Question

我要为一周中的每一天生成div：

var days = new Array("SUN","MON","TUE","WED","THU","FRI","SAT"); 

function generateWeekdays() {
  var d = new Date();
  var weekday = d.getDay();
  var todaysWeekday = days[weekday];

  for (var i = weekday; i < days.length; i++) {
      $('<div>' + days[i] + '</div>').appendTo('#weekdayList');

  }
}


 generateWeekdays();

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<div id="weekdayList"></div>

（例如）今天（星期三）和星期六（最后一个索引）生成该索引，但是如何使它生成一个星期，所以它一直生成到下周的星期三，然后在此之后停止？

Answer 1

您可以使用模块运算符%来使索引保持界限。这将使超出数组长度的所有内容返回，因此您可以从0到数组长度加上当前日期：

var days = new Array("SUN","MON","TUE","WED","THU","FRI","SAT"); 

function generateWeekdays() {
  var d = new Date();
  var weekday = d.getDay();

  for (var i = 0; i < days.length; i++) {
     console.log(days[(i + weekday) % days.length])

  }
}
generateWeekdays()

问题尚不完全清楚，但是如果您想在周三加入 next ，则可以使用for (var i = 0; i < days.length + 1; i++)

进行循环

Answer 2

您可以迭代7次（根据需要一周），而不必在days.length处停止，并进行模（％7）运算以从days数组中获取元素。

var days = new Array("SUN","MON","TUE","WED","THU","FRI","SAT"); 

function generateWeekdays() {
  var d = new Date();
  var weekday = d.getDay();
  var todaysWeekday = days[weekday];

  for (var d = weekday, i = 0; i < 7; d++, i++) {
      $('<div>' + days[d % 7] + '</div>').appendTo('#weekdayList');

  }
}

generateWeekdays();

Answer 3

如何削减日期（从正常工作日开始）并在一周的末尾（那些天）追加（从今天开始，您必须从该日期开始削减）

function getWeekDaysFromToday() {
  let days = ["SUN","MON","TUE","WED","THU","FRI","SAT"];
  return days.concat( days.splice (0,new Date().getDay()));
}
console.log(getWeekDaysFromToday())

Answer 4

在您的for循环中，而不是进行迭代直到days.length，您只想进行7次迭代并循环回到数组的开头。如果您设置了一个将迭代7次的计数器，并且在循环中添加了if语句，如果到达末尾则循环回到数组的开头，那应该可行。

Answer 5

考虑到今天与数组索引有关的内容，如何添加另一个迭代？

var days = new Array("SUN","MON","TUE","WED","THU","FRI","SAT"); 

function generateWeekdays() {
  var d = new Date();
  var weekday = d.getDay();
  var todaysWeekday = days[weekday];

  for (var i = weekday; i < days.length; i++) {
      $('<div>' + days[i] + '</div>').appendTo('#weekdayList');

  }
  
  for(var i = 0; i < days.indexOf(days[weekday]); i++){
     $('<div>' + days[i] + '</div>').appendTo('#weekdayList');
  }
}


 generateWeekdays();

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<div id="weekdayList"></div>

Answer 6

这是我一起提出的快速实施方案。我使用了一个do / while循环，因为您知道您总是至少要执行一次。我还将循环限制为要显示的天数，因此您可以增加显示天数。我还使用了％（模量）与count和weekday来从数组中获取正确的值。

var days = new Array("SUN", "MON", "TUE", "WED", "THU", "FRI", "SAT");

function generateWeekdays() {
  var d = new Date();
  var weekday = d.getDay();
  var todaysWeekday = days[weekday];

  let count = 0;
  do {
    $('<div>' + days[(count + weekday) % days.length] + '</div>').appendTo('#weekdayList');

    count++;
  } while (count < 7)

}

generateWeekdays()

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="weekdayList">

</div>

Answer 7

我对这个问题采取了不同的方法。我没有逐步检查是否已经处理了7天，而是将天数列表连接到自身，所以它将是14天，然后从当天开始，我将7天拼接起来，所以我有一周的时间，对于每个我创建了一个div。

我还对其进行了更改，以收集div并将它们全部附加到末尾，以提高性能。

var days = new Array("SUN","MON","TUE","WED","THU","FRI","SAT"); 

function generateWeekdays() {
  var oneWeekOfDays = days.concat(days).splice(new Date().getDay(), 7).map(function(day){
    return '<div>'+ day +'</div>';
  });
  
  $('#weekdayList').append(oneWeekOfDays);
}

generateWeekdays();

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="weekdayList"></div>

从今天和一周以后产生工作日

7 个答案: