在 for 循环内等待承诺

let currentProduct;

for (let i = 0; i < products.length; i++) { 
    currentProduct = products[i];

    subscription.getAll(products[i]._id)
        .then((subs) => {
            update(subs, currentProduct);
        });
}

我正在使用蓝鸟，方法getAll and update回报承诺。我怎么说“等到两个承诺返回，然后更新 currentProduct 值”？我对 JS 还很陌生...

如果您可以使用，这将很简单async/await:

// Make sure that this code 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 code 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:
    /*
    const subs = await subscription.getAll(products[i]._id);
    await update(subs, currentProduct);
    */
}

或者，如果您只能使用简单的承诺，则可以循环遍历所有产品，并将每个承诺放入.then最后一个循环的。这样，只有当前一个已经解决时，它才会前进到下一个（即使它首先迭代了整个循环）：

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 code 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 code 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))
}

在第二个片段中，循环仅设置整个链，但不执行链内的代码.then立即地。你的getAll直到前面的每个函数依次解决（这就是您想要的）后，函数才会运行。