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

通过添加两个辅助函数,可以简单地解决这类问题。sortByKey接受一个数组和一个函数,该函数应返回一个项目列表,用于与每个数组条目进行比较。

这利用了javascript对简单值的数组进行智能比较的事实,即[2]<[2,0]<[2,1]<[10,0]。

// Two helpers: function cmp(a, b) { if (a > b) { return 1 } else if (a < b) { return -1 } else { return 0 } } function sortByKey(arr, key) { arr.sort((a, b) => cmp(key(a), key(b))) } // A demonstration: let arr = [{a:1, b:2}, {b:3, a:0}, {a:1, b:1}, {a:2, b:2}, {a:2, b:1}, {a:1, b:10}] sortByKey(arr, item => [item.a, item.b]) console.log(JSON.stringify(arr)) // '[{"b":3,"a":0},{"a":1,"b":1},{"a":1,"b":10},{"a":1,"b":2},{"a":2,"b":1},{"a":2,"b":2}]' sortByKey(arr, item => [item.b, item.a]) console.log(JSON.stringify(arr)) // '[{"a":1,"b":1},{"a":2,"b":1},{"a":1,"b":10},{"a":1,"b":2},{"a":2,"b":2},{"b":3,"a":0}]'

我从Python的列表中偷取了这个想法。排序功能。

其他回答

另一种方式

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 sortBy(ar) { return ar.sort((a, b) => a.city === b.city ? b.price.toString().localeCompare(a.price) : a.city.toString().localeCompare(b.city)); } console.log(sortBy(homes));

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)

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'}]));

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>')

我认为这可能是最简单的方法。

https://coderwall.com/p/ebqhca/javascript-sort-by-two-fields

这真的很简单,我尝试了3个不同的键值对,它工作得很好。

这是一个简单的例子,查看链接了解更多细节

testSort(data) {
    return data.sort(
        a['nameOne'] > b['nameOne'] ? 1
        : b['nameOne'] > a['nameOne'] ? -1 : 0 ||
        a['date'] > b['date'] ||
        a['number'] - b['number']
    );
}