如何处理 Promise.race() 中的拒绝而不使 Promise 解析

Question

我想创建一个仅在 Promise 解析时运行s 的函数，并且在它拒绝或达到超时时不执行任何操作。

这是我的想法：

onlyRunIfResolvesInTime().then(function(){
   // only run if resolved
})

不幸的是，以下代码在达到超时时总是抛出 Uncaught (in promise) two 错误（promise2 拒绝）。

// This promise would be replaced with a function 
// that only can resolve under certain conditions, 
// but if it can't in time we want to reject.
const promise1 = new Promise((resolve, reject) => {
    setTimeout(resolve, 500, "one")
})

// This promise is the timeout that rejects if the 
// time limit is reached.
const promise2 = new Promise((resolve, reject) => {
    setTimeout(reject, 100, "two")
})

let onlyRunIfResolvesInTime = function () {
    return Promise.race([promise1, promise2])
}

onlyRunIfResolvesInTime()
    .then(() => {
        console.log("running function")
    })

如果我在 Promise.race() 中发现错误，如下所示

let onlyRunIfResolvesInTime = function () {
    return Promise.race([promise1, promise2])
        .catch(() => { })
}

然后我的 onlyRunIfResolvesInTime 函数总是解析并且运行是 then 函数，而不是在达到超时时什么都不做。

如果 Promise.race() 解决并忽略拒绝，如何让 onlyRunIfResolvesInTime 仅运行？

Answer 1

您可以使用一个变量：

result = promise.then(
        function(v) {
            isRejected = false;
            isPending = false;
            return v; 
        }, 
        function(e) {
            isRejected = true;
            isPending = false;
            throw e; 
        }
    );
function e() {
if(isPending===false){
setTimeout(10, e)
} else if(isRejected===true){
//rejected
} else if(isRejected===false){
//fulfilled
}
}
e()

Answer 2

忽略 onlyRunIfResolvesInTime

的拒绝句柄

onlyRunIfResolvesInTime()
  .then(() => {
    console.log("running function")
  })
  .catch(() => null) // timed out -> do nothing

如何处理 Promise.race() 中的拒绝而不使 Promise 解析

How to handle reject in Promise.race() without making the Promise resolve

javascript

promise