js 取時間差去掉周六周日實現代碼
2024-05-06 14:20:21
供稿：網友
代碼如下:
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;}//標識時間值
if (num7 != "" && value != "") {//兩個時間段可以融合 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; //相差的總天數
var totleWeek = numValue / 7; //差別多少周
var yuDay = numValue % 7; //除了整個星期的天數
var lastDay = 0;
var startWeek = num7.replace(/-/g, "/"); //把“-”改為“/”，其中“-”算時間不準確這個我測試了
//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(); //星期幾
var endWeekDay = 0; //多余的天數有幾天是周六或者周日
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;m.survivalescaperooms.com
}
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; //最終時間
}
}
}
