JavaScript：获取两个日期之间的所有月份？

Question

我有两个这样的日期字符串：

var startDate = '2012-04-01';
var endDate = '2014-11-01';

我想得到一个这样的字符串数组：

var dates = ['2012-04-01', '2012-05-01', '2012-06-01' .... '2014-11-01',];

到目前为止，这就是我所拥有的，但它非常难看：

var startDate = '2012-04-01';
var endDate = '2014-11-01';
var start = new Date(Date.parse(startDate));
var end = new Date(Date.parse(endDate))
var dates = [];
for (var i = start.getFullYear(); i < end.getFullYear() + 1; i++) {
    dates.push(i + '-' + '-01');
}
console.log(dates);

有没有更好的办法？ JSF中。

Answer 1

这应该产生所需的输出：

function dateRange(startDate, endDate) {
  var start      = startDate.split('-');
  var end        = endDate.split('-');
  var startYear  = parseInt(start[0]);
  var endYear    = parseInt(end[0]);
  var dates      = [];

  for(var i = startYear; i <= endYear; i++) {
    var endMonth = i != endYear ? 11 : parseInt(end[1]) - 1;
    var startMon = i === startYear ? parseInt(start[1])-1 : 0;
    for(var j = startMon; j <= endMonth; j = j > 12 ? j % 12 || 11 : j+1) {
      var month = j+1;
      var displayMonth = month < 10 ? '0'+month : month;
      dates.push([i, displayMonth, '01'].join('-'));
    }
  }
  return dates;
}

只需使用您现有的日期格式调用它：

dateRange('2013-11-01', '2014-06-01')
// ["2013-11-01", "2013-12-01", "2014-01-01", "2014-02-01", "2014-03-01", "2014-04-01", "2014-05-01", "2014-06-01", "2014-07-01", "2014-08-01", "2014-09-01", "2014-10-01", "2014-11-01", "2014-12-01"]

Answer 2

您还可以使用出色的 moment.js 库：

var startDate = moment('2012-04-01');
var endDate = moment('2014-11-01');

var result = [];

if (endDate.isBefore(startDate)) {
    throw "End date must be greated than start date."
}      

while (startDate.isBefore(endDate)) {
    result.push(startDate.format("YYYY-MM-01"));
    startDate.add(1, 'month');
}

JSFiddle

Answer 3

如果加载额外的库不是问题，您可以随时尝试很棒的 MomentJS。
提供非常干净和强大的日期操作。

var startDate = moment('2012-04-01');
var endDate = moment('2014-11-01');

var dates = [];
endDate.subtract(1, "month"); //Substract one month to exclude endDate itself

var month = moment(startDate); //clone the startDate
while( month < endDate ) {
    month.add(1, "month");
    dates.push(month.format('YYYY-MM-DD'));
}

console.log(dates);

JSFiddle在这里

Answer 4

 const getMonths = (fromDate, toDate) => { const fromYear = fromDate.getFullYear(); const fromMonth = fromDate.getMonth(); const toYear = toDate.getFullYear(); const toMonth = toDate.getMonth(); const months = []; for(let year = fromYear; year <= toYear; year++) { let monthNum = year === fromYear ? fromMonth : 0; const monthLimit = year === toYear ? toMonth : 11; for(; monthNum <= monthLimit; monthNum++) { let month = monthNum + 1; months.push({ year, month }); } } return months; } const sample = getMonths(new Date('2022-07-28'), new Date('2023-03-20')); console.log(sample); document.write('check the console output');

https://jsfiddle.net/xfayoqvs/

Answer 5

您正在处理“逻辑”跳跃，因此您实际上并不需要计时算法。 所以这是一个简单的计数问题：

var startDate = '2012-04-01';
var endDate = '2014-11-01';
var dates = [];

var d0 = startDate.split('-');
var d1 = endDate.split('-');

for (
    var y = d0[0];
    y <= d1[0];
    y++
) {
    for (
        var m = d0[1];
        m <= 12;
        m++
    ) {
        dates.push(y+"-"+m+"-1");
        if (y >= d1[0] && m >= d1[1]) break;
    };
    d0[1] = 1;
};

console.log(dates);

Answer 6

这是一个仅对特定 YYYY-MM-DD 格式使用字符串操作的解决方案：

 function monthsBetween(...args) { let [a, b] = args.map(arg => arg.split("-").slice(0, 2) .reduce((y, m) => m - 1 + y * 12)); return Array.from({length: b - a + 1}, _ => a++) .map(m => ~~(m / 12) + "-" + ("0" + (m % 12 + 1)).slice(-2) + "-01"); } console.log(monthsBetween('2012-04-01', '2014-11-01'));

