将 didOpen sweetalert2 参数的值传递给 .then

Question

我已经构建了 didOpen 参数并实现了一个 JSON 对象，其中包含值。

问题是...如何将 JSON 对象传递给 Swal.fire({ }) 的 then 部分。then 箭头函数

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
})

Answer 1

解决方案是在 swal 之后包含一些变量，将值传递给该变量并在 .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)
})

将 didOpen sweetalert2 参数的值传递给 .then

passing a value from didOpen sweetalert2 parameter to the .then

javascript

sweetalert2