• [原创]JS端计算一段时间内工作日的天数,排除周末和法定节假日,同时考虑到调休日


    公司项目最近有个需求,要统计人员有多少天没使用工作系统。

    原本只是简单地将当前时间与最后使用时间相减,得到的天数做为未使用的天数。

    结果,领导说这样不行,需要计算得精确些,于是上网搜到不少解决方案,不是通过循环解决,就是都不符合我们的实际情况,便自己写一个。

    项目是jQuery的,所以循环语句也需要自己改一下。希望对同样需求的兄弟姐妹们有所帮助,也希望大家多多拍砖,提点意见哈!

    //法定节假日和调休日的设定
    var
    Holiday = ["2012-01-01", "2012-01-02", "2012-01-03", "2012-01-22", "2012-01-23", "2012-01-24", "2012-01-25", "2012-01-26", "2012-01-27", "2012-01-28", "2012-04-02", "2012-04-03", "2012-04-04", "2012-04-29", "2012-04-30", "2012-05-01", "2012-06-22", "2012-06-23", "2012-06-24", "2012-09-30", "2012-10-01", "2012-10-02", "2012-10-03", "2012-10-04", "2012-10-05", "2012-10-06", "2012-10-07"];




    var
    WeekendsOff = ["2011-12-31", "2012-01-21", "2012-01-29", "2012-03-31", "2012-04-01", "2012-04-28", "2012-09-29"];

    function nearlyWeeks (mode, weekcount, end) {
    /*
    功能:计算当前时间(或指定时间),向前推算周数(weekcount),得到结果周的第一天的时期值;
    参数:
    mode -推算模式('cn'表示国人习惯【周一至周日】;'en'表示国际习惯【周日至周一】)
    weekcount -表示周数(0-表示本周, 1-前一周,2-前两周,以此推算);
    end -指定时间的字符串(未指定则取当前时间);
    */

    if (mode == undefined) mode = "cn";
    if (weekcount == undefined) weekcount = 0;
    if (end != undefined)
    end = new Date(new Date(end).toDateString());
    else
    end = new Date(new Date().toDateString());

    var days = 0;
    if (mode == "cn")
    days = (end.getDay() == 0 ? 7 : end.getDay()) - 1;
    else
    days = end.getDay();

    return new Date(end.getTime() - (days + weekcount * 7) * 24 * 60 * 60 * 1000);
    };

    function getWorkDayCount (mode, beginDay, endDay) {
    /*
    功能:计算一段时间内工作的天数。不包括周末和法定节假日,法定调休日为工作日,周末为周六、周日两天;
    参数:
    mode -推算模式('cn'表示国人习惯【周一至周日】;'en'表示国际习惯【周日至周一】)
    beginDay -时间段开始日期;
    endDay -时间段结束日期;
    */
    var begin = new Date(beginDay.toDateString());
    var end = new Date(endDay.toDateString());

    //每天的毫秒总数,用于以下换算
    var daytime = 24 * 60 * 60 * 1000;
    //两个时间段相隔的总天数
    var days = (end - begin) / daytime + 1;
    //时间段起始时间所在周的第一天
    var beginWeekFirstDay = nearlyWeeks(mode, 0, beginDay.getTime()).getTime();
    //时间段结束时间所在周的最后天
    var endWeekOverDay = nearlyWeeks(mode, 0, endDay.getTime()).getTime() + 6 * daytime;

    //由beginWeekFirstDay和endWeekOverDay换算出,周末的天数
    var weekEndCount = ((endWeekOverDay - beginWeekFirstDay) / daytime + 1) / 7 * 2;
    //根据参数mode,调整周末天数的值
    if (mode == "cn") {
    if (endDay.getDay() > 0 && endDay.getDay() < 6)
    weekEndCount -= 2;
    else if (endDay.getDay() == 6)
    weekEndCount -= 1;

    if (beginDay.getDay() == 0) weekEndCount -= 1;
    }
    else {
    if (endDay.getDay() < 6) weekEndCount -= 1;

    if (beginDay.getDay() > 0) weekEndCount -= 1;
    }

    //根据调休设置,调整周末天数(排除调休日)
    $.each(WLD.Setting.WeekendsOff, function (i, offitem) {
    var itemDay = new Date(offitem.split('-')[0] + "/" + offitem.split('-')[1] + "/" + offitem.split('-')[2]);
    //如果调休日在时间段区间内,且为周末时间(周六或周日),周末天数值-1
    if (itemDay.getTime() >= begin.getTime() && itemDay.getTime() <= end.getTime() && (itemDay.getDay() == 0 || itemDay.getDay() == 6))
    weekEndCount -= 1;
    });
    //根据法定假日设置,计算时间段内周末的天数(包含法定假日)
    $.each(WLD.Setting.Holiday, function (i, itemHoliday) {
    var itemDay = new Date(itemHoliday.split('-')[0] + "/" + itemHoliday.split('-')[1] + "/" + itemHoliday.split('-')[2]);
    //如果法定假日在时间段区间内,且为工作日时间(周一至周五),周末天数值+1
    if (itemDay.getTime() >= begin.getTime() && itemDay.getTime() <= end.getTime() && itemDay.getDay() > 0 && itemDay.getDay() < 6)
    weekEndCount += 1;
    });

    //工作日 = 总天数 - 周末天数(包含法定假日并排除调休日)
    return days - weekEndCount;
    };


     

  • 相关阅读:
    心情记录&考试总结 3.30
    BZOJ 1982 Moving Pebbles
    BZOJ 3759 Hungergame
    51Nod 算法马拉松12 Rikka with sequences
    51Nod 算法马拉松12 移数博弈
    BZOJ 3720 gty的妹子树
    BZOJ 4184 shallot
    BZOJ 3160 万径人踪灭
    好好学习天天向上
    java解析json字符串
  • 原文地址:https://www.cnblogs.com/w3live/p/2345461.html
Copyright © 2020-2023  润新知