一個(gè)異步請求/操作 模擬代碼如下
const request = (url) => new Promise((resolve, reject) => {
setTimeout(() => {
Math.random() > 0.8 ? resolve(url+' ok') : reject(url + ' fail')
}, 1000)
})
request('https://jirengu.com')
.then(data => console.log(data))
.catch(err => console.log(err))
要求：實(shí)現(xiàn)一個(gè)函數(shù)retry 昼扛，能對原函數(shù)進(jìn)行封裝，達(dá)到如下效果：如果如果請求失敗渺鹦，會自動重新發(fā)起請求，最多嘗試n次海铆。
const request = (url) => new Promise((resolve, reject) => {
setTimeout(() => {
Math.random() > 0.8 ? resolve(url+' ok') : reject(url + ' fail')
}, 1000)
})
function retry() {
...
}
let request4 = retry(request, 4)
request4('https://jirengu.com')
.then(data => console.log(data)) //只要成功1次挣惰，進(jìn)入該邏輯
.catch(err => console.log(err)) //連續(xù)失敗4次卧斟，進(jìn)入該邏輯
實(shí)現(xiàn)
const retry = (fn, n=3) => (...args) => {
function run() {
return fn(...args).then(data => {
return Promise.resolve(data)
}).catch(error => {
n--
if(n <= 0) {
return Promise.reject(error)
} else {
return run()
}
})
}
return run()
}
完整測試代碼
let i = 0
const request = (url) => new Promise((resolve, reject) => {
console.log(`第${++i}次執(zhí)行`)
setTimeout(() => {
Math.random() > 0.8 ? resolve(url + ' ok') : reject(url + ' fail')
}, 1000)
})
const retry = (fn, n=3) => (...args) => {
function run() {
return fn(...args).then(data => {
return Promise.resolve(data)
}).catch(error => {
n--
if(n <= 0) {
return Promise.reject(error)
} else {
return run()
}
})
}
return run()
}
let request4 = retry(request, 4)
request4('https://jirengu.com')
.then(data => console.log(data))
.catch(error => console.log(error))
