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

简单的一个:

var someArray = [...];

function generateSortFn(props) {
    return function (a, b) {
        for (var i = 0; i < props.length; i++) {
            var prop = props[i];
            var name = prop.name;
            var reverse = prop.reverse;
            if (a[name] < b[name])
                return reverse ? 1 : -1;
            if (a[name] > b[name])
                return reverse ? -1 : 1;
        }
        return 0;
    };
};

someArray.sort(generateSortFn([{name: 'prop1', reverse: true}, {name: 'prop2'}]));

其他回答

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>')

下面这个简单的解决方案怎么样:

const sortCompareByCityPrice = (a, b) => {
    let comparison = 0
    // sort by first criteria
    if (a.city > b.city) {
        comparison = 1
    }
    else if (a.city < b.city) {
        comparison = -1
    }
    // If still 0 then sort by second criteria descending
    if (comparison === 0) {
        if (parseInt(a.price) > parseInt(b.price)) {
            comparison = -1
        }
        else if (parseInt(a.price) < parseInt(b.price)) {
            comparison = 1
        }
    }
    return comparison 
}

基于这个问题javascript排序数组的多个(数字)字段

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));

这是一个通用的多维排序,允许在每个层次上进行反转和/或映射。

用Typescript编写。对于Javascript,请查看这个JSFiddle

的代码

type itemMap = (n: any) => any;

interface SortConfig<T> {
  key: keyof T;
  reverse?: boolean;
  map?: itemMap;
}

export function byObjectValues<T extends object>(keys: ((keyof T) | SortConfig<T>)[]): (a: T, b: T) => 0 | 1 | -1 {
  return function(a: T, b: T) {
    const firstKey: keyof T | SortConfig<T> = keys[0];
    const isSimple = typeof firstKey === 'string';
    const key: keyof T = isSimple ? (firstKey as keyof T) : (firstKey as SortConfig<T>).key;
    const reverse: boolean = isSimple ? false : !!(firstKey as SortConfig<T>).reverse;
    const map: itemMap | null = isSimple ? null : (firstKey as SortConfig<T>).map || null;

    const valA = map ? map(a[key]) : a[key];
    const valB = map ? map(b[key]) : b[key];
    if (valA === valB) {
      if (keys.length === 1) {
        return 0;
      }
      return byObjectValues<T>(keys.slice(1))(a, b);
    }
    if (reverse) {
      return valA > valB ? -1 : 1;
    }
    return valA > valB ? 1 : -1;
  };
}

用法示例

先按姓排序,再按名排序:

interface Person {
  firstName: string;
  lastName: string;
}

people.sort(byObjectValues<Person>(['lastName','firstName']));

按语言代码的名称排序,而不是按语言代码排序(见地图),然后按降序排序(见反向)。

interface Language {
  code: string;
  version: number;
}

// languageCodeToName(code) is defined elsewhere in code

languageCodes.sort(byObjectValues<Language>([
  {
    key: 'code',
    map(code:string) => languageCodeToName(code),
  },
  {
    key: 'version',
    reverse: true,
  }
]));

只是另一种选择。考虑使用以下效用函数:

/** Performs comparing of two items by specified properties
 * @param  {Array} props for sorting ['name'], ['value', 'city'], ['-date']
 * to set descending order on object property just add '-' at the begining of property
 */
export const compareBy = (...props) => (a, b) => {
  for (let i = 0; i < props.length; i++) {
    const ascValue = props[i].startsWith('-') ? -1 : 1;
    const prop = props[i].startsWith('-') ? props[i].substr(1) : props[i];
    if (a[prop] !== b[prop]) {
      return a[prop] > b[prop] ? ascValue : -ascValue;
    }
  }
  return 0;
};

用法示例(在您的情况下):

homes.sort(compareBy('city', '-price'));

值得注意的是,这个函数可以更一般化,以便能够使用嵌套属性,如'address '。City '或'style.size。宽度”等。