Question

我目前正在开发一个自定义jquery / javascript倒计时器，用于指示客户在发货之前还剩多少时间购买。它很粗糙 - 但它对我有用，我本身不是编码器 - 基本上。

＆＃13;

jQuery(function($) {
  $(document).ready(function() {
    setInterval(function() {
      var now = new Date();
      var day = now.getDay();
      //var day = 6;
      var day2 = (now.getDate() < 10 ? '0' : '') + now.getDate();
      var month = ("0" + (now.getMonth() + 1)).slice(-2);
      var offday = day2 + month;
      var offdayset = false;
      var end;

      if (day >= 1 && day <= 4) {
        end = new Date(now.getYear(), now.getMonth(), day, 15, 30, 0, 0);
      } else if (day == 5) {
        end = new Date(now.getYear(), now.getMonth(), day, 15, 30, 0, 0);
      } else {
        end = new Date(now.getYear(), now.getMonth(), day, 15, 30, 0, 0);
      }

      var timeleft = end.getTime() - now.getTime();
      var diff = new Date(timeleft);
      var weekday = new Array(7);
      weekday[0] = "Zondag";
      weekday[1] = "Maandag";
      weekday[2] = "Dinsdag";
      weekday[3] = "Woensdag";
      weekday[4] = "Donderdag";
      weekday[5] = "Vrijdag";
      weekday[6] = "Zaterdag";
      var shippingday = weekday[now.getDay()];

      /* Declare an array. */
      var offdays = new Array('2303', '2412', '2512', '3112');
      /* Traverse each of value of an array using for loop to 
      check whether the value is exist in array*/
      for (var i = 0; i < offdays.length; i++) {
        if (offdays[i] === offday) {
          //alert('Value exist');
          offdayset = true;
        }
      }

      if (shippingday == "Zaterdag" || shippingday == "Zondag") 
      {
        shippingday = "Maandag";
      } else if ("" + diff.getHours() + ('0' + diff.getMinutes()).slice(-2) <= 1630 && offdayset == false) {
        shippingday = "Vandaag";
      } else {
        shippingday = weekday[now.getDay() + 1];
      }

      $("#datecountdown").html("binnen " + diff.getHours() + "u " + diff.getMinutes() + "min " + diff.getSeconds() + "sec");
      $("#dateshipping").html(shippingday);
      //below are just for testing purposes
      $("#time1").html(offday);
      $("#time2").html(offdayset);
      $("#time3").html(end);

    }, 1000);
  });
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<a href="/verzending-bezorging/" class="shippingtimer" title="*wanneer product op voorraad is">
  <p class="shippingtimer"> Besteld <span id="datecountdown"></span> = <span id="dateshipping"></span> verzonden*</p>
</a>

<span id="time1"></span>
<br>
<span id="time2"></span>
<br>
<span id="time3"></span>

＆＃13;

到目前为止，我的倒计时工作和检查是否是周末;当它的周末（周六或周日）航运将只在周一。但是，我无法通过计时器将其指示为更长的小时数：即星期五星期五16:30之后将是+ 72h，星期六将是+ 48小时，星期日+ 24，直到周一16.30。

任何人都可以伸出援助之手吗？

Answer 1

我重新整理了你的代码，使逻辑更清晰，我想这可能是你想要的。

修改 var shippingDate = new Date（bookingDate.getFullYear（），bookingDate.getMonth（），bookingDate.getDay（），15,30,0,0）; bookingDate.getDay（）应该是bookingDate.getDate（） var shippingDate = new Date（bookingDate.getFullYear（），bookingDate.getMonth（），bookingDate.getDate（），15,30,0,0）;

jQuery(function($) { $(document).ready(function() { var weekday = new Array(7); weekday[0] = "Zondag"; weekday[1] = "Maandag"; weekday[2] = "Dinsdag"; weekday[3] = "Woensdag"; weekday[4] = "Donderdag"; weekday[5] = "Vrijdag"; weekday[6] = "Zaterdag"; //var bookingDate - the date you book var bookingDate = new Date(); //var bookingDay - the day you book var bookingDay = weekday[bookingDate.getDay()]; //var shippingDay - the day you ship var shippingDay = null; //var shippingDate - the date you ship var shippingDate = new Date(bookingDate.getFullYear(), bookingDate.getMonth(), bookingDate.getDate(), 15, 30, 0, 0); //logic to get shippingDay and shippingDate if (bookingDay == "Zaterdag") { shippingDate = shippingDate.setDate(bookingDate.getDate() + 2) shippingDay = weekday[bookingDate.getDay() + 2]; } else if (bookingDay == "Zondag") { shippingDate = shippingDate.setDate(bookingDate.getDate() + 1) shippingDay = weekday[bookingDate.getDay() + 1]; } else { //if we book before 15:30, we still can catch today's shipping //otherwise +1 day if (shippingDate - bookingDate >= 0) { shippingDay = "Vandaag"; } else { shippingDate.setDate(bookingDate.getDate() + 1); shippingDay = weekday[bookingDate.getDay() + 1]; } } //you can deal with offdays with similar logic, I don't know what does shippingday = "Vandaag"; means /*} else if ("" + diff.getHours() + ('0' + diff.getMinutes()).slice(-2) <= 1630 && offdayset == false) { shippingday = "Vandaag"; } */ //I didn't touch your offday logic, you can fill in the gap var now = new Date(); var day = now.getDay(); //var day = 6; var day2 = (now.getDate() < 10 ? '0' : '') + now.getDate(); var month = ("0" + (now.getMonth() + 1)).slice(-2); var offday = day2 + month; var offdayset = false; /* Declare an array. */ var offdays = new Array('2303', '2412', '2512', '3112'); /* Traverse each of value of an array using for loop to check whether the value is exist in array*/ for (var i = 0; i < offdays.length; i++) { if (offdays[i] === offday) { //alert('Value exist'); offdayset = true; } } setInterval(function() { now = new Date(); var timeleft = shippingDate - now; var diff = new Date(timeleft); $("#datecountdown").html("binnen " + diff.getHours() + "u " + diff.getMinutes() + "min " + diff.getSeconds() + "sec"); $("#dateshipping").html(shippingDay); //below are just for testing purposes $("#time1").html(offday); $("#time2").html(offdayset); $("#time3").html("bookingdate: " + bookingDate); $("#time4").html("shippingdate: " + shippingDate); }, 1000); }); });

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> <a href="/verzending-bezorging/" class="shippingtimer" title="*wanneer product op voorraad is"> Besteld = verzonden* </a> 

自定义送货倒数计时器

1 个答案: