下面是我基于施瓦兹变换的解决方案，希望你觉得有用。

function sortByAttribute(array, ...attrs) {
  // generate an array of predicate-objects contains
  // property getter, and descending indicator
  let predicates = attrs.map(pred => {
    let descending = pred.charAt(0) === '-' ? -1 : 1;
    pred = pred.replace(/^-/, '');
    return {
      getter: o => o[pred],
      descend: descending
    };
  });
  // schwartzian transform idiom implementation. aka: "decorate-sort-undecorate"
  return array.map(item => {
    return {
      src: item,
      compareValues: predicates.map(predicate => predicate.getter(item))
    };
  })
  .sort((o1, o2) => {
    let i = -1, result = 0;
    while (++i < predicates.length) {
      if (o1.compareValues[i] < o2.compareValues[i]) result = -1;
      if (o1.compareValues[i] > o2.compareValues[i]) result = 1;
      if (result *= predicates[i].descend) break;
    }
    return result;
  })
  .map(item => item.src);
}

下面是一个如何使用它的例子:

let games = [
  { name: 'Pako',              rating: 4.21 },
  { name: 'Hill Climb Racing', rating: 3.88 },
  { name: 'Angry Birds Space', rating: 3.88 },
  { name: 'Badland',           rating: 4.33 }
];

// sort by one attribute
console.log(sortByAttribute(games, 'name'));
// sort by mupltiple attributes
console.log(sortByAttribute(games, '-rating', 'name'));

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

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>')

function sort(data, orderBy) {
        orderBy = Array.isArray(orderBy) ? orderBy : [orderBy];
        return data.sort((a, b) => {
            for (let i = 0, size = orderBy.length; i < size; i++) {
                const key = Object.keys(orderBy[i])[0],
                    o = orderBy[i][key],
                    valueA = a[key],
                    valueB = b[key];
                if (!(valueA || valueB)) {
                    console.error("the objects from the data passed does not have the key '" + key + "' passed on sort!");
                    return [];
                }
                if (+valueA === +valueA) {
                    return o.toLowerCase() === 'desc' ? valueB - valueA : valueA - valueB;
                } else {
                    if (valueA.localeCompare(valueB) > 0) {
                        return o.toLowerCase() === 'desc' ? -1 : 1;
                    } else if (valueA.localeCompare(valueB) < 0) {
                        return o.toLowerCase() === 'desc' ? 1 : -1;
                    }
                }
            }
        });
    }

使用:

sort(homes, [{city : 'asc'}, {price: 'desc'}])

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"} ]; function sort(data, orderBy) { orderBy = Array.isArray(orderBy) ? orderBy : [orderBy]; return data.sort((a, b) => { for (let i = 0, size = orderBy.length; i < size; i++) { const key = Object.keys(orderBy[i])[0], o = orderBy[i][key], valueA = a[key], valueB = b[key]; if (!(valueA || valueB)) { console.error("the objects from the data passed does not have the key '" + key + "' passed on sort!"); return []; } if (+valueA === +valueA) { return o.toLowerCase() === 'desc' ? valueB - valueA : valueA - valueB; } else { if (valueA.localeCompare(valueB) > 0) { return o.toLowerCase() === 'desc' ? -1 : 1; } else if (valueA.localeCompare(valueB) < 0) { return o.toLowerCase() === 'desc' ? 1 : -1; } } } }); } console.log(sort(homes, [{city : 'asc'}, {price: 'desc'}]));

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 );

以下是@ snowburn的解决方案的通用版本:

var sortarray = [{field:'city', direction:'asc'}, {field:'price', direction:'desc'}];
array.sort(function(a,b){
    for(var i=0; i<sortarray.length; i++){
        retval = a[sortarray[i].field] < b[sortarray[i].field] ? -1 : a[sortarray[i].field] > b[sortarray[i].field] ? 1 : 0;
        if (sortarray[i].direction == "desc") {
            retval = retval * -1;
        }
        if (retval !== 0) {
            return retval;
        }
    }
}


})

这是基于我正在使用的排序例程。我没有测试这个特定的代码，所以它可能有错误，但你知道的。其思想是基于第一个表示差异的字段进行排序，然后停止并转到下一个记录。因此，如果您按三个字段排序，并且compare中的第一个字段足以确定正在排序的两个记录的排序顺序，那么将返回该排序结果并转到下一个记录。

我在5000条记录上测试了它(实际上使用了更复杂的排序逻辑)，它在眨眼之间就做到了。如果实际向客户端加载超过1000条记录，则可能应该使用服务器端排序和过滤。

这段代码不区分大小写，我把它留给读者来处理这个微不足道的修改。