JavaScript排序数组由2个值组成

Question

我有一个数组，我想按“id”和“date”从小到大排序。 我该怎么做才能正确？

示例：

var unsorted = [
    {id: 1, date: "2015-01-18T15:00:00+01:00"}, 
    {id: 1, date: "2015-01-18T14:30:00+01:00"}, 
    {id: 2, date: "2015-01-18T10:00:00+01:00"}, 
    {id: 1, date: "2015-01-18T16:00:00+01:00"}, 
    {id: 3, date: "2015-01-18T14:15:00+01:00"}, 
    {id: 2, date: "2015-01-18T14:00:00+01:00"}
]

应该返回：

var sorted = [
    {id: 1, date: "2015-01-18T14:30:00+01:00"}, 
    {id: 1, date: "2015-01-18T15:00:00+01:00"}, 
    {id: 1, date: "2015-01-18T16:00:00+01:00"}, 
    {id: 2, date: "2015-01-18T10:00:00+01:00"}, 
    {id: 2, date: "2015-01-18T14:00:00+01:00"},
    {id: 3, date: "2015-01-18T14:15:00+01:00"} 
]

Answer 1

以下是使用array.sort的示例：

 var arr = [ {id: 1, date: "2015-01-18T15:00:00+01:00"}, {id: 1, date: "2015-01-18T14:30:00+01:00"}, {id: 2, date: "2015-01-18T10:00:00+01:00"}, {id: 1, date: "2015-01-18T16:00:00+01:00"}, {id: 3, date: "2015-01-18T14:15:00+01:00"}, {id: 2, date: "2015-01-18T14:00:00+01:00"} ]; arr.sort(function(a,b){ if (a.id == b.id) return a.date.localeCompare(b.date); return a.id-b.id; }); // test for (var i in arr) { console.log(arr[i]); } 

结果是：

 Object {id: 1, date: "2015-01-18T14:30:00+01:00"}
 Object {id: 1, date: "2015-01-18T15:00:00+01:00"}
 Object {id: 1, date: "2015-01-18T16:00:00+01:00"}
 Object {id: 2, date: "2015-01-18T10:00:00+01:00"}
 Object {id: 2, date: "2015-01-18T14:00:00+01:00"}
 Object {id: 3, date: "2015-01-18T14:15:00+01:00"}

Answer 2

你可以使用.sort() ：

var unsorted = [
    {id: 1, date: "2015-01-18T15:00:00+01:00"}, 
    {id: 1, date: "2015-01-18T14:30:00+01:00"}, 
    {id: 2, date: "2015-01-18T10:00:00+01:00"}, 
    {id: 1, date: "2015-01-18T16:00:00+01:00"}, 
    {id: 3, date: "2015-01-18T14:15:00+01:00"}, 
    {id: 2, date: "2015-01-18T14:00:00+01:00"}
];

var sorted = unsorted.sort(function(a, b) {
    return a.id == b.id ?
        new Date(a.date) - new Date(b.date) : a.id - b.id;
});

console.log(sorted);

输出：

[ { id: 1, date: '2015-01-18T14:30:00+01:00' },
  { id: 1, date: '2015-01-18T15:00:00+01:00' },
  { id: 1, date: '2015-01-18T16:00:00+01:00' },
  { id: 2, date: '2015-01-18T10:00:00+01:00' },
  { id: 2, date: '2015-01-18T14:00:00+01:00' },
  { id: 3, date: '2015-01-18T14:15:00+01:00' } ]

Answer 3

试一试

var sorted = unsorted.sort(function(a, b) {
    return a.id === b.id ?
      Date.parse(a.date) - Date.parse(b.date) :
      a.id - b.id ;
});

说明

如果id字段相等，我们想要返回date字段的比较。

如果id字段不相等，我们将返回id字段的比较

Answer 4

Array.sort采用带有两个参数的函数来比较数组的两个元素。 如果此函数返回负数，则a放在b之前，如果它返回正数，则a放在b之前，如果它返回0，则它们保持不变。 在这里，我通过id比较它们，如果它们的ID相同，那么我按日期比较它们。

 var unsorted = [{ id: 1, date: "2015-01-18T15:00:00+01:00" }, { id: 1, date: "2015-01-18T14:30:00+01:00" }, { id: 2, date: "2015-01-18T10:00:00+01:00" }, { id: 1, date: "2015-01-18T16:00:00+01:00" }, { id: 3, date: "2015-01-18T14:15:00+01:00" }, { id: 2, date: "2015-01-18T14:00:00+01:00" }]; unsorted.sort(function(a, b) { if (a.id < b.id) return -1; else if (a.id > b.id) return 1; else { if (a.date < b.date) return -1; else if (a.date > b.date) return 1; else return 0; } }); 

Answer 5

分而治之！

首先将输入数组缩减为id => object的映射，即：

 var dataById = unsorted.reduce(function (soFar, value) { // Initialise the array if we haven't processed this // id yet. if (soFar[value.id] === undefined) { soFar[value.id] = []; } // ad this object to Array. soFar[value.id].push(value); return soFar; }, {}); 

现在，您可以通过循环对象的键对每个数组进行排序，请注意，这会修改dataById映射。

 Object.keys(dataById).forEach(function (id) { dataById[id] = dataById[id].sort(); }); 

最后，您可以将所有数据组合在一起，再次迭代地图中的键。 请注意，javascript中的地图（对象）不保证其键的顺序，因此您可能希望在迭代之前首先将ID转储到数组：

 var ids = Object.keys(dataById).sort(); // Reduce the ids into an Array of data. var ids.reduce(function (soFar, value) { return soFar.concat(dataById[id]); }, []); 

不是解决问题的最有效方法，但希望它能为您提供一些思考过程的帮助。