防止在 Ionic 4 App 中按下硬件后退按钮时返回

Question

 this.platform.backButton.subscribe(()=> { const alert = await this.alertController.create({ header: 'Confirm!', message: 'Do you want to go back!!!', buttons: [ { text: 'Yes', handler: () => { // Previous page loaded } }, { text: 'No', handler: () => { //Page should not go back. //This is where i want to write code,if the user clicks No and the back button function should be disabled. //Only when the user presses Yes,the page will go to previous. } } ] }); })

我不知道当用户按下 no 时如何处理，即禁用后退按钮功能或事件。

Answer 1

最后我解决了这个问题。因为从 backButton 发出的事件是一个承诺。如果我不需要回去，我只是拒绝那个承诺。

 this.platform.backButton.subscribe(()=> { const alert = await this.alertController.create({ header: 'Confirm!', message: 'Do you want to go back!!!', buttons: [ { text: 'Yes', handler: () => { // Previous page loaded } }, { text: 'No', handler: () => { reject() } } ] }); })

Answer 2

试试这个方法来防止后退按钮

this.platform.backButton.subscribeWithPriority(9999, () => {
      this.dismiss();
    });