从这个最初的问题,我将如何在多个字段应用排序?

使用这种稍作调整的结构,我将如何排序城市(上升)和价格(下降)?

var homes = [
    {"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"}
    ];

我喜欢的事实是,给出的答案提供了一个一般的方法。在我计划使用这段代码的地方,我将不得不对日期以及其他东西进行排序。“启动”对象的能力似乎很方便,如果不是有点麻烦的话。

我试图把这个答案构建成一个很好的通用示例,但我运气不太好。


当前回答

function sortMultiFields(prop){
    return function(a,b){
        for(i=0;i<prop.length;i++)
        {
            var reg = /^\d+$/;
            var x=1;
            var field1=prop[i];
            if(prop[i].indexOf("-")==0)
            {
                field1=prop[i].substr(1,prop[i].length);
                x=-x;
            }

            if(reg.test(a[field1]))
            {
                a[field1]=parseFloat(a[field1]);
                b[field1]=parseFloat(b[field1]);
            }
            if( a[field1] > b[field1])
                return x;
            else if(a[field1] < b[field1])
                return -x;
        }
    }
}

如果你想按降序排序特定字段,如何使用(在字段前放-(减号)号)

homes.sort(sortMultiFields(["city","-price"]));

使用上面的函数,你可以对带有多个字段的json数组进行排序。根本不需要改变函数体

其他回答

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);
      }
   });

下面是一种可扩展的按多个字段排序的方法。

homes.sort(function(left, right) {
    var city_order = left.city.localeCompare(right.city);
    var price_order = parseInt(left.price) - parseInt(right.price);
    return city_order || -price_order;
});

笔记

A function passed to array sort is expected to return negative, zero, positive to indicate less, equal, greater. a.localeCompare(b) is universally supported for strings, and returns -1,0,1 if a<b,a==b,a>b. Subtraction works on numeric fields, because a - b gives -,0,+ if a<b,a==b,a>b. || in the last line gives city priority over price. Negate to reverse order in any field, as in -price_order Add new fields to the or-chain: return city_order || -price_order || date_order; Date compare with subtraction, because date math converts to milliseconds since 1970.var date_order = new Date(left.date) - new Date(right.date); Boolean compare with subtraction, which is guaranteed to turn true and false to 1 and 0 (therefore the subtraction produces -1 or 0 or 1). var goodness_order = Boolean(left.is_good) - Boolean(right.is_good)This is unusual enough that I'd suggest drawing attention with the Boolean constructor, even if they're already boolean.

按多个字段排序对象数组的最简单方法:

 let homes = [ {"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"}
  ];

homes.sort((a, b) => (a.city > b.city) ? 1 : -1);

输出: “Bevery山” “达拉斯” “达拉斯” “达拉斯” “纽约”

以下是我的简历,请参考,并举例说明:

function msort(arr, ...compFns) {
  let fn = compFns[0];
  arr = [].concat(arr);
  let arr1 = [];
  while (arr.length > 0) {
    let arr2 = arr.splice(0, 1);
    for (let i = arr.length; i > 0;) {
      if (fn(arr2[0], arr[--i]) === 0) {
        arr2 = arr2.concat(arr.splice(i, 1));
      }
    }
    arr1.push(arr2);
  }

  arr1.sort(function (a, b) {
    return fn(a[0], b[0]);
  });

  compFns = compFns.slice(1);
  let res = [];
  arr1.map(a1 => {
    if (compFns.length > 0) a1 = msort(a1, ...compFns);
    a1.map(a2 => res.push(a2));
  });
  return res;
}

let tstArr = [{ id: 1, sex: 'o' }, { id: 2, sex: 'm' }, { id: 3, sex: 'm' }, { id: 4, sex: 'f' }, { id: 5, sex: 'm' }, { id: 6, sex: 'o' }, { id: 7, sex: 'f' }];

function tstFn1(a, b) {
  if (a.sex > b.sex) return 1;
  else if (a.sex < b.sex) return -1;
  return 0;
}

function tstFn2(a, b) {
  if (a.id > b.id) return -1;
  else if (a.id < b.id) return 1;
  return 0;
}

console.log(JSON.stringify(msort(tstArr, tstFn1, tstFn2)));
//output:
//[{"id":7,"sex":"f"},{"id":4,"sex":"f"},{"id":5,"sex":"m"},{"id":3,"sex":"m"},{"id":2,"sex":"m"},{"id":6,"sex":"o"},{"id":1,"sex":"o"}]

这里有一个简单的泛型函数方法。使用数组指定排序顺序。前置减号以指定降序。

var homes = [
    {"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"}
    ];

homes.sort(fieldSorter(['city', '-price']));
// homes.sort(fieldSorter(['zip', '-state', 'price'])); // alternative

function fieldSorter(fields) {
    return function (a, b) {
        return fields
            .map(function (o) {
                var dir = 1;
                if (o[0] === '-') {
                   dir = -1;
                   o=o.substring(1);
                }
                if (a[o] > b[o]) return dir;
                if (a[o] < b[o]) return -(dir);
                return 0;
            })
            .reduce(function firstNonZeroValue (p,n) {
                return p ? p : n;
            }, 0);
    };
}

编辑:在ES6中它甚至更短!

"use strict"; const fieldSorter = (fields) => (a, b) => fields.map(o => { let dir = 1; if (o[0] === '-') { dir = -1; o=o.substring(1); } return a[o] > b[o] ? dir : a[o] < b[o] ? -(dir) : 0; }).reduce((p, n) => p ? p : n, 0); const homes = [{"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}]; const sortedHomes = homes.sort(fieldSorter(['state', '-price'])); document.write('<pre>' + JSON.stringify(sortedHomes, null, '\t') + '</pre>')