JavaScript排序數組由2個值組成

Question

我有一個數組，我想按“id”和“date”從小到大排序。 我該怎么做才能正確？

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"}
]

應該返回：

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]); } 

結果是：

 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);

輸出：

[ { 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

試一試

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

說明

如果id字段相等，我們想要返回date字段的比較。

如果id字段不相等，我們將返回id字段的比較

Answer 4

Array.sort采用帶有兩個參數的函數來比較數組的兩個元素。 如果此函數返回負數，則a放在b之前，如果它返回正數，則a放在b之前，如果它返回0，則它們保持不變。 在這里，我通過id比較它們，如果它們的ID相同，那么我按日期比較它們。

 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！

首先將輸入數組縮減為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; }, {}); 

現在，您可以通過循環對象的鍵對每個數組進行排序，請注意，這會修改dataById映射。

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

最后，您可以將所有數據組合在一起，再次迭代地圖中的鍵。 請注意，javascript中的地圖（對象）不保證其鍵的順序，因此您可能希望在迭代之前首先將ID轉儲到數組：

 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]); }, []); 

不是解決問題的最有效方法，但希望它能為您提供一些思考過程的幫助。