将值从didOpen sweetalert2参数传递到.then
将值从didOpen sweetalert2参数传递到.then
提问于 2020-10-16 03:41:19
我已经构建了didOpen参数，并实现了一个内部带有值的JSON对象。
问题是。如何将JSON对象传递给Swal.fire({ }) .then箭头函数then部分
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
})回答 1
Stack Overflow用户
回答已采纳
发布于 2020-10-16 21:56:38
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)
})页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/64378624
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