如何组合两个反应对象？

Posted 2023-02-22

【中文标题】如何组合两个反应对象？

【英文标题】：How can I combine two react objects?

【发布时间】：2021-07-28 22:26:06

【问题描述】：

const a = [

0: market: "KRW-BTC", korean_name: "비트코인", english_name: "Bitcoin"
1: market: "KRW-ETH", korean_name: "이더리움", english_name: "Ethereum"
2: market: "KRW-NEO", korean_name: "네오", english_name: "NEO"
3: market: "KRW-MTL", korean_name: "메탈", english_name: "Metal"
4: market: "KRW-LTC", korean_name: "라이트코인", english_name: "Litecoin"
5: market: "KRW-XRP", korean_name: "리플", english_name: "Ripple"
6: market: "KRW-ETC", korean_name: "이더리움클래식", english_name: "Ethereum Classic"
7: market: "KRW-OMG", korean_name: "오미세고", english_name: "OmiseGo"
8: market: "KRW-SNT", korean_name: "스테이터스네트워크토큰", english_name: "Status Network Token"
9: market: "KRW-WAVES", korean_name: "웨이브", english_name: "Waves"
10: market: "KRW-XEM", korean_name: "넴", english_name: "NEM"
11: market: "KRW-QTUM", korean_name: "퀀텀", english_name: "Qtum"
12: market: "KRW-LSK", korean_name: "리스크", english_name: "Lisk"
]



const b = [

0: market: "KRW-BTC", trade_date: "20210506", trade_time: "144435", trade_date_kst: "20210506", trade_time_kst: "234435", …
1: market: "KRW-ETH", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436", …
2: market: "KRW-NEO", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436", …
3: market: "KRW-MTL", trade_date: "20210506", trade_time: "144432", trade_date_kst: "20210506", trade_time_kst: "234432", …
4: market: "KRW-LTC", trade_date: "20210506", trade_time: "144433", trade_date_kst: "20210506", trade_time_kst: "234433", …
5: market: "KRW-XRP", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436", …
6: market: "KRW-ETC", trade_date: "20210506", trade_time: "144437", trade_date_kst: "20210506", trade_time_kst: "234437", …
7: market: "KRW-OMG", trade_date: "20210506", trade_time: "144437", trade_date_kst: "20210506", trade_time_kst: "234437", …
8: market: "KRW-SNT", trade_date: "20210506", trade_time: "144434", trade_date_kst: "20210506", trade_time_kst: "234434", …
9: market: "KRW-WAVES", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436", …
10: market: "KRW-XEM", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436", …
11: market: "KRW-QTUM", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436", …
12: market: "KRW-LSK", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436", …
]

有两个这样的，我想根据“市场”合并它们。有什么办法合并吗？？我不确定，因为我是 React 的初学者。

【问题讨论】：

这些是对象数组。 React 是一个前端框架，在这里无关紧要。

您可以使用...obj1, obj2 合并两个对象。不太清楚你的意思based on the 'market'.

@ChrisG 哦，数组，我的错。

@ShivamJha 再次，不。

let combined = ...a, ...b console.log(a, b)。它是一个数组，只是输出没有数组的内容。

【参考方案1】：

类似的东西：

const c = a.map(_a => 
    const foundB = b.find(_b => _a.market===_b.market)
    return foundB ? ..._a,...foundB : _a
)

【讨论】：

非常感谢。谢谢。非常感谢您的帮助。

【参考方案2】：

你可以试试这个。可能有更好的方法来做到这一点。谢谢。

const a = [
  market: "KRW-BTC", korean_name: "비트코인", english_name: "Bitcoin",
  market: "KRW-ETH", korean_name: "이더리움", english_name: "Ethereum",
  market: "KRW-NEO", korean_name: "네오", english_name: "NEO",
  market: "KRW-MTL", korean_name: "메탈", english_name: "Metal",
  market: "KRW-LTC", korean_name: "라이트코인", english_name: "Litecoin",
  market: "KRW-XRP", korean_name: "리플", english_name: "Ripple",
  market: "KRW-ETC", korean_name: "이더리움클래식", english_name: "Ethereum Classic",
  market: "KRW-OMG", korean_name: "오미세고", english_name: "OmiseGo",
  market: "KRW-SNT", korean_name: "스테이터스네트워크토큰", english_name: "Status Network Token",
  market: "KRW-WAVES", korean_name: "웨이브", english_name: "Waves",
  market: "KRW-XEM", korean_name: "넴", english_name: "NEM",
  market: "KRW-QTUM", korean_name: "퀀텀", english_name: "Qtum",
  market: "KRW-LSK", korean_name: "리스크", english_name: "Lisk",
]

const b = [
  market: "KRW-BTC", trade_date: "20210506", trade_time: "144435", trade_date_kst: "20210506", trade_time_kst: "234435",
  market: "KRW-ETH", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436",
  market: "KRW-NEO", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436",
  market: "KRW-MTL", trade_date: "20210506", trade_time: "144432", trade_date_kst: "20210506", trade_time_kst: "234432",
  market: "KRW-LTC", trade_date: "20210506", trade_time: "144433", trade_date_kst: "20210506", trade_time_kst: "234433",
  market: "KRW-XRP", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436",
  market: "KRW-ETC", trade_date: "20210506", trade_time: "144437", trade_date_kst: "20210506", trade_time_kst: "234437",
  market: "KRW-OMG", trade_date: "20210506", trade_time: "144437", trade_date_kst: "20210506", trade_time_kst: "234437",
  market: "KRW-SNT", trade_date: "20210506", trade_time: "144434", trade_date_kst: "20210506", trade_time_kst: "234434",
  market: "KRW-WAVES", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436",
  market: "KRW-XEM", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436",
  market: "KRW-QTUM", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436",
  market: "KRW-LSK", trade_date: "20210506", trade_time: "144436", trade_date_kst: "20210506", trade_time_kst: "234436",
]

const finalArray = a.map(item => (...item, ...b.find(itemOfB => itemOfB.market === item.market)))

console.log(finalArray)

【讨论】：

这基本上是接受的答案（和我的评论）的副本，没有 devrnd 的错误检查。