为了简化操作，可以使用这些辅助函数。

您可以根据需要对任意多个字段进行排序。对于每个排序字段，指定属性名，然后可选地指定-1作为排序方向，以降序排序而不是升序排序。

const data = [ {"h_id":"3","city":"Dallas","state":"TX","zip":"75201","price":"162500"}, {"h_id":"4","city":"Bevery Hills","state":"CA","zip":"90210","price":"319250"}, {"h_id":"6","city":"Dallas","state":"TX","zip":"75000","price":"556699"}, {"h_id":"5","city":"New York","state":"NY","zip":"00010","price":"962500"}, {"h_id":"7","city":"New York","state":"NY","zip":"00010","price":"800500"} ] const sortLexically = (p,d=1)=>(a,b)=>d * a[p].localeCompare(b[p]) const sortNumerically = (p,d=1)=>(a,b)=>d * (a[p]-b[p]) const sortBy = sorts=>(a,b)=>sorts.reduce((r,s)=>r||s(a,b),0) // sort first by city, then by price descending data.sort(sortBy([sortLexically('city'), sortNumerically('price', -1)])) console.log(data)

// custom sorting by city
const sortArray = ['Dallas', 'New York', 'Beverly Hills'];

const sortData = (sortBy) =>
  data
    .sort((a, b) => {
      const aIndex = sortBy.indexOf(a.city);
      const bIndex = sortBy.indexOf(b.city);

      if (aIndex < bIndex) {
        return -1;
      }

      if (aIndex === bIndex) {
        // price descending
        return b.price- a.price;
      }

      return 1;
    });

sortData(sortArray);

这是一个递归算法，按多个字段排序，同时有机会在比较之前格式化值。

var data = [
{
    "id": 1,
    "ship": null,
    "product": "Orange",
    "quantity": 7,
    "price": 92.08,
    "discount": 0
},
{
    "id": 2,
    "ship": "2017-06-14T23:00:00.000Z".toDate(),
    "product": "Apple",
    "quantity": 22,
    "price": 184.16,
    "discount": 0
},
...
]
var sorts = ["product", "quantity", "ship"]

// comp_val formats values and protects against comparing nulls/undefines
// type() just returns the variable constructor
// String.lower just converts the string to lowercase.
// String.toDate custom fn to convert strings to Date
function comp_val(value){
    if (value==null || value==undefined) return null
    var cls = type(value)
    switch (cls){
        case String:
            return value.lower()
    }
    return value
}

function compare(a, b, i){
    i = i || 0
    var prop = sorts[i]
    var va = comp_val(a[prop])
    var vb = comp_val(b[prop])

    // handle what to do when both or any values are null
    if (va == null || vb == null) return true

    if ((i < sorts.length-1) && (va == vb)) {
        return compare(a, b, i+1)
    } 
    return va > vb
}

var d = data.sort(compare);
console.log(d);

如果a和b相等，它将尝试下一个字段，直到没有可用字段。

我喜欢snowburn的方法，但它需要调整来测试城市的等效性，而不是差异。

homes.sort(
   function(a,b){
      if (a.city==b.city){
         return (b.price-a.price);
      } else {
         return (a.city-b.city);
      }
   });

对于你的具体问题，一个非通用的，简单的解决方案:

homes.sort(
   function(a, b) {          
      if (a.city === b.city) {
         // Price is only important when cities are the same
         return b.price - a.price;
      }
      return a.city > b.city ? 1 : -1;
   });

为什么复杂化?只需要整理两次!这是完美的: (只要确保将重要性顺序从低到高颠倒过来就行了):

jj.sort( (a, b) => (a.id >= b.id) ? 1 : -1 );
jj.sort( (a, b) => (a.status >= b.status) ? 1 : -1 );