我有一个具有几个键值对的对象数组，我需要根据'updated_at'对它们进行排序:
[
{
"updated_at" : "2012-01-01T06:25:24Z",
"foo" : "bar"
},
{
"updated_at" : "2012-01-09T11:25:13Z",
"foo" : "bar"
},
{
"updated_at" : "2012-01-05T04:13:24Z",
"foo" : "bar"
}
]
最有效的方法是什么?
当前回答
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
*/
其他回答
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
*/
为了完整起见，这里有一个sortBy的简短泛型实现:
function sortBy(list, keyFunc) {
return list.sort((a,b) => keyFunc(a) - keyFunc(b));
}
sortBy([{"key": 2}, {"key": 1}], o => o["key"])
注意，这里使用了就地排序的数组排序方法。 对于副本，您可以使用arr.concat()或arr.slice(0)或类似的方法来创建副本。
在ES2015支持下，可以通过以下方式完成:
foo.sort((a, b) => a.updated_at < b.updated_at ? -1 : 1)
根据ISO格式的日期进行排序可能代价高昂，除非您将客户端限制为最新和最好的浏览器，这些浏览器可以通过对字符串进行日期解析来创建正确的时间戳。
如果您确定您的输入，并且知道它将始终是yyyy-mm-ddThh:mm:ss和GMT (Z)，您可以从每个成员中提取数字，并将它们作为整数进行比较
array.sort(function(a,b){
return a.updated_at.replace(/\D+/g,'')-b.updated_at.replace(/\D+/g,'');
});
如果日期可以有不同的格式，你可能需要为iso挑战的人添加一些东西:
Date.fromISO: function(s){
var day, tz,
rx=/^(\d{4}\-\d\d\-\d\d([tT ][\d:\.]*)?)([zZ]|([+\-])(\d\d):(\d\d))?$/,
p= rx.exec(s) || [];
if(p[1]){
day= p[1].split(/\D/).map(function(itm){
return parseInt(itm, 10) || 0;
});
day[1]-= 1;
day= new Date(Date.UTC.apply(Date, day));
if(!day.getDate()) return NaN;
if(p[5]){
tz= (parseInt(p[5], 10)*60);
if(p[6]) tz+= parseInt(p[6], 10);
if(p[4]== '+') tz*= -1;
if(tz) day.setUTCMinutes(day.getUTCMinutes()+ tz);
}
return day;
}
return NaN;
}
if(!Array.prototype.map){
Array.prototype.map= function(fun, scope){
var T= this, L= T.length, A= Array(L), i= 0;
if(typeof fun== 'function'){
while(i< L){
if(i in T){
A[i]= fun.call(scope, T[i], i, T);
}
++i;
}
return A;
}
}
}
}
我面对同样的事情，所以我用一个通用的为什么处理这个，我为此建立了一个函数:
/ / example: / /阵列:[姓名:’idan’workerType:(3))、姓名:’stas’workerType: 5),姓名:’kirill’,workerType: (2))] / / keyField:’workerType’ // keysArray:[“4”、“3”、“2”、“5”、“6”]
sortByArrayOfKeys = (array, keyField, keysArray) => {
array.sort((a, b) => {
const aIndex = keysArray.indexOf(a[keyField])
const bIndex = keysArray.indexOf(b[keyField])
if (aIndex < bIndex) return -1;
if (aIndex > bIndex) return 1;
return 0;
})
}
