嗨，我有一个日期对象数组
"["2021-01-05T06:30:00.000Z","2021-01-06T06:30:00.000Z",
"2021-01-20T06:30:00.000Z","2021-02-09T06:30:00.000Z",
"2021-02-23T06:30:00.000Z","2021-02-16T06:30:00.000Z",
"2020-12-08T06:30:00.000Z","2020-12-15T06:30:00.000Z",
"2020-12-02T06:30:00.000Z","2020-12-09T06:30:00.000Z",
"2020-12-16T06:30:00.000Z"]"
我需要格式化成这个
[
{
"month": "12",
"year": "2020",
"dates": [1,14,25]
},
{
"month": "10",
"year": "2020",
"dates": [1]
}
]
如何格式化这样的帮助我。我已经这样做了但没有完成我一直在添加日期。我知道这不是正确的做法。请不要打扰我写的代码我知道这是垃圾。
dateArray.reduce((initial,next)=>{
let result=[]
if(isSameYear(new Date(initial),new Date(next) &&
isSameMonth(new Date(initial),new Date(next))){
result.push({
month:new Date(nex).getMonth(),
year: new Date(next).getFullYear
})
}
})
答案 0 :(得分：2)
您可以根据对象累加器中的 year 和 month 对日期进行分组。
const data = ["2021-01-05T06:30:00.000Z", "2021-01-06T06:30:00.000Z", "2021-01-20T06:30:00.000Z", "2021-02-09T06:30:00.000Z", "2021-02-23T06:30:00.000Z", "2021-02-16T06:30:00.000Z", "2020-12-08T06:30:00.000Z", "2020-12-15T06:30:00.000Z", "2020-12-02T06:30:00.000Z", "2020-12-09T06:30:00.000Z", "2020-12-16T06:30:00.000Z" ],
result = Object.values(data.reduce((r, date) => {
const [year, month, day] = date.substr(0,10).split('-');
const key = `${year}_${month}`;
r[key] = r[key] || {month, year, dates: []};
r[key].dates.push(day);
return r;
},{}));
console.log(result);答案 1 :(得分：1)
也许分两遍
const dateArray = ["2021-01-05T06:30:00.000Z", "2021-01-06T06:30:00.000Z", "2021-01-20T06:30:00.000Z", "2021-02-09T06:30:00.000Z", "2021-02-23T06:30:00.000Z", "2021-02-16T06:30:00.000Z", "2020-12-08T06:30:00.000Z", "2020-12-15T06:30:00.000Z", "2020-12-02T06:30:00.000Z", "2020-12-09T06:30:00.000Z", "2020-12-16T06:30:00.000Z"];
const mapping = dateArray.reduce((initial, next) => {
const month = next.substring(5, 7);
const year = next.substring(0, 4);
const day = next.substring(8, 10);
initial[year] = initial[year] || {};
initial[year][month] = initial[year][month] || [];
initial[year][month].push(parseInt(day, 10));
return initial;
}, {});
const result = []
Object.keys(mapping).forEach(year => {
Object.keys(mapping[year]).forEach(month => {
result.push({
month,
year,
dates: mapping[year][month]
});
});
});
console.log(result);答案 2 :(得分：1)
一个简单的解决方案是使用一个对象按月和年分组，如下所示：
const data = ["2021-01-05T06:30:00.000Z","2021-01-06T06:30:00.000Z",
"2021-01-20T06:30:00.000Z","2021-02-09T06:30:00.000Z",
"2021-02-23T06:30:00.000Z","2021-02-16T06:30:00.000Z",
"2020-12-08T06:30:00.000Z","2020-12-15T06:30:00.000Z",
"2020-12-02T06:30:00.000Z","2020-12-09T06:30:00.000Z",
"2020-12-16T06:30:00.000Z"];
function groupDates(dates) {
const groupedDates = {};
dates.forEach(d => {
const dt = new Date(d);
const date = dt.getDate();
const year = dt.getFullYear();
const month = dt.getMonth() + 1;
const key = `${year}-${month}`;
if (key in groupedDates) {
groupedDates[key].dates = [...groupedDates[key].dates, date];
} else {
groupedDates[key] = {
year,
month,
dates: [date],
};
}
});
return Object.values(groupedDates);
}
