javascript - 在函数中返回对象时使用 for-in/of 的问题

Question

我刚刚在终端上打印了结果“[object Object]'s score is 0”。结果 27 一切正常，直到我将函数分离为返回对象。

1. 如果我必须返回一个对象，我如何获得 27？
2. 如何在 console.log 而不是 [object Object] 上打印“alex”？

const alex = {
  first: [1, 2, 9, 8],
  second: [3],
  third: [0, 0, 0, 1, 3]
};
const gordon = {
  first: [3],
  second: [2, 2, 4, 5, 6, 6, 7, 8]
}

function createPlayer(object) {
  let score = 0;
  return {
    add: function() {
      for (const key in object) {
        for (const item in object[key]) {
          score += object[key][item]
        }
      }
    },
    result: function() {
      return `${object}\'s score is ${score}`
    }
  }
}
createPlayer(alex).add()
console.log(createPlayer(alex).result())

Answer 1

const alex = {
  first: [1, 2, 9, 8],
  second: [3],
  third: [0, 0, 0, 1, 3]
};
const gordon = {
  first: [3],
  second: [2, 2, 4, 5, 6, 6, 7, 8]
}

function createPlayer(object, name) {
  let score = 0;
  return {
    add: function() {
      for (const key in object) {
        for (const item in object[key]) {
          score += object[key][item]
        }
      }
      return this; // <<<<<
    },
    result: function() {
      return `${name}\'s score is ${score}`
    }
  }
}
console.log(createPlayer(alex, 'Alex').add().result())

Answer 2

您不会alex为名为 alex 的对象显示

你可能是这个意思

const alex = {
  Name: "Alex",
  first: [1, 2, 9, 8],
  second: [3],
  third: [0, 0, 0, 1, 3]
};
const gordon = {
  Name: "Gordon",
  first: [3],
  second: [2, 2, 4, 5, 6, 6, 7, 8]
}

function createPlayer(object) {
  let score = 0;
  return {
    add: function() {
      for (const key in object) {
        if (key!=="Name") {
          for (const item in object[key]) {
            score += object[key][item]
          }
        }   
      }
    },
    result: function() {
      return `${object.Name}\'s score is ${score}`
    }
  }
}
const player1 = createPlayer(alex)
player1.add()
console.log(player1.result())