日历展示考勤,怎么判断哪天是休息时间

Question

日历展示考勤，上几天休几天的那种，我不知道怎么判断哪天是休息时间？

 // start_date开始时间，end_date结束时间，以下是开始时间和结束时间的年月日
       var start_year = new Date(start_date).getFullYear();
        var start_month =new Date(start_date).getMonth()+1;
        var start_day = new Date(start_date).getDate();
        var end_year = new Date(end_date).getFullYear();
        var end_month =new Date(end_date).getMonth()+1;
        var end_day = new Date(end_date).getDate();
  var work_no ="3" // 上班天数 
 var rest_no  = "2" // 休息天数。
//totalDay是当月的天数
for (var i = 1; i <= totalDay; i++) {
 // 此循环遍历的当月日期 var date = my_year + "-" + (my_month + 1) + '-' + i;
}

Answer 1

根据班制数据判断，你这没头没尾的。。。。让我一通追问才弄明白

data  = [{
    "end_date": "2023-05-20 10:58:46.0",
    "trusday": "",
    "saturday": "",
    "sf_id": "797795276940116965",
    "person_sex": "男",
    "work_no": "2",
    "frequency_name": "班次二",
    "closing_time": "00:00:00",
    "wednesday": "",
    "friday": "",
    "id": "846475598908882982",
    "person_collection_id": "13a003bfc58d4b6483b834907beb5070",
    "start_date": "2023-04-26 10:34:15.0",
    "monday": "",
    "station_name": "桥西区中转站",
    "work_cycle": "1",
    "create_time": "",
    "rest_no": "1",
    "station_id": "86b1d606baf04dba87230243b6aa2c2a",
    "thursday": "",
    "person_name": "张唐廷",
    "person_age": "34",
    "sunday": "",
    "frequency_id": "797795276940116965",
    "work_time": "04:00:00",
    "organ_id": "57cbf560-7552-44a4-a726-8873a1fd8351"
},
{
    "end_date": "2023-05-30 00:00:00.0",
    "trusday": "0",
    "saturday": "1",
    "sf_id": "843471547225079872",
    "person_sex": "男",
    "work_no": "",
    "frequency_name": "班次三",
    "closing_time": "17:00:00",
    "wednesday": "0",
    "friday": "0",
    "id": "825830464555057319",
    "person_collection_id": "1c0d596547444cfba7b96669846ba568",
    "start_date": "2023-05-01 00:00:00.0",
    "monday": "0",
    "station_name": "河东区中转站",
    "work_cycle": "0",
    "create_time": "",
    "rest_no": "",
    "station_id": "76303b4b78434fe8962b379f29fd53a3",
    "thursday": "0",
    "person_name": "赵六",
    "person_age": "59",
    "sunday": "1",
    "frequency_id": "843471547225079872",
    "work_time": "09:00:00",
    "organ_id": "6ba42ba4-7114-4f84-b877-86a197bea261"
}]

for (var i = 0; i < data.length ; i ++ )
{
    var e = new Date(data[i]['end_date'].replace(/\s+.*$/gi,''));
    var s = new Date(data[i]['start_date'].replace(/\s+.*$/gi,''));
    var days = (e - s) / 1000 / 60 / 60 / 24;
    var b = data[i]['work_no'];
    if (b.length > 0 && parseInt(b) > 0)
    {
        work = parseInt(b);
        rest = parseInt(data[i]['rest_no']);
        arr = Array.from(new Array(days)).map((x,y) => y % (work + rest) < work ? 1 : 0)
    }else{
        week = s.getDay();
        bc = [data[i]['monday'],data[i]['trusday'],0,0,0,data[i]['saturday'],data[i]['sunday']]
        arr = Array.from(new Array(days)).map((x,y) => bc[(y + week) % 7])
    }
    console.log(arr);
}