如何在异步函数中执行javascript等待用户输入
如何在异步函数中执行javascript等待用户输入
提问于 2018-06-25 03:45:53
我遇到了一种情况，需要我弹出一个窗口，让用户在关闭后根据用户输入做出选择，并发出另一个http请求。弹出窗口后，我不知道如何等待。
async function checkRemote(url1, url2) {
var resp
resp = await fetch(url1).then(r => r.json())
if (r.condition1 == 100) {
setState({showPopup: true}) //in a reactjs app
//how do I do await here to wait for the popup being closed
//get the user choice in variable "proceed"
}
if (proceed) {
resp = await fetch(url2)
//do some more work
}
}回答 2
Stack Overflow用户
回答已采纳
发布于 2018-06-25 05:34:01
创建promise，在popup closed事件处理程序中解析它，并在函数中等待它。
var popupClosed = new Promise(function(resolve, reject) {
// create popup close handler, and call resolve in it
});
async function checkRemote(url1, url2) {
var resp
resp = await fetch(url1).then(r => r.json())
if (r.condition1 == 100) {
setState({showPopup: true}) //in a reactjs app
var closed = await popupClosed;
}
if (proceed) {
resp = await fetch(url2)
//do some more work
}
}Stack Overflow用户
发布于 2018-06-26 22:11:21
基于@hikmat-gurbanli的回答，这里有一个可行的解决方案。其思想是保存resolve函数，以便将来某个句柄可以调用它来解除异步函数的阻塞。
const fetch = require('node-fetch')
var savedResolve;
test("http://localhost/prod/te.php");
async function test(url) {
await checkRemote(url)
console.log("completed")
}
var popupClosed = new Promise(function(resolve, reject) {
// create popup close handler, and call resolve in it
console.log("got here")
savedResolve = resolve;
});
async function checkRemote(url1) {
var resp
resp = await fetch(url1).then(r => r.text())
console.log("resp " + resp)
//setState({showPopup: true}) //in a reactjs app
var result = await popupClosed;
console.log("result: ")
console.log(result)
}处理程序可以只调用savedResolve.resolve("Yes")，它将在var result = await popupClosed;行取消阻塞异步函数checkRemote
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/51013412
复制相关文章
相似问题
腾讯云开发者
