如果可以使用async/，这将很简单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 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。