如何对两个嵌套对象进行排序? [英] How to sort two nested objects?

问题描述

排序此嵌套对象时遇到问题；我拥有的对象是:

I have a problem sorting this nested objects; the object I have is:

Array of objects: [Object, Object]

在此数组内有两个对象，这些对象内有1个对象:2016-5:Object 2016-6:Object

inside this array there are two objects and inside those objects are 1 object: 2016-5:Object 2016-6:Object

在两个对象中的每个内部都是只有1个数字的数组:

Inside every of the two objects are arrays with only 1 number inside:

shift0:Array[1]
shift1:Array[1]
shift2:Array[1]
shift3:Array[1]

并且在数组内部只有一个这样的数字(所有数组的索引均为0):

and inside the array is only one number like so (all with index 0 of the array):

shift0:Array[1] -> 26
shift1:Array[1] -> 42
shift2:Array[1] -> 53
shift3:Array[1] -> 31

我希望能够对数字进行排序-26、42、53、31

I want to be able to sort the numbers - 26, 42, 53, 31

看起来像

var source = [{'2016-5': [{shift0: [26]}, 
                       {shift1: [42]}, 
                       {shift2: [53]}, 
                       {shift3: [31]}]},
           {'2016-6':  [{shift0: [33]}, 
                        {shift1: [15]}, 
                        {shift2: [13]}, 
                        {shift3: [66]}]}
 ];

我想要的结果应该是:

var source = [{'2016-5': [{shift0: [26]}, 
                           {shift3: [31]}, 
                           {shift1: [42]}, 
                           {shift2: [53]}]},
               {'2016-6':  [{shift2: [13]}, 
                            {shift1: [15]}, 
                            {shift0: [33]}, 
                            {shift3: [66]}]}
     ];

推荐答案

尝试一下

var source =  [{'2016-5': [{shift0: [26]}, 
                       {shift1: [42]}, 
                       {shift2: [53]}, 
                       {shift3: [31]}]},
           {'2016-6':  [{shift0: [33]}, 
                        {shift1: [15]}, 
                        {shift2: [13]}, 
                        {shift3: [66]}]}
 ];

source.forEach( function(obj){
   Object.keys(obj).forEach( function(key){
      console.log(obj[key]);
      obj[key].sort( function(a,b){
         return a[ Object.keys(a)[0] ]  - b[ Object.keys(b)[0] ] ;
      })
   });
});

document.body.innerHTML += JSON.stringify(source, 0, 4 );

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