Question

我有一个数组，我想按“id”和“date”从小到大排序。我怎么能正确地做到这一点？

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

您可以使用.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 ;
});

解释

如果id字段相等，我们希望返回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接受一个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 =＆gt;的映射。对象，即：

＆＃13;

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;
}, {});

＆＃13;

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

＆＃13;

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

＆＃13;

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：

＆＃13;

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]);
}, []);

＆＃13;

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。

JavaScript排序数组由2个值组成

5 个答案: