javascript异步sweetAlert

Question

所以我必须尝试创建一个异步函数，在返回之前等待用户输入，但是，我不太确定如何去做：

  async createAlert() {
      return await swal({
          title: 'Are you sure?',
          text: "You won't be able to revert this!",
          type: 'warning',
          showCancelButton: true,
          confirmButtonText: 'Yes, delete it!',
          cancelButtonText: 'No, cancel!',
          reverseButtons: true
      }).then(function (result) {
          //user has answered we want to return the result
      })
  }

此 jquery 创建以下弹出窗口：

当用户按下任一按钮时 (then ) 部分代码，在这里我想返回该结果

任何人都可以指出我正确的方向吗？

Answer 1

像这样尝试：

async createAlert() {
    try{
      let result = await swal({
          title: 'Are you sure?',
          text: "You won't be able to revert this!",
          type: 'warning',
          showCancelButton: true,
          confirmButtonText: 'Yes, delete it!',
          cancelButtonText: 'No, cancel!',
          reverseButtons: true
      });
      // SUCCESS
      return result;
    }catch(e){
        // Fail!
        console.error(e);
    }
}