T_w520
2015-07-06 06:53
采纳率: 0%
浏览 4.4k

PHP计算一个时间段中的工作日天数

目前我有一个js版本的,我想做成PHP版本的,希望大家能帮忙。
结果就是一个时间段中,计算工作日天数,排除法定节假日,周末,还要加上调休。
js的代码我先贴上来,请大家帮忙。
因菜鸟,目前没有C币,sorry!

 <script src="/Public/Js/jquery.js" type="text/javascript"></script>
<script>
var Holiday = ["2015-07-13", "2015-07-14", "2015-07-15", "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 = ["2015-07-11", "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;'en'表示国际习惯)

  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;'en'表示国际习惯)

  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();
    //alert(beginWeekFirstDay);
  //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(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(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;

  }


function getDate(strDate) {
            var date = eval('new Date(' + strDate.replace(/\d+(?=-[^-]+$)/,
             function (a) { return parseInt(a, 10) - 1; }).match(/\d+/g) + ')');
            return date;
        }
    var r = getWorkDayCount('cn',getDate('2015-07-01'),getDate('2015-07-06'));
    alert(r);
</script>
  • 写回答
  • 好问题 提建议
  • 关注问题
  • 收藏
  • 邀请回答

2条回答 默认 最新

  • T_w520 2015-07-06 07:26

    没人回答呀,求指教呀。哥哥们!

    评论
    解决 无用
    打赏 举报
  • threenewbee 2015-07-06 17:01
    评论
    解决 无用
    打赏 举报

相关推荐 更多相似问题