从didOpen sweetalert2参数传递值到.then

问题描述

我已经构建了didOpen参数，并实现了一个内部包含值的JSON对象。

问题是...如何将JSON对象传递给Swal.fire（{}）的随后部分。然后使用箭头功能

Swal.fire({
  html: ` HERE IS ALL THE ID's STUFF `,didopen() {
    let itemPOST = {
      /* the json i trying to catch */ }

    // do bunch of stuff and save it into itemPOST

  } //==>end of didOpen
}).then(() => {
  //console log here for itemPOST
})

解决方法

解决方案是在swal之后包含一些变量，将值传递给该变量，然后在.then（）箭头函数中调用它

let giveMeTheValue;  //<==    variable to receive the value
Swal.fire({
  html: ` HERE IS ALL THE ID's STUFF `,didOpen() {
    let itemPOST = { /* the json with data */ }

    // do bunch of stuff and save it into itemPOST

    giveMeTheValue = itemPOST   //<== giving the data to pass to then()
  } 
}).then(() => {
  console.log(giveMeTheValue)
})