Javascript 返回对象值作为另一个对象值出现的次数

Javascript 返回对象值作为另一个对象值出现的次数,javascript,arrays,function,javascript-objects,higher-order-functions,Javascript,Arrays,Function,Javascript Objects,Higher Order Functions,accounts数组中account对象的示例: const accounts = [ { id: "5f446f2ecfaf0310387c9603", picture: "https://api.adorable.io/avatars/75/esther.tucker@zillacon.me", age: 25, name: { first: "Esther", last:

accounts数组中account对象的示例:

const accounts = [
  {
    id: "5f446f2ecfaf0310387c9603",
    picture: "https://api.adorable.io/avatars/75/esther.tucker@zillacon.me",
    age: 25,
    name: {
      first: "Esther",
      last: "Tucker",
    },
    company: "ZILLACON",
    email: "esther.tucker@zillacon.me",
    registered: "Thursday, May 28, 2015 2:51 PM",
  },
books数组中的book对象示例:

const books = [
  {
    id: "5f447132d487bd81da01e25e",
    title: "sit eiusmod occaecat eu magna",
    genre: "Science",
    authorId: 8,
    borrows: [
      {
        id: "5f446f2e2cfa3e1d234679b9",
        returned: false,
      },
      {
        id: "5f446f2ed3609b719568a415",
        returned: true,
      },
      {
        id: "5f446f2e1c71888e2233621e",
        returned: true,
      },
      {
        id: "5f446f2e6059326d9feb9a68",
        returned: true,
      },
      {
        id: "5f446f2ede05a0b1e3394d8b",
        returned: true,
      },
      {
        id: "5f446f2e4081699cdc6a2735",
        returned: true,
      },
      {
        id: "5f446f2e3900dfec59489477",
        returned: true,
      },
      {
        id: "5f446f2e6059326d9feb9a68",
        returned: true,
      },
      {
        id: "5f446f2e409f8883af2955dd",
        returned: true,
      },
      {
        id: "5f446f2e3900dfec59489477",
        returned: true,
      },
      {
        id: "5f446f2eae901a82e0259947",
        returned: true,
      },
      {
        id: "5f446f2ef2ab5f5a9f60c4f2",
        returned: true,
      },
      {
        id: "5f446f2ea6b68cf6f85f6e28",
        returned: true,
      },
      {
        id: "5f446f2eed18105706d6ca19",
        returned: true,
      },
      {
        id: "5f446f2eae901a82e0259947",
        returned: true,
      },
      {
        id: "5f446f2e91c2af00cb74e82b",
        returned: true,
      },
      {
        id: "5f446f2e5aa2bb5545a0f8a6",
        returned: true,
      },
      {
        id: "5f446f2ea508b6a99c3e42c6",
        returned: true,
      },
      {
        id: "5f446f2e50cc2da9cd80efdb",
        returned: true,
      },
      {
        id: "5f446f2e0b3e2ff72fc503e7",
        returned: true,
      },
      {
        id: "5f446f2e91c2af00cb74e82b",
        returned: true,
      },
      {
        id: "5f446f2ef795e593cd3cd19d",
        returned: true,
      },
      {
        id: "5f446f2e2f35653fa80bf490",
        returned: true,
      },
      {
        id: "5f446f2e7b9cd304fed3a8bc",
        returned: true,
      },
      {
        id: "5f446f2ed9aac23c0340aab2",
        returned: true,
      },
    ],
  },
我需要我的函数返回一个数字,该数字表示帐户ID在任何书籍的
借阅
数组中出现的次数

以下是我所拥有的:

function getTotalNumberOfBorrows(account, books) {
  const accId = account.id;
  let idBorrowed = books.filter((book) => accId === book.borrows.id);
  return idBorrowed.length;
}

我得到了0分,而我应该在考试中得到2分。请注意,我刚刚学习完高级函数,希望在需要时使用查找、筛选、映射、减少和分解对象。感谢您的帮助/建议。

借用
是一个数组,您需要对其进行迭代以获得每个借用的
id
属性

函数获取借方总数(账户、账簿){
const accId=account.id;
设total=0;
books.forEach(book=>book.borrows.forEach(borrow=>accId==borrow.id&&total++);
返回总数;

}
借用
是一个数组,需要对其进行迭代以获得每个借用的
id
属性

函数获取借方总数(账户、账簿){
const accId=account.id;
设total=0;
books.forEach(book=>book.borrows.forEach(borrow=>accId==borrow.id&&total++);
返回总数;

}
如果需要使用“高级”功能,另一种可能的方法

function getTotalNumberOfBorrows(account, books) {
  const { id: accId } = account;

  return books.reduce((accumulator, book) => {
    return (
      accumulator +
      book.borrows
        .filter(borrow => borrow.id === accId)
        .reduce((accumulatorBorrows, borrow) => accumulatorBorrows + 1, 0)
    );
  }, 0);
}

工作

如果需要使用“高级”功能,另一种可能的方法

function getTotalNumberOfBorrows(account, books) {
  const { id: accId } = account;

  return books.reduce((accumulator, book) => {
    return (
      accumulator +
      book.borrows
        .filter(borrow => borrow.id === accId)
        .reduce((accumulatorBorrows, borrow) => accumulatorBorrows + 1, 0)
    );
  }, 0);
}

正在工作

我不确定是否遵循了此操作,运行此操作时会出现wrapSafe错误。我也没有见过&用那种方式,但我肯定我不知道它的所有用途。我只在检查条件和另一个条件时见过它。它只是
的简写,如果我有打字错误,
book。借用
应该是
book。借用
明白了。谢谢你的帮助!我不确定我是否遵循了这一点,并且在运行此程序时会出现wrapSafe错误。我也没有见过&用那种方式,但我肯定我不知道它的所有用途。我只在检查条件和另一个条件时见过它。它只是
的简写,如果我有打字错误,
book。借用
应该是
book。借用
明白了。谢谢你的帮助!