>[danger] 合并對象_排序 ---- > 思路1: 雙層循環, 挨個比, 最后排序 --- > 思路2: 合并數組, 向map添加, id為key, 合并對象, 最后排序 --- 題意: 現有如下數組, 根據ID值相同合并對象, 以age排序, 打印數組結果 ~~~ let arr = [ { "id": 1, "text": "大嫂" }, { "id": 3, "text": "三嫂" }, { "id": 2, "text": "二嫂" } ]; let brr = [ { "id": 1, "age": 30 }, { "id": 2, "age": 17 }, { "id": 3, "age": 23 } ]; ~~~ 1. 方式1: 循環a數組, 與b數組每個判斷id是否相等 ~~~ let resultArr = []; arr.map(oneObj=>{ brr.map(twoObj=>{ if (oneObj['id'] === twoObj['id']){ resultArr.push(Object.assign({}, oneObj, twoObj)); } }) }); console.log(resultArr.sort((a, b)=>a['age'] - b['age'])); ~~~ 2. 方式2: 循環a數組, 去b數組find到id相等的元素 (實際上還是雙層循環) ~~~ let resultArr = []; arr.map(oneObj => { let target = brr.find(twoObj => oneObj['id'] === twoObj['id']); resultArr.push(Object.assign({}, oneObj, target)); }); console.log(resultArr.sort((a, b)=>a['age'] - b['age'])); ~~~ 3. 方式3: 循環所有元素, id為key, 合并對象為value (這種最快) ~~~ let map = new Map(); arr.concat(brr).forEach(obj => { map.set(obj.id, Object.assign(map.get(obj.id) || {}, obj)); }); console.log(Array.from(map.values()).sort((a, b)=>a['age'] - b['age'])); ~~~