如何提高嵌套for回圈的性能

這是我的 for 回圈，它運行 2 個列舉，并將它們都發送到服務器，獲取一個值，并通過另一個 for 回圈計算該值。很確定它可以改進。這是代碼：

  const paths = [];
  for await (let category of Object.values(CategoriesEnum)) {
    for await (let format of Object.values(FormatsEnum)) {
      const totalPosts = await getPagesCount(category, format);
      const totalPages = Math.ceil(totalPosts.offsetPagination.total / 12);
      for (let page = 1; page <= totalPages; page  ) {
        paths.push({ params: { category: category , format: format page: page } });
      }
    }
  }
return paths;

我的主要目標是減少時間，盡管我知道服務器將獲得相同數量的查詢，因此差異不會很大。謝謝。

uj5u.com熱心網友回復：

您可以使用 [Promise.all][1]。

這是示例代碼段。

let promises = [];
let paths = [];

Object.values(CategoriesEnum).forEach(category => {
    Object.values(FormatsEnum).forEach(format => {
        promises.push(getPagesCount(category, format))
    })
});

const totalPostCounts = await Promise.all(promises);



totalPostCounts.forEach(totalPosts => {
    const totalPages = Math.ceil(totalPosts.offsetPagination.total / 12);
    for (let page = 1; page <= totalPages; page  ) {
        paths.push({ params: { category: category, format: format, page: page } });
    }
});

return paths;




[1]: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/all

使用相同方法的另一種方法

let promises = [];
let paths = [];


/*
Gather Api calls
*/
Object.values(CategoriesEnum).forEach(category => {
    Object.values(FormatsEnum).forEach(format => {
        promises.push(getPath(category, format))
    })
});

// wait for the call
const pathsArray = await Promise.all(promises);

// flat the array
pathsArray.forEach(pathArray => {
    paths = paths.concat(pathArray)
});


async function getPath(category, format) {
    let paths = [];

    const totalPosts = await getPagesCount(category, format);
    const totalPages = Math.ceil(totalPosts.offsetPagination.total / 12);
    for (let page = 1; page <= totalPages; page  ) {
        paths.push({ params: { category: category, format: format, page: page } });
    }

    return paths;
}

uj5u.com熱心網友回復：

一開始，我將邏輯分為 3 個部分：

1. 準備異步呼叫
2. 并行解決異步呼叫
3. 應用我的業務邏輯

const pagesPromises = []
for  (let category of Object.values(CategoriesEnum)) {
  for  (let format of Object.values(FormatsEnum)) {
    pagesPromises.push(getPagesCount(category, format))
  }
}

const totalPostsResults = await Promise.all(pagesPromises)

const paths = [];
totalPostsResults.forEach(totalPosts => {
  const totalPages = Math.ceil(totalPosts.offsetPagination.total / 12);
  for (let page = 1; page <= totalPages; page  ) {
    paths.push({ params: { category: category , format: format page: page } });
  }
})

return paths;

uj5u.com熱心網友回復：

正如其他答案所暗示的那樣，您可以使用Promise.all()。無需太多重構，您就可以撰寫：

const getStaticPaths = async () => {
  const paths = [];
  const promises = []; // : Array<Promise<void>> in ts

  Object.values(CategoriesEnum).forEach((category) => {
    Object.values(FormatsEnum).forEach((format) => {
      promises.push(async () => {

        const totalPosts = await getPagesCount(category, format); // <-- weird name though, I'd prefer `getPostsCount` instead
        const totalPages = Math.ceil(totalPosts.offsetPagination.total / 12);
        for (const page = 1; page <= totalPages;   page)
          paths.push({ params: { category: category, format: format, page: page } });

      });
    });
  });

  await Promise.all(promises);
  return { paths };
};

export { getStaticPaths };

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