根据 id 比较两个数组并根据值返回一个新数组

Question

尝试遍历两个数组并找到每个数组的代码，我可能会遗漏某些部分。我需要用这个值构造一个新数组，每个id可以在arrayB中多次出现。根据第一个数组中的 id，我们必须匹配第二个数组（arrayA）中的 id 并获取代码

let arrayA=[
    {"breadcrumb":{id: "abdc4051"}, type:"details"},
    {"breadcrumb":{id: "abdc4052"}, type:"details"},

    let arrayB=[
    {"breadcrumb": {id: "abdc4051",code: "mike", length:"short"}},
    {"breadcrumb": {id: "abdc4051", code: "pohan", length:"long"}}, {"breadcrumb": {id: "abdc4052", code: "junior", length:"short"}}]

    let arrayC = [];
    // output expected
[{"id":"abdc4051", shortLength: "mike", longLength:"pohan"}, {"id":"abdc4052", shortLength: "Junior", longLength:"-"}]
    // tried this
    function isBiggerThan10(element, index, array) {
      return element > 10;
    }

    arrayA.forEach(function(element){
      arrayC.push({
      id:element.id,
      firstName:(arrayB.find(
        e => e.attributes.code==="mike")).breadCrumbs.shortLength,
      lastName:(arrayB.find(
        e => e.code==="pohan")).breadCrumbs.longlength
      })
    });

    console.log(arrayC);

Answer 1

这是使用内置数组方法的一种解决方案。步骤是：

对于arrayA 中的每个项目，执行以下操作：

• 查找arrayB 中具有相同ID 的所有项目（使用.filter）。
• 将所有这些结果合并到一个对象中，如果不存在名字和姓氏，则使用默认的“-”（使用.reduce）。
• 将其放入结果数组中（首先在 arrayA 上使用 .map 处理）。

let arrayA = [
    {"breadcrumb":{id: "abdc4051"}, type:"details"},
    {"breadcrumb":{id: "abdc4052"}, type:"details"},
]

let arrayB = [
    {"breadcrumb": {id: "abdc4051", firstName: "mike"}},
    {"breadcrumb": {id: "abdc4051", lastName: "pohan"}},
    {"breadcrumb": {id: "abdc4052", firstName: "junior"}},
]

// output expected
// [
//   {"id":"abdc4051", firstName: "mike", lastName:"pohan"},
//   {"id":"abdc4052", firstName: "Junior", lastName:"-"},
// ]

const result = arrayA.map(itemA => {
    return arrayB
        .filter(itemB => itemB.breadcrumb.id === itemA.breadcrumb.id)
        .reduce((combo, item) => ({...combo, ...item.breadcrumb}), {firstName: "-", lastName: "-"})
});

console.log(result);

编辑：根据已编辑的问题，您可以修改 reduce 函数以查看组合是否设置了 firstCode。如果是，则将下一个code 放在键lastCode 下，否则保持为firstCode。这将基于它们在arrayB 中出现的顺序的第一个/最后一个代码：

let arrayA = [
    {"breadcrumb":{id: "abdc4051"}, type:"details"},
    {"breadcrumb":{id: "abdc4052"}, type:"details"},
]

let arrayB = [
    {"breadcrumb": {id: "abdc4051", code: "mike"}},
    {"breadcrumb": {id: "abdc4051", code: "pohan"}},
    {"breadcrumb": {id: "abdc4052", code: "junior"}},
]

// output expected
// [
//   {"id":"abdc4051", firstCode: "mike", lastCode:"pohan"},
//   {"id":"abdc4052", firstCode: "Junior", lastCode:"-"},
// ]

const result = arrayA.map(itemA => {
    return arrayB
        .filter(itemB => itemB.breadcrumb.id === itemA.breadcrumb.id)
        .reduce((combo, item) => ({...combo, [combo.firstCode === "-" ? "firstCode" : "lastCode"]: item.breadcrumb.code, id: itemA.breadcrumb.id}), {firstCode: "-", lastCode: "-"})
});

console.log(result);

编辑2：每秒编辑，您可以再次修改reduce以满足您的需要：

let arrayA = [
    {"breadcrumb":{id: "abdc4051"}, type:"details"},
    {"breadcrumb":{id: "abdc4052"}, type:"details"},
]

let arrayB = [
    {"breadcrumb": {id: "abdc4051", code: "mike", length: "short"}},
    {"breadcrumb": {id: "abdc4051", code: "pohan", length: "long"}},
    {"breadcrumb": {id: "abdc4052", code: "junior", length: "short"}},
]

// output expected
// [
//   {"id":"abdc4051", shortLength: "mike", longLength:"pohan"},
//   {"id":"abdc4052", shortLength: "junior", longLength:"-"},
// ]

const result = arrayA.map(itemA => {
    return arrayB
        .filter(itemB => itemB.breadcrumb.id === itemA.breadcrumb.id)
        .reduce((combo, item) => ({...combo, [item.breadcrumb.length + "Length"]: item.breadcrumb.code, id: itemA.breadcrumb.id}), {shortLength: "-", longLength: "-"})
});

console.log(result);

Answer 2

以arrayB 开始一个链。使用_.map()通过id获取breadcrumb、_.groupBy()的内容。使用_.pick() 和arrayA 的ID 来获取您想要的组。使用 _.transform() 将每个组映射到所需的表单：

const arrayA=[{"breadcrumb":{"id":"abdc4051"},"type":"details"},{"breadcrumb":{"id":"abdc4052"},"type":"details"}];
const arrayB=[{"breadcrumb":{"id":"abdc4051","code":"mike","length":"short"}},{"breadcrumb":{"id":"abdc4051","code":"pohan","length":"long"}},{"breadcrumb":{"id":"abdc4052","code":"junior","length":"short"}}];
    
const result = _(arrayB)
  .map('breadcrumb') // unwrap breadcrumb
  .groupBy('id')
  .pick(arrayA.map((o) => _.get(o, 'breadcrumb.id'))) // get all groups that match arrayA ids
  .map((g, key) => _.transform(g, (acc, v) => { // transform each group to the requested form
    acc[`${v.length}Length`] = v.code;
  }, { key, shortLength: '-', longLength: '-' }))
  .value(); 
  
console.log(result);

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