Answer 7

使用 moment.js 获取给定日期和现在之间月份的所有第一天的示例。

   var getMonths = function (startDate) {
    var dates = [];
    for (var year = startDate.year(); year <= moment().year(); year++) {
        var endMonth = year != moment().year() ? 11 : moment().month();
        var startMonth = year === startDate.year() ? startDate.month() : 0;
        for (var currentMonth = startMonth; currentMonth <= endMonth; currentMonth = currentMonth > 12 ? currentMonth % 12 || 11 : currentMonth + 1) {
            var month = currentMonth + 1;
            var displayMonth = month < 10 ? '0' + month : month;
            dates.push([year, displayMonth, '01'].join('-'));
        }
    }
    return dates;
};

Answer 8

上述所有解决方案都以 O(n^2) 时间复杂度运行，效率不高。 请参阅以下 O(n) 时间复杂度的解决方案：

 function getAllMonths(start, end){ let startDate = new Date(start); let startYear = startDate.getFullYear(); let startMonth = startDate.getMonth()+1; let endDate = new Date(end); let endYear = endDate.getFullYear(); let endMonth = endDate.getMonth()+1; let countMonth = 0; let countYear = 0; let finalResult = []; for(let a=startYear; a<=endYear; a++){ if(startYear<endYear){ if(countYear==0){ countMonth += 12-startMonth; }else if(countYear>0){ countMonth += 12; } countYear+=1; startYear++; }else if(startYear==endYear){ countMonth+=endMonth; } } for(let i=startMonth; i<=countMonth+startMonth; i++){ finalResult.push(startDate.getFullYear()+(Math.floor(i/12)) + "-" + Math.round(i%13) + "-" + "01"); } return finalResult; } getAllMonths('2016-04-01', '2018-01-01');

可能会共享一个更简单的代码

Answer 9

这是另一个解决方案，使用 Date 对象：

const enumerateMonths = (from, to) => {
  const current = new Date(from)
  current.setUTCDate(1)
  current.setUTCHours(0, 0, 0, 0)
  const toDate = new Date(to)
  const months = []
  while (current.getTime() <= toDate.getTime()) {
    months.push(current.getUTCFullYear() + "-" + `${current.getUTCMonth() + 1}`.padStart(2, "0"))
    current.setUTCMonth(current.getUTCMonth() + 1)
  }
  return months
}

此解决方案假定您提供 Date 对象或 ISO 8601 字符串。 请注意，ISO 8601 日期不一定必须包含小时-分钟-秒部分。 "2012-01-14"是有效的 ISO 8601 日期。

Answer 10

仍然不是一个非常优雅的答案，但可以得到你想要的字符串数组：

var startDate = '2012-04-01';
var endDate = '2014-11-01';
var start = new Date(startDate);
var end = new Date(endDate);
var dates = [];
for (var i = start.getFullYear(); i < end.getFullYear() + 1; i++) {
    for (var j = 1; j <= 12; j++) {
      if (i === end.getFullYear() && j === end.getMonth() + 3) {
        break;
      }
      else if (i === 2012 && j < 4){
        continue;
      }
      else if (j < 10) {
        var dateString = [i, '-', '0' + j, '-','01'].join('');
        dates.push(dateString)
        }
      else {
        var dateString = [i, '-', j, '-','01'].join('');
        dates.push(dateString);
        }
    }
}
console.log(dates);

jsfiddle 链接在这里：http: //jsfiddle.net/8kut035a/

Answer 11

这是我的解决方案，借助数学和 O(n)

determineMonthInInterval(startDate, endDate) {

    let startYear = startDate.getFullYear();
    let endYear = endDate.getFullYear();

    let startMonth = startDate.getMonth() + 1;
    let endMonth = endDate.getMonth() + 1;

    let monthAmount = (endMonth - startMonth) + 1 + (12 * (endYear - startYear));

    let dates = [];
    let currMonth = startMonth;
    let currYear = startYear;
    for( let i=0; i<monthAmount; i++){
        
        let date = new Date(currYear + "/"+currMonth+"/1");
        dates.push(date);

        currYear = startYear + Math.floor((startMonth+i) / 12);
        currMonth = (currMonth) % 12 +1;

    }

    return dates;
}

Answer 12

这是另一种选择：

getRangeOfMonths(startDate: Date, endDate: Date) {
    const dates = new Array<string>();
    const dateCounter = new Date(startDate);
    // avoids edge case where last month is skipped
    dateCounter.setDate(1);
    while (dateCounter < endDate) {
      dates.push(`${dateCounter.getFullYear()}-${dateCounter.getMonth() + 1}`);
      dateCounter.setMonth(dateCounter.getMonth() + 1);
    }
    return dates;
  }