289

我一直试图表现出“你想退出吗？” 用户尝试退出 Activity 时的对话框类型。

但是我找不到合适的 API 挂钩。 Activity.onUserLeaveHint()最初看起来很有希望，但我找不到阻止 Activity 完成的方法。

4

11 回答 11

400

在 Android 2.0+ 中，这看起来像：

@Override
public void onBackPressed() {
    new AlertDialog.Builder(this)
        .setIcon(android.R.drawable.ic_dialog_alert)
        .setTitle("Closing Activity")
        .setMessage("Are you sure you want to close this activity?")
        .setPositiveButton("Yes", new DialogInterface.OnClickListener()
    {
        @Override
        public void onClick(DialogInterface dialog, int which) {
            finish();    
        }

    })
    .setNegativeButton("No", null)
    .show();
}

在早期版本中，它看起来像：

@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
    //Handle the back button
    if(keyCode == KeyEvent.KEYCODE_BACK) {
        //Ask the user if they want to quit
        new AlertDialog.Builder(this)
        .setIcon(android.R.drawable.ic_dialog_alert)
        .setTitle(R.string.quit)
        .setMessage(R.string.really_quit)
        .setPositiveButton(R.string.yes, new DialogInterface.OnClickListener() {

            @Override
            public void onClick(DialogInterface dialog, int which) {

                //Stop the activity
                YourClass.this.finish();    
            }

        })
        .setNegativeButton(R.string.no, null)
        .show();

        return true;
    }
    else {
        return super.onKeyDown(keyCode, event);
    }

}
于 2010-02-13T15:56:21.793 回答
190 
@Override
public void onBackPressed() {
    new AlertDialog.Builder(this)
           .setMessage("Are you sure you want to exit?")
           .setCancelable(false)
           .setPositiveButton("Yes", new DialogInterface.OnClickListener() {
               public void onClick(DialogInterface dialog, int id) {
                   ExampleActivity.super.onBackPressed();
               }
           })
           .setNegativeButton("No", null)
           .show();
}
于 2011-08-30T08:00:22.987 回答
33

已修改@user919216 代码.. 并使其与 WebView 兼容

@Override
public void onBackPressed() {
    if (webview.canGoBack()) {
        webview.goBack();

    }
    else
    {
     AlertDialog.Builder builder = new AlertDialog.Builder(this);
builder.setMessage("Are you sure you want to exit?")
       .setCancelable(false)
       .setPositiveButton("Yes", new DialogInterface.OnClickListener() {
           public void onClick(DialogInterface dialog, int id) {
                finish();
           }
       })
       .setNegativeButton("No", new DialogInterface.OnClickListener() {
           public void onClick(DialogInterface dialog, int id) {
                dialog.cancel();
           }
       });
AlertDialog alert = builder.create();
alert.show();
    }

}
于 2012-03-28T05:55:53.437 回答
24

我宁愿双击后退按钮退出，而不是退出对话框。

在这个解决方案中，它在第一次返回时会显示一个 toast，警告再次按下将关闭应用程序。在本例中不到 4 秒。

private Toast toast;
private long lastBackPressTime = 0;

@Override
public void onBackPressed() {
  if (this.lastBackPressTime < System.currentTimeMillis() - 4000) {
    toast = Toast.makeText(this, "Press back again to close this app", 4000);
    toast.show();
    this.lastBackPressTime = System.currentTimeMillis();
  } else {
    if (toast != null) {
    toast.cancel();
  }
  super.onBackPressed();
 }
}

令牌来自：http ://www.androiduipatterns.com/2011/03/back-button-behavior.html

于 2014-01-16T09:05:15.787 回答
22

如果您不确定对“返回”的调用是否会退出应用程序，或者是否会将用户带到另一个活动，您可以将上述答案包装在检查中，isTaskRoot()。如果您的主要活动可以多次添加到后台堆栈，或者您正在操作后台堆栈历史记录，则可能会发生这种情况。 

if(isTaskRoot()) {
    AlertDialog.Builder builder = new AlertDialog.Builder(this);
    builder.setMessage("Are you sure you want to exit?")
       .setCancelable(false)
       .setPositiveButton("Yes", new DialogInterface.OnClickListener() {
           public void onClick(DialogInterface dialog, int id) {
                YourActivity.super.onBackPressed;
           }
       })
       .setNegativeButton("No", new DialogInterface.OnClickListener() {
           public void onClick(DialogInterface dialog, int id) {
                dialog.cancel();
           }
       });
    AlertDialog alert = builder.create();
    alert.show();

} else {
    super.onBackPressed();
}
于 2013-11-27T00:43:15.367 回答
7

在中国，大部分App都会通过“点击两次”确认退出：

boolean doubleBackToExitPressedOnce = false;

