問題描述

例如舊數(shù)據(jù)：var old = [

{ id: 1, name: ’css’, type: ’html’},{ id: 2, name: ’css’, type: ’html’}, { id: 3, name: ’javacript’, type: ’code’},{ id: 4, name: ’javacript’, type: ’code’}

]想得到的 var new = [

{ id: [1,2], name: ’css’, type: ’html’}, { id: [3,4], name: ’javacript’, type: ’code’},

]希望把相同name的對(duì)象合并，并且把對(duì)應(yīng)的id放到一個(gè)數(shù)組

問題解答

var hash = {};var i = 0;var res = [];old.forEach(function(item) { var name = item.name; hash[name] ? res[hash[name] - 1].id.push(item.id) : hash[name] = ++i && res.push({id: [item.id],name: name,type: item.type })});console.log(JSON.stringify(res))回答2：

從下面的數(shù)組 old 里

var old = [ {id: 1,name: ’css’,type: ’html’ }, {id: 2,name: ’css’,type: ’html’ }, {id: 3,name: ’javacript’,type: ’code’ }, {id: 4,name: ’javacript’,type: ’code’ }]

得到 new

var new = [ {id: [1,2],name: ’css’,type: ’html’ }, {id: [3,4],name: ’javacript’,type: ’code’ }]實(shí)現(xiàn)

var isEqual = (a, b) => a.name === b.name && b.type === b.type; var create = e => { e.id = [e.id]; return e; }var getNew = old => old.reduce((acc, cur) => { let hasItem = acc.some(e => {let temp = isEqual(e, cur); if (temp) e.id.push(cur.id); return temp; });if (!hasItem) acc.push(create(cur))return acc; }, []);