如何在节点 js 中对两个泛型对象进行条件检查？

Question

在下面的代码中，我尝试计算每个主题的总分数并将它们存储在标记对象中。我正在检查 edu 中的每个 obj 是否是基于主题名称（名称）和主题级别（级别）的信息的一部分。如果匹配，我会将所有内容放入数组中。

在这里，每次将主题添加到信息时，在函数“func”中，我都必须添加另一个“else if”来检查。我不想总是添加这个条件检查。

我想让它变得通用，比如说，我的信息变成：

let info = {"math": 1, "science": 2}
or
let info = {"math": 1, "science": 2, "history": 1, arts: 1}

我不想删除或添加条件检查。我希望在不依赖的情况下完成响应（edu）和信息之间的检查。仅根据信息进行检查。

这是在节点js中。我该怎么做？我虽然想到了 switch，但这和“if else”是一样的。

let res = {
  edu: [
    {
      name: "math",
      level: 1,
      marks: 50,
      part: 1,
    },
    {
      name: "math",
      level: 1,
      marks: 57,
      part: 2,
    },
    {
      name: "science",
      level: 2,
      marks: 70,
      part: 1,
    },
  ],
  age: 20,
  name: "abc",
}

let info = { math: 1, science: 2, history: 1 }
const func = () => {
  var marks = {}
  let subject = Object.keys(info)
  subject.forEach((subjName, index) => {
    marks[subjName] = 0
  })
  console.log("marks = ", marks)
  var consolidatedInfo = []

  res.edu.forEach(function (subj) {
    if (subj.name === Object.keys(info)[0] && subj.level === info[Object.keys(info)[0]]) {
      marks[subj.name] += subj.marks

      const eachSubjPartInfo = {
        name: subj.name,
        level: subj.level,
        marks: subj.marks,
        part: subj.part,
      }
      consolidatedInfo.push(eachSubjPartInfo)
      console.log("marks: ", marks)
    } else if (subj.name === Object.keys(info)[1] && subj.level === info[Object.keys(info)[1]]) {
      marks[subj.name] += subj.marks

      const eachSubjPartInfo = {
        name: subj.name,
        level: subj.level,
        marks: subj.marks,
        part: subj.part,
      }
      consolidatedInfo.push(eachSubjPartInfo)
      console.log("marks: ", marks)
    } else if (subj.name === Object.keys(info)[2] && subj.level === info[Object.keys(info)[2]]) {
      marks[subj.name] += subj.marks

      const eachSubjPartInfo = {
        name: subj.name,
        level: subj.level,
        marks: subj.marks,
        part: subj.part,
      }
      consolidatedInfo.push(eachSubjPartInfo)
      console.log("marks: ", marks)
    }
  })

  console.log("consolidatedInfo = ", consolidatedInfo)
  console.log("marks = ", marks)
}

func()

Answer 1

你可以在每次迭代中使用逻辑

info[name] === level

来动态检查并且只在满足时添加。无需转换为数组然后检查

我使用数组

reduce

。为了创建每个主题的初始对象，我使用了 0 分

Object.fromEntries(Object.keys(info).map(k => [k,0]))

let res = {    "edu": [        {            "name": "math",            "level": 1,            "marks": 50,"part": 1        },{"name": "math",            "level": 1,            "marks": 57,"part": 2},{"name": "science",            "level": 2,            "marks": 70,"part": 1},    ],    "age": 20,    "name": "abc"}

let info = {"math": 1, "science": 2, "history": 1}

const marksAccumulator = Object.fromEntries(Object.keys(info).map(k => [k,0]))

const result = res.edu.reduce((acc,{name,level,marks,part}) => {
  if(info?.[name] === level){
    acc.marksAccumulator[name] += marks
    acc.consolidated.push({name,level,marks,part})
  }
  return acc
},{marksAccumulator,consolidated:[]})

console.log(result)

从

result

您可以获得所需的数据，如

result.consolidated

等