群和和的ES6实现
群和和的ES6实现
提问于 2018-03-26 06:04:22
我有一个具有以下属性的对象数组: material_no、material_name、qty。
let data = [
{ material_no: '1001', material_name: 'Material 1', qty: 100 },
{ material_no: '1001', material_name: 'Material 1', qty: 50 },
{ material_no: '1002', material_name: 'Material 2', qty: 44 },
{ material_no: '1003', material_name: 'Material 3', qty: 125 },
{ material_no: '1002', material_name: 'Material 2', qty: 59 },
{ material_no: '1004', material_name: 'Material 4', qty: 999 },
{ material_no: '1005', material_name: 'Material 5', qty: 80 },
{ material_no: '1005', material_name: 'Material 5', qty: 66 }
]如何返回按其material/material和qty的和分组的对象数组，这些对象具有相同的material_no/material_name？
[
{ material_no: '1001', material_name: 'Material 1', qty: 150 },
{ material_no: '1002', material_name: 'Material 2', qty: 103 },
{ material_no: '1003', material_name: 'Material 3', qty: 125 },
{ material_no: '1004', material_name: 'Material 4', qty: 999 },
{ material_no: '1005', material_name: 'Material 5', qty: 146 }
]回答 4
Stack Overflow用户
回答已采纳
发布于 2018-03-26 06:29:12
可以将数组缩减为Map，该数组通过material_no存储总项。然后将地图值提取到数组中。
let data = [
{ material_no: '1001', material_name: 'Material 1', qty: 100 },
{ material_no: '1001', material_name: 'Material 1', qty: 50 },
{ material_no: '1002', material_name: 'Material 2', qty: 44 },
{ material_no: '1003', material_name: 'Material 3', qty: 125 },
{ material_no: '1002', material_name: 'Material 2', qty: 59 },
{ material_no: '1004', material_name: 'Material 4', qty: 999 },
{ material_no: '1005', material_name: 'Material 5', qty: 80 },
{ material_no: '1005', material_name: 'Material 5', qty: 66 }
]
const sums = [
...data.reduce(
(map, item) => {
const { material_no: key, qty } = item;
const prev = map.get(key);
if(prev) {
prev.qty += qty
} else {
map.set(key, Object.assign({}, item))
}
return map
},
new Map()
).values()
]
console.log(sums)
Stack Overflow用户
发布于 2018-03-26 10:51:42
您可以这样使用.reduce()：
let data = [{ material_no: '1001', material_name: 'Material 1', qty: 100 },{ material_no: '1001', material_name: 'Material 1', qty: 50 },{ material_no: '1002', material_name: 'Material 2', qty: 44 },{ material_no: '1003', material_name: 'Material 3', qty: 125 },{ material_no: '1002', material_name: 'Material 2', qty: 59 },{ material_no: '1004', material_name: 'Material 4', qty: 999 },{ material_no: '1005', material_name: 'Material 5', qty: 80 },{ material_no: '1005', material_name: 'Material 5', qty: 66 }];
let result = Object.values(
data.reduce((a, c) => (
a[c.material_no] = a[c.material_no] ?
(a[c.material_no].qty += c.qty, a[c.material_no]) :
c, a), {}
)
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }
有用资源：
Stack Overflow用户
发布于 2018-03-26 06:51:56
data
.sort((a, b) => a.material_no - b.material_no)
.map(({material_no, material_name, qty}, index, arr) => ({
material_no,
material_name,
qty: qty + (material_no === (arr[index - 1] || {}).material_no && arr[index - 1].qty)
}))
.filter(({material_no}, index, arr) => material_no !== (arr[index + 1] || {}).material_no);它很慢，只是为了好玩:)
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/49484910
复制相关文章
相似问题
腾讯云开发者
