我有一组具有以下属性的对象: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material_no / material_name分组的对象数组以及具有相同material_no / material_name的数量的总和?
[
{ 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 }
]
答案 0 :(得分:4)
您可以将数组减少到Map
,以material_no
存储总项目数。然后只需将地图值提取到数组。
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)
答案 1 :(得分:1)
您可以像这样使用.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; }

有用的资源:
答案 2 :(得分:0)
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);
它很慢而且只是为了好玩:)
答案 3 :(得分:0)
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 accumulation = data.reduce((total, val, index)=>{
let foundItemIndex = total.findIndex((obj)=>obj.material_no == val.material_no);
if(foundItemIndex < 0) total.push(val)
else total[foundItemIndex].qty += val.qty;
return total;
}, []);
console.log(accumulation);