Question

使用Javascripts .sort（），我有一个具有＆＃39;类别＆＃39;的对象列表。财产和子类别＆＃39;属性。到目前为止，我使用item1.category().localeCompare(item2.category())按类别排序列表，但在每个类别组中，我希望按子类别对对象进行排序。

到目前为止，我的阵列分类为：

[
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat2'}
]

我需要它像这样：

[
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat2'}
]

此外，我只能使用一种（）并且不能做任何事情，因为它作为参数传递给另一个函数。

Answer 1

只需在sort函数中添加一个额外的谓词：

const arr = [ 
  { "field1": "a2", "field2": "b2" },  
  { "field1": "a1", "field2": "b2" },
  { "field1": "a1", "field2": "b1" }
];

arr.sort((a,b) => {
  return a.field1.localeCompare(b.field1) || a.field2.localeCompare(b.field2);
});
// [ {"field1":"a1","field2":"b1"},
//   {"field1":"a1","field2":"b2"},
//   {"field1":"a2","field2":"b2"}]

Answer 2

您可以在子类别上添加另一个localeCompare，例如：

arr.sort(function (a,b) {
    return a.category.localeCompare(b.category)
        || a.subcategory.localeCompare(b.subcategory);
})

Answer 3

只需扩展您的排序功能。

 categories.sort(function(a, b) {
   if(a.category < b.category){
     return -1;  
   }else if(a.category > b.category){
     return 1;
   }else{
     if(a.subcategory < b.subcategory){
       return -1
     }else if(a.subcategory > b.subcategory){
       return 1;
     }
   }
 });

请看这里的工作小提琴：http://jsfiddle.net/Sergey_Mell/htv6zs8g/

Answer 4

试试这个：

var values = [
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat1', 'subcategory':'subcat1'},
  {'category':'cat1', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat2'},
  {'category':'cat2', 'subcategory':'subcat1'},
  {'category':'cat2', 'subcategory':'subcat2'}
];
console.log(values.sort(function (a, b) {
  var tmp_a = a.category + a.subcategory;
  var tmp_b = b.category + b.subcategory;

  if (tmp_a < tmp_b) {
    return -1;
  }
  if (tmp_a > tmp_b) {
    return 1;
  }
  // a must be equal to b
  return 0;
}));

Answer 5

这对我有用：

arr.sort(function(a, b) {
    if (a.category != b.category) {
        return (a.category > b.category) ? 1 : -1;
    } else {
        return (a.subcategory > b.subcategory) ? 1 : -1;
    }
    return 0;
});

检查 category 是否不同。如果它不同，那么就按它排序。否则（如果类别是相同的），它按子类别排序。

在数组中，如何使用.sort（）将基于一个属性的对象组合在一起？

5 个答案: