javascript 异步 sweetAlert

Question

所以我必须尝试创建一个在返回之前等待用户输入的异步函数，但是，我不太确定该怎么做：

  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) 部分代码将被执行，在这里我想返回该结果

谁能指出我正确的方向？

Answer 1

这样试试：

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);
    }
}