按自定义变量排序数组

Question

$("#show_times_available_" + pos).find("td span").parent().each(function(){ 
    array.push($(this).attr("id")); 
    array = array.sort();
});

我的数组获取所有元素并获取它们的 ID 并将其推送到一个数组中，因此我的数组最终将成为；

[ "mon_0600-0", "sun_0700-0", "thu_0600-0", "thu_0700-0", "tue_0300-0", "wed_0700-0" ];

每次将新元素推入数组时，我尝试将这些元素排序为（周一、周二、周三等）。这样我的数组最终会变成;

[ "mon_0600-0", "tue_0300-0", "wed_0700-0", "thu_0600-0", "thu_0700-0", "sun_0700-0" ];

使用基本的sort() 函数会按字母顺序排列，我知道sort() 函数可以接收另一个函数。我只是不确定如何在 vanilla javascript 或 jQuery 中进行设置。是否有 CASE THEN 或使用 when() 和 then() 对这些进行排序的方法？我已经在 google 和 SO 周围搜索过，但什么都没有。

Answer 1

如您所说，您需要将自定义函数传递给sort()。在这里，您需要将天与时间分开，然后返回日期部分和时间部分的比较（如果天数相同）：

var arr = ["mon_0600-0", "sun_0700-0", "thu_0600-0", "thu_0700-0", "tue_0300-0", "wed_0700-0"];

// store the order of the days
var days = ["mon", "tue", "wed", "thu", "fri", "sat", "sun"];

var arrSorted = arr.sort(function(a, b) {
  // a and b are two times to compare

  // split each into days and times, e.g ["mon", "0600-0"]
  var aSplit = a.split("_");
  var bSplit = b.split("_");
  
  // get the index of the days to compare
  var dayA = days.indexOf(aSplit[0]);
  var dayB = days.indexOf(bSplit[0]);
  
  // if days are the same, compare the times using normal String comparison
  // if days are different, return the comparison of their position in the array
  return dayA == dayB ?
  	aSplit[1].localeCompare(bSplit[1])
  	: dayA - dayB;
});

console.log(arrSorted);

Answer 2

这里有一个自定义排序功能

// your array
array = [ "mon_0600-0", "sun_0700-0", "thu_0600-0", "thu_0700-0", "tue_0300-0", "wed_0700-0" ];

// the order stored in an object
daynames={
    mon:1,
    tue:2,
    wed:3,
    thu:4,
    fri:5,
    sat:6,
    sun:7
}

// the custom sort function which gets the value for the dayname-key and compares
function SortByDayName(a, b){

  // get the value of dayname + the time 
  var aName = daynames[a.substr(0,3)] + a.substr(4,7);;
  var bName = daynames[b.substr(0,3)] + b.substr(4,7);; 
  return ((aName < bName) ? -1 : ((aName > bName) ? 1 : 0));
}

// apply sort
array.sort(SortByDayName);

// output:
["mon_0600-0", "tue_0300-0", "wed_0700-0", "thu_0600-0", "thu_0700-0", "sun_0700-0"]

Answer 3

我建议你使用一个以排序方式包含天数的数组：

var days = ["mon", "tue", "wed", "thu", "fri", "sat", "sun"];

然后您可以使用索引对数据进行排序：

array = array.sort(function (x, y) {
    return days.indexOf(x.slice(0,3)) - days.indexOf(y.slice(0,3));
});

Answer 4

当您使用.sort 时，您可以指定一个进行比较的函数。

你的函数必须接收两个参数，比如compare(a,b)，如果a在前则返回小于零，如果两者必须占据相同位置则返回零，如果b必须在a之后则返回大于零.

var array = ["X123", "X0072", "X04"];

var compareFunction = function(a,b) {
  var _a = parseInt(a.substring(1));
  var _b = parseInt(b.substring(1));
  return _a-_b;
}

$('#original').text(JSON.stringify(array));

array.sort();
$('#sort').text(JSON.stringify(array));

array.sort(compareFunction);
$('#customSort').text(JSON.stringify(array));

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

<p>Original:</p>
<pre id="original"></pre>

<p>Sort:</p>
<pre id="sort"></pre>

<p>Custom sort:</p>
<pre id="customSort"></pre>

Answer 5

你可以使用这个函数来做到这一点

var array = ["mon_0600-0", "sun_0700-0", "thu_0600-0", "thu_0700-0", "tue_0300-0", "wed_0700-0"],
  sort = ["mon", "tue", "wed", "thu", "fri", "sat", "sun"];

var sortondays = function() {
  var newArray = [];
  for (i = 0; i < sort.length; i++) {
    var t = sort[i];
    for (x = 0; x < array.length; x++) {
      if (array[x].startsWith(t)) {
        newArray.push(array[x])
      }
    }
  }
  return newArray;
}
console.log(sortondays());

Answer 6

这个问题对于Sorting with map 来说是完美的，因为一天的拆分和查找只针对一个元素进行一次，而不是针对每个排序回调。

// the array to be sorted
var data = ["mon_0600-0", "sun_0700-0", "thu_0600-0", "thu_0700-0", "tue_0300-0", "wed_0700-0"];

// temporary array holds objects with position and sort-value
var mapped = data.map(function (el, i) {
    var value = el.split('_'), day;
    return {
        index: i,
        day: { mon: 1, tue: 2, wed: 3, thu: 4, fri: 5, sat: 6, sun: 7 }[value[0]] || 0,
        hour: value[1]
    };
})

// sorting the mapped array containing the reduced values
mapped.sort(function (a, b) {
    return a.day - b.day || a.hour.localeCompare(b.hour);
});

// container for the resulting order
var result = mapped.map(function (el) {
    return data[el.index];
});

document.write('<pre>' + JSON.stringify(result, 0, 4) + '</pre>');