将数组中的对象拼接到新数组中答案

【问题标题】：splice objects from arrays into new array将数组中的对象拼接到新数组中
【发布时间】：2020-09-28 00:24:53
【问题描述】：

我有两个对象数组，我想将它们组合成一个新数组，根据previewNumber 值作为.splice 中使用的索引，将每个对象从一个数组拼接到另一个数组。我用过.map，但它只循环arrayOne的长度，下面是一些示例代码，任何帮助将不胜感激

const arrayOne = [
  {
    "title": "A Project",
    "slug": "a-project",
    "previewNumber": 2,
  },
  {
    "title": "New Project",
    "slug": "new-project",
    "previewNumber": 4,
  }
]

const arrayTwo = [
  {
    "title": "Project 5546",
    "slug": "project-5546",
  },
  {
    "title": "Project 456",
    "slug": "project-456",
  },
    {
    "title": "Rand Project 467",
    "slug": "rand-project-467",
  },
  {
    "title": "Random Project 245",
    "slug": "random-project-245",
  },
    {
    "title": "Example Project",
    "slug": "example-project",
  },
]


  const newArray = arrayOne.map((item) =>
    arrayTwo.splice(item.previewNumber, 0, item)
  );
  
  console.log(newArray)
  
  const desiredOutput = [
  {
    "title": "Project 5546",
    "slug": "project-5546",
  },
  {
    "title": "Project 456",
    "slug": "project-456",
  },
    {
    "title": "A Project",
    "slug": "a-project",
    "previewNumber": 2,
  },
    {
    "title": "Rand Project 467",
    "slug": "rand-project-467",
  },
  {
    "title": "Random Project 245",
    "slug": "random-project-245",
  },
    {
    "title": "New Project",
    "slug": "new-project",
    "previewNumber": 4,
  },
    {
    "title": "Example Project",
    "slug": "example-project",
  },
]

【问题讨论】：

只要const mergedArr = [...arrayOne, ...arrayTwo]。从所需的输出中，我了解到您正在寻找合并两个数组。
我意识到我错过了一个关键细节，我想使用previewNumber 值将arrayOne 中的对象拼接到arrayTwo 顺序中，但创建一个新数组而不是改变arrayTwo
arrayOne 是否按previewNumber 排序？
@LLawliet 不，我想在与arrayTwo 结合时使用previewNumber 的值作为新数组中的索引。

标签： javascript

【解决方案1】：

您可以使用flatMap 实现此目的。这是一个实现。

const arrayOne = [ { "title": "A Project", "slug": "a-project", "previewNumber": 2, }, { "title": "New Project", "slug": "new-project", "previewNumber": 4, }];

const arrayTwo = [ { "title": "Project 5546", "slug": "project-5546", }, { "title": "Project 456", "slug": "project-456", }, { "title": "Rand Project 467", "slug": "rand-project-467", }, { "title": "Random Project 245", "slug": "random-project-245", }, { "title": "Example Project", "slug": "example-project", }];

const result = arrayTwo.flatMap((p,i)=>{
   position = arrayOne.find(k=>k.previewNumber===i);
   return position ? [position, p] : p
});

console.log(result);

或者您可以将 arrayOne 转换为对象。像这样的：

const arrayOne = [ { "title": "A Project", "slug": "a-project", "previewNumber": 2, }, { "title": "New Project", "slug": "new-project", "previewNumber": 4, }];

const mapped = Object.fromEntries(arrayOne.map(p=>[p.previewNumber,p]));

const arrayTwo = [ { "title": "Project 5546", "slug": "project-5546", }, { "title": "Project 456", "slug": "project-456", }, { "title": "Rand Project 467", "slug": "rand-project-467", }, { "title": "Random Project 245", "slug": "random-project-245", }, { "title": "Example Project", "slug": "example-project", }];

const result = arrayTwo.flatMap((p,i)=>mapped[i] ? [mapped[i], p] : p);

console.log(result);

【讨论】：

【解决方案2】：

首先，制作arrayTwo 的newArray 副本。之后，您可以通过从previewNumber + index拼接来迭代arrayOne添加到newArray

const newArray = [...arrayTwo]
arrayOne.forEach((el, index) => {
  newArray.splice(el.previewNumber + index, 0, el)
})

console.log(newArray)

完整演示

const arrayOne = [
  {
    title: "A Project",
    slug: "a-project",
    previewNumber: 2,
  },
  {
    title: "New Project",
    slug: "new-project",
    previewNumber: 4,
  },
]

const arrayTwo = [
  {
    title: "Project 5546",
    slug: "project-5546",
  },
  {
    title: "Project 456",
    slug: "project-456",
  },
  {
    title: "Rand Project 467",
    slug: "rand-project-467",
  },
  {
    title: "Random Project 245",
    slug: "random-project-245",
  },
  {
    title: "Example Project",
    slug: "example-project",
  },
]

const newArray = [...arrayTwo]
arrayOne.forEach((el, index) => {
  newArray.splice(el.previewNumber + index, 0, el)
})

console.log(newArray)

【讨论】：

@gorak 感谢您的纠正。包含减一是多余的