你可以创建一个闭包并这样传递它 这是我的例子

$.get('https://data.seattle.gov/resource/3k2p-39jp.json?$limit=10&$where=within_circle(incident_location, 47.594972, -122.331518, 1609.34)', 
  function(responce) {

    var filter = 'event_clearance_group', //sort by key group name
    data = responce; 

    var compare = function (filter) {
        return function (a,b) {
            var a = a[filter],
                b = b[filter];

            if (a < b) {
                return -1;
            } else if (a > b) {
                return 1;
            } else {
                return 0;
            }
        };
    };

    filter = compare(filter); //set filter

    console.log(data.sort(filter));
});

这是一个稍微修改过的@David Brainer-Bankers的答案，按字母顺序排序，或按数字排序，并确保以大写字母开头的单词不会排在以小写字母开头的单词之上(例如“apple,Early”将按此顺序显示)。

function sortByKey(array, key) {
    return array.sort(function(a, b) {
        var x = a[key];
        var y = b[key];

        if (typeof x == "string")
        {
            x = (""+x).toLowerCase(); 
        }
        if (typeof y == "string")
        {
            y = (""+y).toLowerCase();
        }

        return ((x < y) ? -1 : ((x > y) ? 1 : 0));
    });
}

我在这里已经回答了一个非常类似的问题:对对象数组进行排序的简单函数

对于这个问题，我创建了一个小函数，可以做你想做的事情:

function sortByKey(array, key) {
    return array.sort(function(a, b) {
        var x = a[key]; var y = b[key];
        return ((x < y) ? -1 : ((x > y) ? 1 : 0));
    });
}

这是另一种更数学的方式，做同样的事情，但更短:

arr.sort(function(a, b){
    var diff = new Date(a.updated_at) - new Date(b.updated_at);
    return diff/(Math.abs(diff)||1);
});

或者在光滑的lambda箭头样式中:

arr.sort((a, b) => {
    var diff = new Date(a.updated_at) - new Date(b.updated_at);
    return diff/(Math.abs(diff)||1);
});

这个方法可以使用任何数字输入

这样我们就可以传递一个用于排序的键函数

Array.prototype.sortBy = function(key_func, reverse=false){
    return this.sort( (a, b) => {
        var keyA = key_func(a),
            keyB = key_func(b);
        if(keyA < keyB) return reverse? 1: -1;
        if(keyA > keyB) return reverse? -1: 1;
        return 0;
    }); 
}

例如，如果我们有

var arr = [ {date: "01/12/00", balls: {red: "a8",  blue: 10}},
            {date: "12/13/05", balls: {red: "d6" , blue: 11}},
            {date: "03/02/04", balls: {red: "c4" , blue: 15}} ]

我们可以

arr.sortBy(el => el.balls.red)
/* would result in
[ {date: "01/12/00", balls: {red: "a8", blue: 10}},
  {date: "03/02/04", balls: {red: "c4", blue: 15}},
  {date: "12/13/05", balls: {red: "d6", blue: 11}} ]
*/

or

arr.sortBy(el => new Date(el.date), true)   // second argument to reverse it
/* would result in
[ {date: "12/13/05", balls: {red: "d6", blue:11}},
  {date: "03/02/04", balls: {red: "c4", blue:15}},
  {date: "01/12/00", balls: {red: "a8", blue:10}} ]
*/

or

arr.sortBy(el => el.balls.blue + parseInt(el.balls.red[1]))
/* would result in
[ {date: "12/13/05", balls: {red: "d6", blue:11}},    // red + blue= 17
  {date: "01/12/00", balls: {red: "a8", blue:10}},    // red + blue= 18
  {date: "03/02/04", balls: {red: "c4", blue:15}} ]   // red + blue= 19
*/

你可以使用array。sort。

这里有一个例子: Var arr = [{ “updated_at”:“2012 - 01 - 01 t06:25:24z”, “foo”:“酒吧” }， ｛ “updated_at”:“2012 - 01 - 09年t11:25:13z”, “foo”:“酒吧” }， ｛ “updated_at”:“2012 - 01 - 05 t04:13:24z”, “foo”:“酒吧” ｝ ］ 加勒比海盗。排序(函数(a, b) { var keyA = new Date(a.updated_at)， keyB = new Date(b.updated_at); //比较两个日期 if (keyA < keyB)返回-1; if (keyA > keyB)返回1; 返回0; })； console.log (arr);