我正在为一个销售应用程序开发backEnd部分,其中我有两个Objects数组,一个得到卖家的名字,另一个得到每月的销售额。我的上级告诉我,使用包含销售额的数组中的数据,将人名数组从销售额最多的数组排序到最小(从最高到最低)的数组,基本上是对关键的=>数量进行求和。我不知道怎么做,我试着用一种减少方法来处理每个销售商的销售,但是我想不出如何比较它们来重组Array。守则是:
const sellers= [{id:1,name: 'juan', age: 23},{id:2,name: 'adrian', age: 32},{id:3,name: 'apolo', age: 45}];
const sales= [{equipo: 'frances', sellerId: 2, quantity: 234},{equipo: 'italiano', sellerId: 3, quantity: 24},{equipo: 'polaco', sellerId: 1, quantity: 534},{equipo: 'frances', sellerId: 2, quantity: 1234},{equipo: 'frances', sellerId: 3, quantity: 2342}];这就是我已经尝试过的:
const bestSellers= () => {
sales.reduce((sum, value) => ( value.sellerId == 1 ? sum + value.area : sum), 0); }最终结果应该如下所示:
const sellers= [{id:3,name: 'apolo', age: 45},{id:2,name: 'adrian', age: 32},{id:1,name: 'juan', age: 23}]发布于 2022-05-20 02:02:32
你在这里想做两件事。
在我的排序函数中,您可以看到,我正在过滤卖方的所有销售。
一旦我有一个销售者的销售,我使用减少方法,把他们的销售数量之和为一个容易使用的数字。
然后,我比较以前的卖家数量和当前的卖家数量,用排序方法重新订购。
我鼓励您阅读所使用方法的文档,以便您了解每一步所发生的事情。
const sellers = [{
id: 1,
name: 'juan',
age: 23
}, {
id: 2,
name: 'adrian',
age: 32
}, {
id: 3,
name: 'apolo',
age: 45
}];
const sales = [{
equipo: 'frances',
sellerId: 2,
quantity: 234
}, {
equipo: 'italiano',
sellerId: 3,
quantity: 24
}, {
equipo: 'polaco',
sellerId: 1,
quantity: 534
}, {
equipo: 'frances',
sellerId: 2,
quantity: 1234
}, {
equipo: 'frances',
sellerId: 3,
quantity: 2342
}];
const expected = [{
id: 3,
name: 'apolo',
age: 45
}, {
id: 2,
name: 'adrian',
age: 32
}, {
id: 1,
name: 'juan',
age: 23
}]
const result = sellers.sort((a, b) => {
totalA = sales.filter(sale => sale.sellerId === a.id).reduce((acc, val) => acc + val.quantity, 0)
totalB = sales.filter(sale => sale.sellerId === b.id).reduce((acc, val) => acc + val.quantity, 0)
return totalB - totalA
})
// Check we get what we expect
console.log(JSON.stringify(expected) === JSON.stringify(result))
发布于 2022-05-20 02:09:01
首先,将sales数组缩小为每个卖方的数量之和,对此列表进行排序,然后从sellers数组映射到相应的卖方。
const sellers= [{id:1,name: 'juan', age: 23},{id:2,name: 'adrian', age: 32},{id:3,name: 'apolo', age: 45}];
const sales= [{equipo: 'frances', sellerId: 2, quantity: 234},{equipo: 'italiano', sellerId: 3, quantity: 24},{equipo: 'polaco', sellerId: 1, quantity: 534},{equipo: 'frances', sellerId: 2, quantity: 1234},{equipo: 'frances', sellerId: 3, quantity: 2342}];
const bestSellers = (sellers, sales) => {
let res = sales.reduce((result, sale) => {
// Check if seller is already added in result
const obj = result.find(s => s.sellerId === sale.sellerId);
if (obj) {
// If seller already added then increase it's quantity
obj.quantity += sale.quantity;
} else {
// If seller not added, add seller in result
result.push({ sellerId: sale.sellerId, quantity: sale.quantity });
}
return result;
}, []);
// Sort result by quantity in decreasing order
res.sort((a,b) => b.quantity - a.quantity);
// Map each sale to the seller
res = res.map(s => sellers.find(seller => seller.id === s.sellerId));
return res;
}
const result = bestSellers(sellers, sales);
console.log(result);
发布于 2022-05-20 02:14:33
你可以这样做:
const
sellers =
[ { id: 1, name: 'juan', age: 23 }
, { id: 2, name: 'adrian', age: 32 }
, { id: 3, name: 'apolo', age: 45 }
]
, sales =
[ { equipo: 'frances', sellerId: 2, quantity: 234 }
, { equipo: 'italiano', sellerId: 3, quantity: 24 }
, { equipo: 'polaco', sellerId: 1, quantity: 534 }
, { equipo: 'frances', sellerId: 2, quantity: 1234 }
, { equipo: 'frances', sellerId: 3, quantity: 2342 }
];
sellers.forEach( ({id},i,all) => // add a sum attribute
{
all[i].sum = sales
.filter(({sellerId})=>sellerId===id)
.reduce((sum,{quantity})=>sum+quantity,0)
});
sellers
.sort( (a,b) => b.sum - a.sum)
.forEach((seller,i,all) => delete seller.sum ) // remove the sum
console.log( sellers ).as-console-wrapper {max-height: 100% !important;top: 0;}
.as-console-row::after {display: none !important;}
或者:
const
sellers =
[ { id: 1, name: 'juan', age: 23 }
, { id: 2, name: 'adrian', age: 32 }
, { id: 3, name: 'apolo', age: 45 }
]
, sales =
[ { equipo: 'frances', sellerId: 2, quantity: 234 }
, { equipo: 'italiano', sellerId: 3, quantity: 24 }
, { equipo: 'polaco', sellerId: 1, quantity: 534 }
, { equipo: 'frances', sellerId: 2, quantity: 1234 }
, { equipo: 'frances', sellerId: 3, quantity: 2342 }
]
// compute all the sum in a new Sums object:
const Sums = sales.reduce((sum,{sellerId : id, quantity}) =>
( sum[id] ??= 0, sum[id] += quantity , sum ), {} )
sellers .sort( (a,b) => Sums[b.id] - Sums[a.id])
console.log( sellers ).as-console-wrapper {max-height: 100% !important;top: 0;}
.as-console-row::after {display: none !important;}
https://stackoverflow.com/questions/72312675
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