如果可以使用
async/，这将很简单
await。
// Make sure that this pre is inside a function declared using// the `async` keyword.let currentProduct;for (let i = 0; i < products.length; i++) { currentProduct = products[i]; // By using await, the pre will halt here until // the promise resolves, then it will go to the // next iteration... await subscription.getAll(products[i]._id) .then((subs) => { // Make sure to return your promise here... return update(subs, currentProduct); }); // You could also avoid the .then by using two awaits: }或者，如果您只能使用简单的承诺，则可以遍历所有产品，并将每个承诺置于
.then最后一个循环中。这样，仅当前一个问题解决时，它才会前进到下一个问题（即使它将首先迭代整个循环）：
let currentProduct;let promiseChain = Promise.resolve();for (let i = 0; i < products.length; i++) { currentProduct = products[i]; // Note that there is a scoping issue here, since // none of the .then pre runs till the loop completes, // you need to pass the current value of `currentProduct` // into the chain manually, to avoid having its value // changed before the .then pre accesses it. const makeNextPromise = (currentProduct) => () => { // Make sure to return your promise here. return subscription.getAll(products[i]._id) .then((subs) => { // Make sure to return your promise here. return update(subs, currentProduct); }); } // Note that we pass the value of `currentProduct` into the // function to avoid it changing as the loop iterates. promiseChain = promiseChain.then(makeNextPromise(currentProduct))}在第二个代码段中，循环仅建立了整个链，但没有
.then立即执行内部代码。您的
getAll功能要等到之前的每个功能依次解决后才能运行。