console.log(groupDates(data));答案 3 :(得分：0)
当您对一般事物进行分组时，将它们分组为一个对象会更容易。原因是您不必在数组中搜索要追加的匹配结果，只需查找要连接的键即可。
这是一种构建对象的解决方案，按月份和年份构建的字符串键分组，然后映射该对象的值以构建您要查找的数组，方法是将字符串键拆分为它们的重要值部分。
const dates = ["2021-01-05T06:30:00.000Z","2021-01-06T06:30:00.000Z","2021-01-20T06:30:00.000Z","2021-02-09T06:30:00.000Z","2021-02-23T06:30:00.000Z","2021-02-16T06:30:00.000Z","2020-12-08T06:30:00.000Z","2020-12-15T06:30:00.000Z","2020-12-02T06:30:00.000Z","2020-12-09T06:30:00.000Z","2020-12-16T06:30:00.000Z"];
const grouped = dates.reduce((accumulator, date) => {
const parsed = new Date(date);
const year = parsed.getFullYear();
const month = parsed.getMonth();
const groupKey = `${month},${year}`;
accumulator[groupKey] = accumulator[groupKey] || {dates: []};
accumulator[groupKey].dates.push(parsed.getDay());
return accumulator;
}, {});
const result = Object.entries(grouped).map(([key, dates]) => {
const parts = key.split(',');
return {
month: parts[0],
year: parts[1],
dates: dates
};
});
console.log(result);
答案 4 :(得分：0)
这是一个不使用任何库的纯 javascript 解决方案。它基于一个简单的 O(n^2) 运行时。但是如果你喜欢用一些库来做二分查找，你可以把它减少到 O(nlogn)。
诀窍是把这个任务变成更小的任务，就像我用函数 getMonthYear（将字符串转换为对象）、比较和添加日期一样：
data = ["2021-01-05T06:30:00.000Z","2021-01-06T06:30:00.000Z","2021-01-20T06:30:00.000Z","2021-02-09T06:30:00.000Z","2021-02-23T06:30:00.000Z","2021-02-16T06:30:00.000Z","2020-12-08T06:30:00.000Z","2020-12-15T06:30:00.000Z","2020-12-02T06:30:00.000Z","2020-12-09T06:30:00.000Z","2020-12-16T06:30:00.000Z"];
function categorize(data) {
// 2021-01-05T06:30:00.000Z => {month:"01", year:"2021", date:"05"}
function getMonthYear(str) {
var datePart = str.toString().trim().split("T")[0];
var datePartArr = datePart.split("-");
return {month:datePartArr[1], year:datePartArr[0], date:datePartArr[2]};
}
// testing
//var ans = getMonthYear("2021-01-06T06:30:00.000Z");
//console.log(ans);
// comparing two items to see if they have the same year and month
function compare(item1, item2) {
return (item1.month == item2.month) && (item1.year == item2.year);
}
// testing
//var ans = compare({month:"04", year:"2021"}, {month:"03", year:"2021"});
//console.log(ans);
// adding a date to the list of dates
function addDate(dateList, dateNumber) {
for(var i in dateList) {
if (dateList[i] == dateNumber) return;
}
dateList.push(dateNumber);
}
// testing
/*var ans = [2,4];
addDate(ans, 4);
console.log(ans);*/
// Now lets build the answer by looping over
// --------------------------------------------
var list = []; // the final answer list
data.forEach(function(str){
var item = getMonthYear(str);
var itemMatched = false;
// now lopping over the list to see if it has any match
for(var i in list) {
if (compare(item, list[i])) { // matched found
itemMatched = true;
addDate(list[i].date, item.date);
break;
}
}
// item had no match, add it as a new item to list
if (!itemMatched) {
list.push({
month: item.month,
year: item.year,
date: [item.date]
});
}
});
return list;
}
var ans = categorize(data);
console.log(ans);
这是指向 jsfiddle
的链接