Question

我正在使用以下代码作为后退按钮，但如果我按下用户设备的后退按钮，它无法退出应用程序

  private Boolean exit = false;

        @Override
        public void onBackPressed() {
            if (exit) {
                this.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);

            }

        }

Answer 1

private Boolean exit = false;

        @Override
        public void onBackPressed() {
            if (exit) {
                this.finishAffinity();
            } 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);

            }

        }

Answer 2

使用此...

private long lastBackPressTime = 0;

     @Override
    public void onBackPressed() {

        if (this.lastBackPressTime < System.currentTimeMillis() - 4000) {

            //Add Snhackbar or Toast, whatever you want
            this.lastBackPressTime = System.currentTimeMillis();

        } else {
            if (toast != null) {
                toast.cancel();
            }
            moveTaskToBack(true);

        }

    }

Answer 3

这可能有用。请改用finishAffinity()。

  boolean doubleBackToExitPressedOnce = false;

  public void onBackPressed(){
        if(doubleBackToExitPressedOnce)
        {
            finishAffinity();
            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;

            }
        },3500);
    }

Answer 4

保持布尔值toClose = false; 并且在onBackPressed（）方法中使用此代码

if (toClose)
    super.onBackPressed();
    else
    {
        toClose=true;
        Utils.showToast(context,getString(R.string.back_again));
        new Handler().postDelayed(new Runnable() {
            @Override
            public void run() {
                toClose=false;
            }
        },2000);
    }

Answer 5

像这样更改你的代码。

 private Boolean exit = false;

            @Override
            public void onBackPressed() {
                if (exit) {
                    this.finishAffinity();
                } 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);

                }

            }

Answer 6

这对你有用全局定义

private boolean doubleBackToExitProceed = false;
@Override
  public void onBackPressed() {
    if (doubleBackToExitProceed) {
      finish();
      return;
    }
    doubleBackToExitProceed = true;
    Toast.makeText(this, "Press Back again to Exit.", Toast.LENGTH_SHORT).show();

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

      @Override
      public void run() {
        doubleBackToExitProceed = false;
      }
    }, 2000);// time in milliseconds.
  }

Answer 7

使用它可以在按下后退按钮时关闭应用程序。

@Override
    public void onBackPressed() {

            Intent intent = new Intent(Intent.ACTION_MAIN);
            startMain.addCategory(Intent.CATEGORY_HOME);
            startActivity(intent);
        }

Android后退按钮应用程序未关闭

7 个答案: