給定以下物件陣列：
[{
"id": 1,
"name": "random_name1",
"published_at": "2021-01-16T08:52:24.408Z",
},
{
"id": 2,
"name": "random_name2",
"published_at": "2022-02-16T08:52:24.408Z",
},
{
"id": 3,
"name": "random_name3",
"published_at": "2020-04-16T08:52:24.408Z",
},
{
"id": 4,
"name": "random_name4",
"published_at": "2020-04-16T08:52:24.408Z",
},
{
"id": 5,
"name": "random_name5",
"published_at": "2022-05-16T08:52:24.408Z",
}
]
我需要按年和月將一組嵌套物件（降序）中的專案分組，結果應該是：
[
{
year: '2022',
months: [
{
month: '5',
items: [
{
id: '5',
name: 'random_name5'
}
]
},
{
month: '2',
items: [
{
id: '2',
name: 'random_name2'
}
]
}
]
},
{
year: '2021',
months: [
{
month: '1',
items: [
{
id: '1',
name: 'random_name1'
}
]
},
{
month: '2',
items: [
{
id: '2',
name: 'random_name2'
}
]
}
]
},
{
year: '2020',
months: [
{
month: '4',
items: [
{
id: '3',
name: 'random_name3'
},
{
id: '4',
name: 'random_name4'
}
]
}
]
}
];
我嘗試了以下方法：
items = [...new Set(items.map((item) => parseInt(item.published_at.split('-')[0])))].map((year) => [
{
year: year,
months: [
...new Set(
items
.filter((item) => parseInt(item.published_at.split('-')[0]) === year)
.map((item) => parseInt(item.published_at.split('-')[1]))
)
].map((month) => [
{
month: month,
items: items.filter(
(item) => parseInt(item.published_at.split('-')[0]) === year && parseInt(item.published_at.split('-')[1]) === month
)
}
])
}
]);
return items
上述解決方案的問題在于，它會像這樣創建一個二維陣列（月份也是二維的）：
[
[ { year: 2022, months: [Array] } ],
[ { year: 2021, months: [Array] } ],
[ { year: 2020, months: [Array] } ],
[ { year: 2019, months: [Array] } ],
[ { year: 2018, months: [Array] } ]
]
如何解決這個問題？
uj5u.com熱心網友回復：
如果您獲得唯一的年月串列，您可以使用它來映射您的物件
const items = [{ "id": 1,"name": "random_name1","published_at": "2021-01-16T08:52:24.408Z", },
{ "id": 2, "name": "random_name2", "published_at": "2022-02-16T08:52:24.408Z",},
{ "id": 3, "name": "random_name3","published_at": "2020-04-16T08:52:24.408Z",},
{"id": 4, "name": "random_name4", "published_at": "2020-04-16T08:52:24.408Z",},
{ "id": 5, "name": "random_name5", "published_at": "2022-05-16T08:52:24.408Z",}]
let uniqueYearMonths = [... new Set(items.map(x => x.published_at.substring(0,7)))];
let results = [... new Set(items.map(x => x.published_at.substring(0,4)))]
.map(year => ({
year: year,
months: uniqueYearMonths
.filter(ym => ym.startsWith(year))
.map(ym => ({
month: ym.substring(5,7),
items: items
.filter(item => item.published_at.startsWith(ym))
.map(item => ({
id: item.id,
name: item.name
}))
}))
}));
console.log(results);uj5u.com熱心網友回復：
給定陣列 as data，您可以使用陣列方法做一些事情map和reduce。
像這樣：
const groupedByYear = data.map((e) => ({ ...e, published_at: new Date(e.published_at) }))
.reduce((acc, e) => {
const year = e.published_at.getFullYear();
const month = e.published_at.getMonth() 1;
if (!acc[year]) acc[year] = { year };
if (!acc[year][month]) acc[year][month] = [];
acc[year][month] = e;
return acc;
}, {})
const result = Object.values(groupedByYear).reduce((acc, e) => {
const { year, ...months } = e;
acc.push({ year: year, months: months });
return acc;
}, [])
這是一個例子，可能不是最好的方法。它僅旨在向您展示資料轉換的路徑。
首先data.map能夠對日期進行操作。然后reduce對資料進行分組（這里使用物件）。然后從物件值創建一個陣列以匹配您想要的輸出。
與您展示的解決方案相比，您可以限制迭代陣列的次數。避免在陣列上迭代太多時間以獲得更好的性能始終是一個好主意。
轉載請註明出處，本文鏈接：https://www.uj5u.com/net/455821.html
標籤：javascript 数组 排序 约会时间 目的