anim.pull_in_left, R.anim.push_out_right);
        } else {
            mIsBackAlreadyPressed = true;
            Snackbar.make(drawer, R.string.press_back_twice, Snackbar.LENGTH_SHORT).show();

            new Handler().postDelayed(new Runnable() {
                @Override
                public void run() {
                    mIsBackAlreadyPressed = false;
                }
            }, 2000);
        }
    }

希望这有帮助。快乐编码。

Answer 2

试试这段代码：

import android.app.Activity;
import android.os.Bundle;
import android.view.Menu;
import android.widget.Toast;

public class MainActivity extends Activity {
private static final int TIME_DELAY = 2000;
private static long back_pressed;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
}

@Override
public void onBackPressed() {
    if (back_pressed + TIME_DELAY > System.currentTimeMillis()) {
        super.onBackPressed();
    } 
    else {
        AlertDialog.Builder alertDialogBuilder = new AlertDialog.Builder(this);
        alertDialogBuilder.setMessage("Are you sure, You wanted to exit");
        alertDialogBuilder.setPositiveButton("yes", 
        new DialogInterface.OnClickListener() {

        @Override
        public void onClick(DialogInterface arg0, int arg1) {
          finish();
        }
       });

        alertDialogBuilder.setNegativeButton("No",new
       DialogInterface.OnClickListener() {
       Override
       public void onClick(DialogInterface dialog, int which) {
        dismiss();
         }
       });

       AlertDialog alertDialog = alertDialogBuilder.create();
       alertDialog.show();
      }    
    }
    back_pressed = System.currentTimeMillis();
}

Answer 3

你能试试吗

 private boolean exit=false;//declare in public

public void onBackPressed() {
    // TODO Auto-generated method stub
    if (exit) {
        Intent intent = new Intent(Intent.ACTION_MAIN);
        intent.addCategory(Intent.CATEGORY_HOME);
        intent.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
        intent.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TASK);
        startActivity(intent);
        finish();
    } else {
        Toast.makeText(this, "Tap again to exit.", Toast.LENGTH_SHORT)
                .show();
        exit = true;
        new Handler().postDelayed(new Runnable() {
            @Override
            public void run() {
                exit = false;
            }
        }, 3 * 1000);
    }
}

Answer 4

试试这个：

    private Boolean close_app = false;
@Override
    public void onBackPressed() {
        if (close_app) {
            //finish(); // pressed twice
            backButtonHandler();
        } else {
            Toast.makeText(this, "Press Back again to Exit.",
                    Toast.LENGTH_SHORT).show();
            close_app = true;
            new Handler().postDelayed(new Runnable() {
                @Override
                public void run() {
                    close_app = false;
                }
            }, 3 * 1000);
        }
    }

这里Handler处理后退，只显示Toast，如果在3秒内再次按下，则会显示退出应用程序的对话框

Answer 5

如果按回去，那么你可以去任何你想要的地方

@Override
public void onBackPressed() {
    super.onBackPressed();
    Intent intent = new Intent(Exit.this,Home.class);
    startActivity(intent);
    finish();
}

如果你现在双击后退按钮，你可以退出尝试下面的代码

private Boolean exit = false;
@Override
public void onBackPressed() {
    if (exit) {
        finish(); // finish activity
    } else {
        Toast.makeText(this, "Press Back again to Exit.",
                Toast.LENGTH_SHORT).show();
        exit = true;
        new Handler().postDelayed(new Runnable() {
            @Override
            public void run() {
                exit = false;
            }
        }, 3 * 1000);

    }

}

如何在按下单按钮的同时返回上一屏幕以及如何在按下后退按钮两次时退出应用程序？

5 个答案: