javascript - 在javascript中对天数进行排序

Question

我有一个数组。该数组可以包含 1 到 7 个唯一的日期名称字符串。日期名称将按从周一到周日的顺序排列。- 例如：

[“周二”、“周四”、“周日”]

我想使用 javascript 对该数组进行排序，以便从今天开始排序。

即：如果今天是星期五，那么排序后的数组应该是

[“周日”、“周二”、“周四”]

如果今天是星期四，那么排序的数组应该是

[“周四”、“周日”、“周二”]

任何人都可以帮忙吗？

Answer 1

function sort_days(days) {

要获取今天的星期几，请使用new Date().getDay(). 这假设Sunday = 0, Monday = 1, ..., Saturday = 6.

    var day_of_week = new Date().getDay();

要生成星期几的列表，然后切片名称列表：

    var list = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"];
    var sorted_list = list.slice(day_of_week).concat(list.slice(0,day_of_week));

（今天是星期五，所以sorted_list是['Fri','Sat','Sun','Mon','Tue','Wed','Thu']）

最后，排序，使用indexOf：

    return days.sort(function(a,b) { return sorted_list.indexOf(a) > sorted_list.indexOf(b); });
}

把它们放在一起：

function sort_days(days) {
    var day_of_week = new Date().getDay();
    var list = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"];
    var sorted_list = list.slice(day_of_week).concat(list.slice(0,day_of_week));
    return days.sort(function(a,b) { return sorted_list.indexOf(a) > sorted_list.indexOf(b); });
}

Answer 2

const days = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"];

const sortDays = function (a, b) {
  a = days.indexOf(a);
  b = days.indexOf(b);
  return a < b ? 0 : 1;
};

const myArrayOfDays = ["Tuesday", "Saturday", "Monday", "Thursday"].sort(sortDays);
// returns ["Monday", "Tuesday", "Thursday", "Saturday"];

Answer 3

这是我想出的一个功能：

function sortDays(days) {
  var daysOfWeek = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"];
  var today = new Date().getDay();
  for (var i=0;i<today;i++) daysOfWeek.push(daysOfWeek.shift());
  return daysOfWeek.filter(function(d) { return days.indexOf(d) >= 0; });
}

总体思路是通过根据今天的日期从开始到结束旋转元素来重新安排一周中的日子。然后，您使用该排序来重新排序您的输入数组以匹配。我没有实际排序，而是daysOfWeek根据输入数组的内容过滤了数组。

我不确定Array.filter支持的效果如何，因此您可能希望将其更改为通用 for 循环，而不是取决于您想要支持的浏览器。

这是一个jsfiddle，您也可以在其中玩。

Array.sort或者，您可以按照类似的策略使用内置方法：

var daysOfWeek = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"];

{
    var today = new Date().getDay();
    for (var i=0;i<today;i++) daysOfWeek.push(daysOfWeek.shift());
}

function daysOfWeekSorter(x,y) {
    return daysOfWeek.indexOf(x)-daysOfWeek.indexOf(y);
}

var myDays = ["Tue", "Thu", "Sun"];
myDays.sort(daysOfWeekSorter);

这是另一个可以玩的小提琴。

Answer 4

以防万一我们设法获得或失去一天，我已经建立了我的不需要硬编码的日期列表:) 希望有一天我们能在周六和周日之间获得额外的 24 小时！

function anyDayNow( dys ) {
  var ret = [], cur = new Date(), today = cur.getUTCDay(), txt;
  do { 
    txt = cur.toUTCString().split(',')[0];
    dys.indexOf(txt)!=-1 && ret.push(txt);
    cur.setUTCDate( cur.getUTCDate() + 1 ); 
  } while ( cur.getUTCDay() != today );
  return ret;
}

console.log( anyDayNow( ["Tue", "Thu", "Sun"] ) );

Answer 5

这是一个简单的方法，只使用数组的indexOf、splice filter和concat函数，不需要循环：

function sortMyArray(toSort) {
    var today = new Date().toUTCString().substr(0, 3), //get today as 3 letter string
        list = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"], // days list
        before = list.splice(0, list.indexOf(today)); // splice what is before today in the list

    list = list.concat(before); // concat the list with what was spliced

    return list.filter(function (item) { return toSort.indexOf(item) !== -1}); // return the sorted list with only the asked days
}

利用

console.log(sortMyArray(["Tue", "Thu", "Sun"]));

Answer 6

我也选择了一个filter选项。

const inOrderDays = arr => {
  const list = ["Sun","Mon","Tue","Wed","Thu","Fri","Sat"];
  return list.filter(each => arr.includes(each));
}

Answer 7

这是解决问题的一种有趣的方法——如果这也相当高效（即没有复杂的对象等），也不会感到惊讶

const sortDays = days => {
    let arr = ['', '', '', '', '', '', '']
    days.forEach(day => {
        if (day === 'Sun') arr[0] = 'Sun'
        if (day === 'Mon') arr[1] = 'Mon'
        if (day === 'Tue') arr[2] = 'Tue'
        if (day === 'Wed') arr[3] = 'Wed'
        if (day === 'Thu') arr[4] = 'Thu'
        if (day === 'Fri') arr[5] = 'Fri'
        if (day === 'Sat') arr[6] = 'Sat'
    })
    return arr.filter(str => str !== '')
}

Answer 8

function sortDaysByToday(ds){
  var days = {Sun: 0, Mon: 1, Tue: 2, Wed: 3, Thu: 4, Fri: 5, Sat: 6},
      today = new Date().getDay()

  return ds.sort(function(a,b){
    return (days[a] < today ? days[a] + 7 : days[a])
         - (days[b] < today ? days[b] + 7 : days[b])
  })
}

Answer 9

如果您想对“星期一”开始的一天进行排序，这就是我的回答。您可以通过更改变daySort量值来更改开始日期。

你只需要给这个方法一个未排序的工作日数组，它就会对其进行排序。

sortDays(unsortedDays) {
  let daysSort = ['monday', 'tuesday', 'wednesday', 'thursday', 'friday',
                  'satuday', 'sunday'];
  let sortedDays = [];
  daysSort.forEach((value) => {
    if (unsortedDays.includes(value)) {
      sortedDays.push(value);
    }
    });
  return sortedDays;
}

Answer 10

如果您有一个以一天为键的对象数组，您可以使用此版本的@SheetJS 答案

    const sortDays = (days, timezone) => {
      const dayOfWeek = 6;
      const list = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday'];
      const sortedList = list.slice(dayOfWeek).concat(list.slice(0, dayOfWeek));
      return days.sort((a, b) => {
        if (sortedList.indexOf(a.day) > sortedList.indexOf(b.day)) return 1;
        if (sortedList.indexOf(a.day) < sortedList.indexOf(b.day)) return -1;
        return 0;
      });
    };

    const days = [
    {"_id":"Z378zCrqGM5XNbsXK","color":"#F47373","day":"Friday","hour":7,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"SY83MsxwEyKYrZvxx","color":"#ea5030","day":"Friday","hour":12,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"Cy4SwenuDJqsSu8Wd","color":"#5a9830","day":"Friday","hour":22,"minute":53,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"MDboigebEAokYiuJv","color":"#F47373","day":"Monday","hour":7,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"PzhT93JKkJSbmuLqc","color":"#5a9830","day":"Monday","hour":22,"minute":53,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"kNuPToeSoJ3j8d6wW","color":"#F47373","day":"Saturday","hour":7,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"44NrPF8byhktY3w4K","color":"#5a9830","day":"Saturday","hour":22,"minute":53,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"BxwYYBKPWWEodtkbs","color":"#F47373","day":"Sunday","hour":7,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"9kHsucTj9JZJtyxos","color":"#37D67A","day":"Sunday","hour":9,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"5fz3tAHfHARiafuBg","color":"#ea5030","day":"Sunday","hour":12,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"ZeC8Y8YLGKrK7q3g7","color":"#5a9830","day":"Sunday","hour":22,"minute":53,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"vHjA9hcfLPCp3CBQQ","color":"#F47373","day":"Thursday","hour":7,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"Fmd4xccrPHqDyrhRx","color":"#37D67A","day":"Thursday","hour":8,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"Zijtdb8Cv68cPBc3L","color":"#ea5030","day":"Thursday","hour":12,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"yDf5tj3NQCZXT3iWa","color":"#5a9830","day":"Thursday","hour":22,"minute":53,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"ve9vuxcb6ZkLZgfFq","color":"#F47373","day":"Tuesday","hour":7,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"s5mbpz9oyzjtzXwCt","color":"#0f5b30","day":"Tuesday","hour":21,"minute":23,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"aXZSoJ3cQiA9Hocwa","color":"#5a9830","day":"Tuesday","hour":22,"minute":53,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"FqqaRBPKd3RjHzQEz","color":"#F47373","day":"Wednesday","hour":7,"minute":45,"workspaceId":"6oaKfaAc7GhGYohA9"},{"_id":"e8A5LACfXYfGtacJA","color":"#5a9830","day":"Wednesday","hour":22,"minute":53,"workspaceId":"6oaKfaAc7GhGYohA9"}]
    
    console.log(sortDays(days));