Javascript 如何通过同时执行两个异步函数在收到响应值后执行函数
在同时执行asynchronous1和asynchronous2之后,我希望在接收到这两个响应值后执行LastFunction。您可以得到如下结果:Javascript 如何通过同时执行两个异步函数在收到响应值后执行函数,javascript,vue.js,Javascript,Vue.js,在同时执行asynchronous1和asynchronous2之后,我希望在接收到这两个响应值后执行LastFunction。您可以得到如下结果: export default { data: () =>({ data1: [], data2: [], lastData: [] }), mounted() { asynchronous1(val, (data)=>{ return this.data1 = data }
export default {
data: () =>({
data1: [],
data2: [],
lastData: []
}),
mounted() {
asynchronous1(val, (data)=>{
return this.data1 = data
})
asynchronous2(val, (data)=>{
return this.data2 = data
})
function lastFuction() {
this.lastData = this.data1.concat(this.data2)
}
},
}
假设asynchronous1和asynchronous2不返回承诺-这很可能是因为它们使用回调函数 你可以这样做(凌乱) 但它会起作用的 更好的解决方案是使用异步/等待和承诺
export default {
data: () =>({
data1: [],
data2: [],
lastData: []
}),
mounted() {
let lastFunction = (() => {
let count = 0;
return () => {
count++;
if (count === 2) {
this.lastData = this.data1.concat(this.data2);
}
};
})();
asynchronous1(val, (data) => {
this.data1 = data;
lastFunction();
})
asynchronous2(val, (data) => {
this.data2 = data;
lastFunction();
})
},
}
我可能会这么说:
export default {
data: () =>({
data1: [],
data2: [],
lastData: []
}),
async mounted() {
const p1 = new Promise(resolve => {
asynchronous1(val, (data) => {
this.data1 = data;
resolve();
})
});
const p2 = new Promise(resolve => {
asynchronous2(val, (data) => {
this.data2 = data;
resolve();
})
});
await Promise.all([p1, p2]);
this.lastData = this.data1.concat(this.data2);
},
}
这里的区别是,在两个异步函数完成之前,this.data1/this.data2不会被填充-这可能不符合您的喜好
你可以
export default {
data: () =>({
data1: [],
data2: [],
lastData: []
}),
async mounted() {
const promisify = fn => (...args) => new Promise(resolve => fn(...args, resolve));
const p1 = promisify(asynchronous1);
const p2 = promisify(asynchronous2);
const [r1, r2] = await Promise.all([p1(val), p2(val)]);
this.data1 = r1;
this.data2 = r2;
this.lastData = [...r1, ...r2];
},
}
而不是
const [r1, r2] = await Promise.all([
p1(val).then(r => this.data1 = r),
p2(val).then(r => this.data2 = r)
]);
const [r1, r2] = await Promise.all([asynchronous1(val), asynchronous2(val)]);
这将使它以与原始代码相同的方式填充-我只是讨厌使用async/await和。然后。。。但在某些情况下,它比纯异步/等待更好——在我看来
当然,如果asynchronous1/2可以在不指定回调函数的情况下返回承诺(现在越来越普遍),那么代码可能会更简单
i、 e
而不是
const [r1, r2] = await Promise.all([
p1(val).then(r => this.data1 = r),
p2(val).then(r => this.data2 = r)
]);
const [r1, r2] = await Promise.all([asynchronous1(val), asynchronous2(val)]);
您不能同时执行两个函数。@Pointy-抱歉,这意味着asynchronous2在返回Asynchronous1之前执行,前提是
asynchronous
返回一个承诺lastFunction
将永远不会执行,因为您返回的是firstoops,忘记清理了-根本不需要返回:p@Dan谢谢你的提醒你通常写的比这更好的答案。更好的方法是将两个异步函数包装在一个承诺中。@danh-开始:pThanks@Dan-这就是湿代码的问题(即非干代码)
const p1 = promisify(asynchronous1);
const p2 = promisify(asynchronous2);
const [r1, r2] = await Promise.all([p1(val), p2(val)]);