Question

我想让javascript返回从上个星期五到星期一的日期。现在，如果当天是在星期五和星期一之间，那么我想在星期五之前回到今天。这是我到目前为止所做的，但我遗漏了一些东西。

function getPrevFriday(){
    var friday = 5;
    var currentTime = new Date();
     var month = currentTime.getMonth() + 1;
     var day = currentTime.getDay();
     var date = currentTime.getDate();
     var year = currentTime.getFullYear();
     var lastFriday = date - (friday + (7 - day)) % 7+1;
     return month + "/" + lastFriday + "/" + year;  
};
function getPrevMonday(){
    var friday = 5;
    var currentTime = new Date();
     var month = currentTime.getMonth() + 1;
     var day = currentTime.getDay();
     var date = currentTime.getDate();
     var year = currentTime.getFullYear();
     var lastFriday = date - (friday + (7 - day)) % 7+4;
     return month + "/" + lastFriday + "/" + year;
};
function returnPrevWeekend(){
    var currentTime = new Date();
     var month = currentTime.getMonth() + 1;
     var day = currentTime.getDay();
     var year = currentTime.getFullYear();
     function returnToday(){
         return month + "/" + day + "/" + year;
     }
     $('#from, .startDate').val(getPrevFriday());
    if (day < 5 && day > 0) {
        $('#to, .endDate').val(getPrevMonday());
        $('.date-from').html(getPrevFriday() + '&nbsp;-<br>&nbsp;' + getPrevMonday());
    }else {
        $('#to, .endDate').val(returnToday());
        $('.date-from').html(getPrevFriday() + '&nbsp;-<br>&nbsp;' + returnToday());
        }
};

Answer 1

类似的东西：

var today = new Date();
var noOfDays = 0;
// Today is friday, saturyda or sunday
if (today.getDay() >= 5) {
    noOfDays = today.getDay() - 5;
} else {
    // Otherwise
    noOfDays = (2 + today.getDay());
}

var days = [];
for(var i = 0; i < 4; i++) {
    days.push(new Date(today.getFullYear(), today.getMonth(), today.getDate() - noOfDays + i));
}
console.log(days) // Friday - Monday

Answer 2

如果周末在几个月之间意识到这个功能正在破裂，我想出了一个类似的解决方案，但是会全面运作：

function returnPrevWeekend(){
    var today = new Date();
    day = today.getDay();

    if (today.getDay() == 0) { //if today is sunday
        x = 2; // go back to friday!
    }
    else if (today.getDay() == 1) { //if today is monday
        x = 3; // go back to friday!
        y = 0; // don't go back. return today!
    }
    else if (today.getDay() == 2) {//if today is tuesday
        x = 4; // go back to friday!
        y = 1; // go back to monday!
    }
    else if (today.getDay() == 3) {//if today is wednesday
        x = 5; // go back to friday!
        y = 2; // go back to monday!
    }
    else if (today.getDay() == 4) {//if today is thursday
        x = 6; // go back to friday!
        y = 3; // go back to monday!
    }
    else if (today.getDay() == 5) {//if today is friday
        x = 0; // today is friday!
        y = 0; // don't go back. return today!
    }
    else if (today.getDay() == 6) {
        x = 1; // go back to friday!
        y = 0; // don't go back. return today!
    }

    d1 = new Date(); // today!
    d2 = new Date(); // today!
    friday = new Date(d1.setDate(d1.getDate() - x));
    lastFriday = friday.getDate();
    fridayYear = new Date(friday.setFullYear(friday.getFullYear()));
    fridayYear = fridayYear.getFullYear();
    fridayMonth = new Date(friday.setMonth(friday.getMonth()));
    fridayMonth = friday.getMonth()+1;
    monday = new Date(d2.setDate(d2.getDate() - y));
    lastMonday = monday.getDate();
    mondayYear = new Date(monday.setFullYear(monday.getFullYear()));
    mondayYear = mondayYear.getFullYear();
    mondayMonth = new Date(monday.setMonth(monday.getMonth()));
    mondayMonth = monday.getMonth()+1;

    lastFriday = fridayMonth + "/" + lastFriday + "/" + fridayYear;
    lastMonday = fridayMonth + "/" + lastMonday + "/" + mondayYear;

    $('#from, .startDate').val(lastFriday);
    $('#to, .endDate').val(lastMonday);
    $('.date-from').html("<span class='abbrDateFrom'>" + lastFriday + "</span>" + '<p>thru</p>'+"<span class='abbrDateTo'>" + lastMonday +"</span>");


};

返回上周末的日期

2 个答案: