javascript - 在 Ionic 4 应用程序中按下硬件后退按钮时防止返回

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“否​​”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 promise 。如果我不需要返回，我就拒绝这个 promise 。

    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()
              }
            }
         ]
      });
    })