ionic3监听android返回按钮事件实现双击退出功能

有时候我们有这样的需求，就是当按下手机的返回按钮，在指定的页面双击退出，而在其他页面点击一次返回到上个页面；下面我能看看ionic3监听android返回按钮事件实现双击退出功能。

1.首先需要引入Platform

 import {Platform, ToastController, Nav, IonicApp} from 'ionic-angular';

2.需要注入 Platform

constructor(public ionicApp: IonicApp, public platform: Platform, public toastCtrl: ToastController) {
     platform.ready().then(() => {
       StatusBar.styleDefault();
       Splashscreen.hide();
       this.registerBackButtonAction();//注册返回按键事件
     });
   }

3.监听this.platform.registerBackButtonAction

import {Component, ViewChild} from '@angular/core';
 import {Platform, ToastController, Nav, IonicApp} from 'ionic-angular';
 import {StatusBar, Splashscreen} from 'ionic-native';
 import {TabsPage} from '../pages/tabs/tabs';

 @Component({
   templateUrl: 'app.html'
 })
 export class MyApp {
   rootPage = TabsPage;
   backButtonPressed: boolean = false;  //用于判断返回键是否触发
   @ViewChild('myNav') nav: Nav;

   constructor(public ionicApp: IonicApp, public platform: Platform, public toastCtrl: ToastController) {
     platform.ready().then(() => {
       StatusBar.styleDefault();
       Splashscreen.hide();
       this.registerBackButtonAction();//注册返回按键事件
     });
   }

   registerBackButtonAction() {
     this.platform.registerBackButtonAction(() => {
       //如果想点击返回按钮隐藏toast或loading或Overlay就把下面加上
       // this.ionicApp._toastPortal.getActive() || this.ionicApp._loadingPortal.getActive() || this.ionicApp._overlayPortal.getActive()
       let activePortal = this.ionicApp._modalPortal.getActive();
       if (activePortal) {
         activePortal.dismiss().catch(() => {});
         activePortal.onDidDismiss(() => {});
         return;
       }
       let activeVC = this.nav.getActive();
       let tabs = activeVC.instance.tabs;
       let activeNav = tabs.getSelected();
       return activeNav.canGoBack() ? activeNav.pop() : this.showExit()
     }, 1);
   }

   //双击退出提示框
   showExit() {
     if (this.backButtonPressed) { //当触发标志为true时，即2秒内双击返回按键则退出APP
       this.platform.exitApp();
     } else {
       this.toastCtrl.create({
         message: '再按一次退出应用',
         duration: 2000,
         position: 'top'
       }).present();
       this.backButtonPressed = true;
       setTimeout(() => this.backButtonPressed = false, 2000);//2秒内没有再次点击返回则将触发标志标记为false
     }