This is what works:
const limit = 1000
// fetchMyProducts(page, limit, flag)
return fetchMyProducts(1, 1, true)
.then(function (products) {
return fetchMyProducts(2, limit, false)
}).then(function (totalProducts) {
return fetchMyProducts(3, limit, false)
}).then(function (totalProducts) {
return fetchMyProducts(4, limit, false)
}).then(function (totalProducts) {
return fetchMyProducts(5, limit, false)
}).then(function (totalProducts) {
return fetchMyProducts(6, limit, false)
}).then(function (totalProducts) {
return fetchMyProducts(7, limit, false)
})
I am trying to get all the products in our system through fetch. The problem is, at the moment, I know how many products there are, but in 1 year / 3 years... who know??
I am trying to loop over a fetch dynamically and get all the products.
I have tried this, however it doesn't seem to get called at all.
return fetchMyProducts(1, 1, true)
.then(function (numberOfProducts) {
let pages = Math.ceil(numberOfProducts / 1000) + 1;
console.log(pages);
return getAllProducts = () => {
for (let i = 1; i < pages; i++) {
const element = array[i];
return fetchMyProducts(2, limit, false)
}
}
}).then(... something else)
Is there a way to loop over a fetch promise and return something when it's finished, then continue on doing something else?
You are looking for
const limit = 1000
let chain = Promise.resolve();
for (let i=1; i<8; i++) {
chain = chain.then(function(products) {
return fetchMyProducts(i, limit, false)
});
}
return chain;
which dynamically builds the promise chain that you spelled out.
For a more simple and efficient solution, consider using async
/ await
:
const limit = 1000
for (let i=1; i<8; i++) {
const products = await fetchMyProducts(i, limit, false);
}
return;
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.