获取两个日期相差的星期数

思路

1.分别计算两个日期所在周对应的周一的日期，比如2022-12-21，为周三，求对应的周一日期，则为2022-12-19，再把求得的两个日期进行天数相减，再除以7，即为两个日期相差的星期数。

原生date实现：

function getDifferWeeksBetweenTwoDates1(str1, str2) {
    //1.转换为date对象
    let date1 = new Date(str1);
    let date2 = new Date(str2);

    //2.由星期推算出该日所在周的星期一与该日相差天数
    let diffDaysToMon1 = date1.getDay() - 1;
    let diffDaysToMon2 = date2.getDay() - 1;
    if(diffDaysToMon1 < 0){
        diffDaysToMon1 = 6;
    }
    if(diffDaysToMon2 < 0){
        diffDaysToMon2 = 6;
    }

    //3.获取参考计算日
    let referDateDay1 = new Date(date1).setDate(date1.getDate() - diffDaysToMon1);
    let referDateDay2 = new Date(date2).setDate(date2.getDate() - diffDaysToMon2);

    let diffDays = parseInt(Math.abs(referDateDay1 - referDateDay2) / 1000 / 60 / 60 / 24);
    
    //4.计算相差周数
    let diffWeeks = diffDays / 7;

    return diffWeeks;
}

moment实现:

function getDifferWeeksBetweenTwoDates2(str1,str2) {
    //1.转换为moment对象
    let moment1 = moment(str1);
    let moment2 = moment(str2);

    //2.由星期推算出该日所在周的星期一与该日相差天数
    let diffDays1 = moment1.format("E") - 1;
    let diffDays2 = moment2.format("E") - 1;

    //3.获取参考计算日
    let referMoment1 = moment1.subtract(diffDays1, "days");
    let referMoment2 = moment2.subtract(diffDays2, "days");

    //4.计算相差周数
    let diffWeeks = Math.abs(Math.floor(referMoment1.diff(referMoment2, "days", true) / 7));

    return diffWeeks;
}

2.设日期a为x1,与周一差为k1,日期b为x2,与周一差为k2,周一对应日期分别为y1,y2,它们相差天数为z。
则有 z = x2 - k2 - (x1 - k1) (x2>=x1) , z = -1 * ( x2 - k2 - (x1 - k1)) (x2 < x1)

function getDifferWeeksBetweenTwoDates3(str1, str2) {
    //1.转换为date对象
    let date1 = new Date(str1);
    let date2 = new Date(str2);

    //2.由星期推算出该日所在周的星期一与该日相差天数
    let diffDaysToMon1 = date1.getDay() - 1;
    let diffDaysToMon2 = date2.getDay() - 1;
    if(diffDaysToMon1 < 0){
        diffDaysToMon1 = 6;
    }
    if(diffDaysToMon2 < 0){
        diffDaysToMon2 = 6;
    }

    let time1 = date1.getTime();
    let time2 = date2.getTime();

    let diffDays =  parseInt((time1 - time2) / 1000 / 60 / 60 / 24);
    let referMonDiffDays = (time1 >= time2 ?  1 : -1) * (diffDays - diffDaysToMon1 + diffDaysToMon2);
    
    //4.计算相差周数
    let diffWeeks =referMonDiffDays / 7;

    return diffWeeks;
}

测试

const testCase = [
    ["2022-01-01","2022-01-01",0],
    ["2022-01-02","2022-01-04",1],
    ["2022-01-01","2021-12-11",3],
    ["2018-03-12","2015-04-12",153],
    ["2022-12-26","2023-01-03",1],
]

function testCaseFunc() {
    testCase.forEach(dates=>{
        const [startDate,endDate,result] = dates;
        console.assert(getDifferWeeksBetweenTwoDates1(startDate,endDate) === result,"failed");
        console.assert(getDifferWeeksBetweenTwoDates2(startDate,endDate) === result,"failed");
        console.assert(getDifferWeeksBetweenTwoDates3(startDate,endDate) === result,"failed");
    })
}


testCaseFunc();

性能比较

const testCase = [
    ["2022-01-01","2022-01-01",0],
    ["2022-01-02","2022-01-04",1],
    ["2022-01-01","2021-12-11",3],
    ["2018-03-12","2015-04-12",153],
    ["2022-12-26","2023-01-03",1],
]
const testPerformanceCount = 10000;

function testPerformanceFunc() {
    console.time("getDifferWeeksBetweenTwoDates1");
    for(let i=0;i<testCount;i++ ){
        testCase.forEach(dates=>{
            const [startDate,endDate] = dates;
            getDifferWeeksBetweenTwoDates1(startDate,endDate);
        })
    }
    console.timeEnd("getDifferWeeksBetweenTwoDates1");

    console.time("getDifferWeeksBetweenTwoDates2");
    for(let i=0;i<testCount;i++ ){
        testCase.forEach(dates=>{
            const [startDate,endDate] = dates;
            getDifferWeeksBetweenTwoDates2(startDate,endDate);
        })
    }
    console.timeEnd("getDifferWeeksBetweenTwoDates2");

    console.time("getDifferWeeksBetweenTwoDates3");
    for(let i=0;i<testCount;i++ ){
        testCase.forEach(dates=>{
            const [startDate,endDate] = dates;
            getDifferWeeksBetweenTwoDates3(startDate,endDate);
        })
    }
    console.timeEnd("getDifferWeeksBetweenTwoDates3");
}

testPerformanceFunc();

使用方法2的话不用先计算出两个日期对应的周一日期，性能较方法1快了2倍,如果使用moment的话,则快了20倍。