笔记 | 一道使用js合并数组的题目

[
    {
        code: 1,
        data: [
            { title: 'haha', size: 123 },
            { title: 'xixi', size: 456 }
        ]
    },
    {
        code: 2,
        data: [
            { title: 'cscs', size: 4344 },
            { title: 'vvv', size: 555 }
        ]
    }
]

[
  { code: 1, title: 'haha', size: 123 },
  { code: 1, title: 'xixi', size: 456 },
  { code: 2, title: 'cscs', size: 4344 },
  { code: 2, title: 'vvv', size: 555 }
]

// 首先写成这样子
const result = data.reduce((acc, cur) => {
 
    return acc
    }
    , [])

        code: 1,
        data: [
            { title: 'haha', size: 123 },
            { title: 'xixi', size: 456 }
        ]
    }
// 即要求以上结构变成
  { code: 1, title: 'haha', size: 123 },
  { code: 1, title: 'xixi', size: 456 },

 cur.data.forEach(item => {
        acc.push({
          code: cur.code,
          ...item // 即展开数组（
        })
    })

const result = data.reduce((acc, cur) => {
    cur.data.forEach(item => {
        acc.push({
          code: cur.code,
          ...item
        })
    })
    return acc
    }
    , [])

data.map(item => item.data.map(v =>({code: item.code, ...v}))).flat()

const result = data.reduce((acc, cur) => {
    return acc.concat(cur.data.map((item)=>{
      item.code = cur.code
      return item
    }))
    }
    , [])

const result = data.reduce((acc, cur) => 
// 箭头函数不用 return 值 :D
// https://developer.mozilla.org/zh-CN/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce

    acc.concat(

        cur.data.map((item) => ({ // 多加一个括号，是为了把返回值当做一个参数返回
            ...item,
            code: cur.code
        })
        )

    )

    , [])

result.sort((a, b) => a.title > b.title ? 1 : -1)