javascript 按最近的开始日期时间排序以显示最远的距离和另一个属性

Question

我必须按开始日期时间进行排序，并引用具有多个属性的当前日期时间。例如，我有这个列表:

<ul>
    <li data-startDate="09-04-2019 15:00" data-inplay=false>Name1</li>
    <li data-startDate="09-04-2019 15:30" data-inplay=false>Name2</li>
    <li data-startDate="09-04-2019 15:20" data-inplay=false>Name3</li>
    <li data-startDate="09-04-2019 16:00" data-inplay=false>Name4</li>
    <li data-startDate="09-04-2019 15:00" data-inplay=true> Name5</li>
    <li data-startDate="09-04-2019 15:10" data-inplay=false>Name6</li>
    <li data-startDate="09-04-2019 15:05" data-inplay=true> Name7</li>
    <li data-startDate="08-04-2019 12:05" data-inplay=false>Name8</li>
</ul>

并且当前日期时间是 09-04-2019 15:15，所需的输出应该是:

<!--
  Let currentDate = 09-04-2019 15:15
  Sort priority:
    1. inplay is true
    2. startDate is greater than currentDate
    3. startDate is lesser than currentDate
-->
<ul>
  <!-- inplay == true -->
  <li data-startDate="09-04-2019 15:05" data-inplay=true>Name7</li>
  <li data-startDate="09-04-2019 15:00" data-inplay=true>Name5</li>

  <!-- startDate > currentDate -->
  <li data-startDate="09-04-2019 15:20" data-inplay=false>Name3</li>
  <li data-startDate="09-04-2019 15:30" data-inplay=false>Name2</li>
  <li data-startDate="09-04-2019 16:00" data-inplay=false>Name4</li>

  <!-- startDate < currentDate -->
  <li data-startDate="09-04-2019 15:10" data-inplay=false>Name6</li>
  <li data-startDate="09-04-2019 15:00" data-inplay=false>Name1</li>
  <li data-startDate="08-04-2019 12:05" data-inplay=false>Name8</li>
</ul>

我试过了，但不包括 inplay 参数，而且顺序也不完全正确

var list = $("ul > li");
var target = new Date().getTime();
var sortByStartTime = function (a, b) {
    var ret = 0;
    var startDateA = $(a).data("startdate");
    var startDateB = $(b).data("startdate");
    var dateA = startDateA.split(" ")[0].split("-");
    var timeA = startDateA.split(" ")[1].split(":");
    var dateB = startDateB.split(" ")[0].split("-");
    var timeB = startDateB.split(" ")[1].split(":");
    startDateA = new Date(dateA[2], dateA[1] - 1, dateA[0], timeA[0], timeA[1]);
    startDateB = new Date(dateB[2], dateB[1] - 1, dateB[0], timeB[0], timeB[1]);
    var dA = Math.abs(startDateA.getTime() - target);
    var dB = Math.abs(startDateB.getTime() - target);
    if (dA < dB) {
        ret = -1;
    } else if (dA > dB) {
        ret = 1;
    }  else {
        ret = 0;
    }
    return ret;
};

list.sort(sortByStartTime);

for (var i = 0; i < list.length; i = i + 1) {
    list[i].parentNode.appendChild(list[i]);
}

基本上，列表应该按最近的开始日期时间(相对于当前日期时间)到最远的顺序排列，但如果有一些内容，它们应该位于列表的顶部，按最近的开始日期时间(相对于当前日期时间)排列时间)到最远。我希望我解释得足够清楚。感谢您提供任何帮助，谢谢。

Answer 1

排序逻辑:

1. 根据数据输入真假排序，使用数字简化排序逻辑(降序排序)
2. 如果差异相反零的一侧先保持正值，后保持负值(降序排列)
3. 如果与当前时间不同因为两个日期都在 0 的同一侧，按它们的绝对值排序值(升序排列)

var list = Array.from(document.querySelectorAll("ul > li"));
var target = new Date('09-04-2019 15:15');
var boolValue = { true: 1, false: 0 };
var sortByStartTime = function (a, b) {
    var diffA = new Date(a.dataset.startdate) - target;
    var diffB = new Date(b.dataset.startdate) - target;
    var inplayA = boolValue[a.dataset.inplay];
    var inplayB = boolValue[b.dataset.inplay];
    return inplayB - inplayA || (
        diffA < 0 === diffB < 0 ? Math.abs(diffA) - Math.abs(diffB) : diffB - diffA
    );
};

list.sort(sortByStartTime);

for (var i = 0; i < list.length; i = i + 1) {
    list[i].parentNode.appendChild(list[i]);
}

<ul>
  <li data-startDate="09-04-2019 15:00" data-inplay=false>Name1</li>
  <li data-startDate="09-04-2019 15:30" data-inplay=false>Name2</li>
  <li data-startDate="09-04-2019 15:20" data-inplay=false>Name3</li>
  <li data-startDate="09-04-2019 16:00" data-inplay=false>Name4</li>
  <li data-startDate="09-04-2019 15:00" data-inplay=true> Name5</li>
  <li data-startDate="09-04-2019 15:10" data-inplay=false>Name6</li>
  <li data-startDate="09-04-2019 15:05" data-inplay=true> Name7</li>
  <li data-startDate="08-04-2019 12:05" data-inplay=false>Name8</li>
</ul>

PS: replace new Date('09-04-2019 15:15'); with new Date(); to get what you wanted. I hardcoded the date to show that results are as per your requirement.