js takes the time difference and removes Saturday and Sunday implementation code_javascript skills

function GetDayNum(no,type) {
if (type = = "35") {
var num7 = document.getElementById("6").value;
var value = document.getElementById("7").value;
var valueR = num7;
if (no == "7") {valueR = value;}//Identify the time value
if (num7 != "" && value != "") {//Two time periods can be merged BUG1
var numValue = 0;
var numValue = 0;
var day = 24 * 60 * 60 * 1000;
var dateArr = value.split("-");
var checkDate = new Date();
checkDate.setFullYear(dateArr[0], dateArr[1] - 1, dateArr[2]);
var checkTime = checkDate.getTime();
var dateArr2 = num7.split ("-");
var checkDate2 = new Date();
checkDate2.setFullYear(dateArr2[0], dateArr2[1] - 1, dateArr2[2]);
var checkTime2 = checkDate2. getTime();
var numValue = (checkTime - checkTime2) / day; //Total number of days difference

var totleWeek = numValue / 7; //How many weeks difference
var yuDay = numValue % 7; //Except the number of days in the entire week
var lastDay = 0;
var startWeek = num7.replace(/-/g, "/"); //Change "-" to "/", where "-" is not accurate in calculating time. I tested this
//var date = new Date(parseInt(startWeek[0]), parseInt(startWeek[1]), parseInt(startWeek[2]));
var date = Date.parse(startWeek);
var dateNew = new Date(date);
var weekDay = dateNew.getDay(); //Day of the week
var endWeekDay = 0; //Excessive How many days are Saturday or Sunday
if ((weekDay == 6 && yuDay >= 2) || (weekDay == 7 && yuDay >= 1) || (weekDay == 5 && yuDay >= 3) || (weekDay == 4 && yuDay >= 4) || (weekDay == 3 && yuDay >= 5) || (weekDay == 2 && yuDay >= 6) || ( weekDay == 1 && yuDay >= 7)) {
endWeekDay = 2;www.jb51.net
}
if ((weekDay == 6 && yuDay < 1) || (weekDay = = 7 && yuDay < 5) || (weekDay == 5 && yuDay < 2) || (weekDay == 4 && yuDay < 3) || (weekDay == 3 && yuDay < 4) || ( weekDay == 2 && yuDay < 5) || (weekDay == 1 && yuDay < 6)) {
endWeekDay = 1;
}
// if (numValue == 0) { numValue = 1; }
numValue = numValue - (totleWeek * 2) - endWeekDay; //Final time

}
}
}