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
数据导入
[
{
"gameStatus": "1",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-20 11:32:04"
},
{
"gameStatus": "0",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 18:08:24"
},
{
"gameStatus": "2",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 18:35:40"
},
{
"gameStatus": "0",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 10:42:53"
},
{
"gameStatus": "2",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-20 10:54:09"
},
{
"gameStatus": "0",
"userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
"created_at": "2018-12-19 18:46:22"
},
{
"gameStatus": "1",
"userId": "7118ed61-d8d9-4098-a81b-484158806d21",
"created_at": "2018-12-20 10:50:48"
}
]
由高到低
arr.sort(function(a, b){
var keyA = new Date(a.updated_at),
keyB = new Date(b.updated_at);
// Compare the 2 dates
if(keyA < keyB) return -1;
if(keyA > keyB) return 1;
return 0;
});
例如Asc Order
[
{
"gameStatus": "0",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 10:42:53"
},
{
"gameStatus": "0",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 18:08:24"
},
{
"gameStatus": "2",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 18:35:40"
},
{
"gameStatus": "0",
"userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
"created_at": "2018-12-19 18:46:22"
},
{
"gameStatus": "1",
"userId": "7118ed61-d8d9-4098-a81b-484158806d21",
"created_at": "2018-12-20 10:50:48"
},
{
"gameStatus": "2",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-20 10:54:09"
},
{
"gameStatus": "1",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-20 11:32:04"
}
]
降序为
arr.sort(function(a, b){
var keyA = new Date(a.updated_at),
keyB = new Date(b.updated_at);
// Compare the 2 dates
if(keyA > keyB) return -1;
if(keyA < keyB) return 1;
return 0;
});
描述顺序示例
[
{
"gameStatus": "1",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-20 11:32:04"
},
{
"gameStatus": "2",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-20 10:54:09"
},
{
"gameStatus": "1",
"userId": "7118ed61-d8d9-4098-a81b-484158806d21",
"created_at": "2018-12-20 10:50:48"
},
{
"gameStatus": "0",
"userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
"created_at": "2018-12-19 18:46:22"
},
{
"gameStatus": "2",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 18:35:40"
},
{
"gameStatus": "0",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 18:08:24"
},
{
"gameStatus": "0",
"userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
"created_at": "2018-12-19 10:42:53"
}
]
其他回答
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));
});
}
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;
}
}
}
}
使用array .sort()对数组进行排序 克隆数组使用展开运算符(…)使函数纯 按所需键排序(updated_at) 将日期字符串转换为日期对象 Array.sort()的工作原理是，如果current项和next项是一个可以执行无节奏操作的数字/对象，则从current项和next项中减去两个属性
const input = [
{
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',
}
];
const sortByUpdatedAt = (items) => [...items].sort((itemA, itemB) => new Date(itemA.updated_at) - new Date(itemB.updated_at));
const output = sortByUpdatedAt(input);
console.log(input);
/*
[ { 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' } ]
*/
console.log(output)
/*
[ { updated_at: '2012-01-01T06:25:24Z', foo: 'bar' },
{ updated_at: '2012-01-05T04:13:24Z', foo: 'bar' },
{ updated_at: '2012-01-09T11:25:13Z', foo: 'bar' } ]
*/
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);
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));
});