@Override
public void onBackPressed() {
    if (doubleBackToExitPressedOnce) {
        super.onBackPressed();
        return;
    }

    this.doubleBackToExitPressedOnce = true;
    Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT).show();

    new Handler().postDelayed(new Runnable() {

        @Override
        public void run() {
            doubleBackToExitPressedOnce=false;                       
        }
    }, 2000);
}
于 2017-08-25T00:36:57.793 回答
6

首先super.onBackPressed();从onbackPressed()方法中删除，而不是下面的代码：

@Override
public void onBackPressed() {
    AlertDialog.Builder builder = new AlertDialog.Builder(this);
    builder.setMessage("Are you sure you want to exit?")
           .setCancelable(false)
           .setPositiveButton("Yes", new DialogInterface.OnClickListener() {
               public void onClick(DialogInterface dialog, int id) {
                    MyActivity.this.finish();
               }
           })
           .setNegativeButton("No", new DialogInterface.OnClickListener() {
               public void onClick(DialogInterface dialog, int id) {
                    dialog.cancel();
               }
           });
    AlertDialog alert = builder.create();
    alert.show();

}
于 2017-09-22T09:52:00.203 回答
5

使用 Lambda：

    new AlertDialog.Builder(this).setMessage(getString(R.string.exit_msg))
        .setTitle(getString(R.string.info))
        .setPositiveButton(getString(R.string.yes), (arg0, arg1) -> {
            moveTaskToBack(true);
            finish();
        })
        .setNegativeButton(getString(R.string.no), (arg0, arg1) -> {
        })
        .show();

您还需要在 gradle.build 中设置级别语言以支持 java 8：

compileOptions {
       targetCompatibility 1.8
       sourceCompatibility 1.8
}
于 2016-10-27T14:59:49.360 回答
4 
Just put this code in your first activity 

@Override
    public void onBackPressed() {
        if (drawerLayout.isDrawerOpen(GravityCompat.END)) {
            drawerLayout.closeDrawer(GravityCompat.END);
        }
        else {
// if your using fragment then you can do this way
            int fragments = getSupportFragmentManager().getBackStackEntryCount();
            if (fragments == 1) {
new AlertDialog.Builder(this)
           .setMessage("Are you sure you want to exit?")
           .setCancelable(false)
           .setPositiveButton("Yes", new DialogInterface.OnClickListener() {
               public void onClick(DialogInterface dialog, int id) {
                    finish();
               }
           })
           .setNegativeButton("No", null)
           .show();


            } else {
                if (getFragmentManager().getBackStackEntryCount() > 1) {
                    getFragmentManager().popBackStack();
                } else {

           super.onBackPressed();
                }
            }
        }
    }
于 2018-03-13T12:55:06.243 回答
2

我喜欢@Glee 方法并将其与下面的片段一起使用。

@Override
public void onBackPressed() {
    if(isTaskRoot()) {
        new ExitDialogFragment().show(getSupportFragmentManager(), null);
    } else {
        super.onBackPressed();
    }
}

使用片段的对话框：

public class ExitDialogFragment extends DialogFragment {

    @Override
    public Dialog onCreateDialog(Bundle savedInstanceState) {
        return new AlertDialog.Builder(getActivity())
            .setTitle(R.string.exit_question)
            .setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    getActivity().finish();
                }
            })
            .setNegativeButton(android.R.string.no, new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    getDialog().cancel();
                }
            })
            .create();
    }
}
于 2016-04-08T12:57:12.327 回答
1

另一种选择是在第一次后按时显示Toast/Snackbar要求再次按返回 Exit，这比显示AlertDialog确认用户是否要退出应用程序的侵入性要小得多。

您可以使用DoubleBackPress Android Library几行代码来实现此目的。显示类似行为的示例 GIF。

首先，将依赖项添加到您的应用程序：

dependencies {
    implementation 'com.github.kaushikthedeveloper:double-back-press:0.0.1'
}

接下来，在您的 Activity 中，实现所需的行为。

// set the Toast to be shown on FirstBackPress (ToastDisplay - builtin template)
// can be replaced by custom action (new FirstBackPressAction{...})
FirstBackPressAction firstBackPressAction = new ToastDisplay().standard(this);

// set the Action on DoubleBackPress
DoubleBackPressAction doubleBackPressAction = new DoubleBackPressAction() {
    @Override
    public void actionCall() {
        // TODO : Exit the application
        finish();
        System.exit(0);
    }
};

// setup DoubleBackPress behaviour : close the current Activity
DoubleBackPress doubleBackPress = new DoubleBackPress()
        .withDoublePressDuration(3000)     // msec - wait for second back press
        .withFirstBackPressAction(firstBackPressAction)
        .withDoubleBackPressAction(doubleBackPressAction);

最后，将此设置为后按时的行为。

@Override
public void onBackPressed() {
    doubleBackPress.onBackPressed();
}
于 2018-03-11T07:27:07.257 回答