我有一个JavaScript对象数组:

var objs = [ 
    { first_nom: 'Lazslo', last_nom: 'Jamf'     },
    { first_nom: 'Pig',    last_nom: 'Bodine'   },
    { first_nom: 'Pirate', last_nom: 'Prentice' }
];

如何在JavaScript中按last_nom的值对它们进行排序?

我知道排序(a,b),但这似乎只适用于字符串和数字。是否需要向对象添加toString()方法?


当前回答

我一直在各种项目中使用这个实用程序,而且效果很好。它也非常模块化:

传递要排序的键的名称选择排序是升序还是降序

按KeyUtil.js排序对象数组

// Sort array of objects by key
// ------------------------------------------------------------
const sortArrayOfObjsByKey = (array, key, ascdesc) =>
  array.sort((a, b) => {
    const x = a[key];
    const y = b[key];
    if (ascdesc === 'asc') {
      return x < y ? -1 : x > y ? 1 : 0;
    }
    if (ascdesc === 'desc') {
      return x > y ? -1 : x < y ? 1 : 0;
    }
    return null;
  });

按KeyUtil.test.js排序对象数组

import sortArrayOfObjsByKey from './sortArrayOfObjsByKeyUtil';

const unsortedArray = [
  {
    _id: '3df55221-ce5c-4147-8e14-32effede6133',
    title: 'Netlife Design',
    address: {
      PostalAddress: {
        streetAddress: 'Youngstorget 3',
        addressLocality: 'Oslo',
        addressRegion: null,
        postalCode: '0181',
        addressCountry: 'Norway',
      },
    },
    geopoint: { lat: 59.914322, lng: 10.749272 },
  },
  {
    _id: 'cd00459f-3755-49f1-8847-66591ef935b2',
    title: 'Home',
    address: {
      PostalAddress: {
        streetAddress: 'Stockfleths gate 58A',
        addressLocality: 'Oslo',
        addressRegion: null,
        postalCode: '0461',
        addressCountry: 'Norway',
      },
    },
    geopoint: { lat: 59.937316, lng: 10.751862 },
  },
];

const sortedArray = [
  {
    _id: 'cd00459f-3755-49f1-8847-66591ef935b2',
    title: 'Home',
    address: {
      PostalAddress: {
        streetAddress: 'Stockfleths gate 58A',
        addressLocality: 'Oslo',
        addressRegion: null,
        postalCode: '0461',
        addressCountry: 'Norway',
      },
    },
    geopoint: { lat: 59.937316, lng: 10.751862 },
  },
  {
    _id: '3df55221-ce5c-4147-8e14-32effede6133',
    title: 'Netlife Design',
    address: {
      PostalAddress: {
        streetAddress: 'Youngstorget 3',
        addressLocality: 'Oslo',
        addressRegion: null,
        postalCode: '0181',
        addressCountry: 'Norway',
      },
    },
    geopoint: { lat: 59.914322, lng: 10.749272 },
  },
];

describe('sortArrayOfObjsByKey', () => {
  it(`sort array by 'title' key, ascending`, () => {
    const testInput = sortArrayOfObjsByKey(unsortedArray, 'title', 'asc');
    const testOutput = sortedArray;
    expect(testInput).toEqual(testOutput);
  });
});

其他回答

let propName = 'last_nom';

let sorted_obj = objs.sort((a,b) => {
    if(a[propName] > b[propName]) {
        return 1;
    }
    if (a[propName] < b[propName]) {
        return -1;
    }
    return 0;
}

//This works because the js built-in sort function allows us to define our
//own way of sorting, this funny looking function is simply telling `sort` how to
//determine what is larger. 
//We can use `if(a[propName] > b[propName])` because string comparison is already built into JS
//if you try console.log('a' > 'z' ? 'a' : 'z')
//the output will be 'z' as 'a' is not greater than 'z'
//The return values 0,-1,1 are how we tell JS what to sort on. We're sorting on the last_nom property of the object. 
//When sorting a list it comes down to comparing two items and how to determine which one of them is "larger". 
//We need a way to tell JS how to determine which one is larger. 
//The sort defining function will use the case that returns a 1 to mean that a > b
//and the case that returns -1 to mean that a < b

这是我的看法:

order参数是可选的,对于升序,默认为“ASC”。

它适用于重音字符,并且不区分大小写。

注意:它排序并返回原始数组。

function sanitizeToSort(str) {
  return str
    .normalize('NFD')                   // Remove accented and diacritics
    .replace(/[\u0300-\u036f]/g, '')    // Remove accented and diacritics
    .toLowerCase()                      // Sort will be case insensitive
  ;
}

function sortByProperty(arr, property, order="ASC") {
  arr.forEach((item) => item.tempProp = sanitizeToSort(item[property]));
  arr.sort((a, b) => order === "ASC" ?
      a.tempProp > b.tempProp ?  1 : a.tempProp < b.tempProp ? -1 : 0
    : a.tempProp > b.tempProp ? -1 : a.tempProp < b.tempProp ?  1 : 0
  );
  arr.forEach((item) => delete item.tempProp);
  return arr;
}

一小条

函数cleaniteToSort(str){返回str.normalize('NFD')//删除重音字符.replace(/[\u0300-\u036f]/g,“”)//删除变音符号.to小写();}函数sortByProperty(arr,property,order=“ASC”){arr.forEach((item)=>item.tempProp=消毒排序(item[property]));arr.sort((a,b)=>顺序==“ASC”?a.tempProp>b.tempProp?1:a.tempProp<b.tempProp-1 : 0:a.tempProp>b.tempProp-1:a.tempProp<b.tempProp?1 : 0);arr.forEach((item)=>删除item.tempProp);返回arr;}常量rockStars=[{name:“Axl”,姓:“Rose”},{name:“埃尔顿”,姓:“John”},{name:“Paul”,姓氏:“McCartney”},{name:“楼”,姓:“里德”},{name:“freddie”,//使用小写/大写姓氏:“mercury”},{name:“Ámy”,//也适用于重音字符姓氏:“酒庄”}];sortByProperty(rockStars,“name”);console.log(“按名称A-Z排序:”);rockStars.forEach((item)=>console.log(item.name+“”+item.lastname));sortByProperty(rockStars,“姓氏”,“DESC”);console.log(“\n按姓氏Z-A排序:”);rockStars.forEach((item)=>console.log(item.lastname+“,”+item.name));

Lodash(Undercore.js的超集)。

不为每一个简单的逻辑添加一个框架是很好的,但是依赖于经过良好测试的实用程序框架可以加快开发并减少错误数量。

Lodash生成了非常干净的代码,并促进了更具功能性的编程风格。一眼望去,代码的意图就一目了然了。

OP的问题可以简单地解决为:

const sortedObjs = _.sortBy(objs, 'last_nom');

更多信息?例如,我们有以下嵌套对象:

const users = [
  { 'user': {'name':'fred', 'age': 48}},
  { 'user': {'name':'barney', 'age': 36 }},
  { 'user': {'name':'wilma'}},
  { 'user': {'name':'betty', 'age': 32}}
];

我们现在可以使用_.properties速记user.age来指定应该匹配的属性的路径。我们将根据嵌套的年龄属性对用户对象进行排序。是的,它允许嵌套属性匹配!

const sortedObjs = _.sortBy(users, ['user.age']);

想要反转吗?没问题。使用_反向。

const sortedObjs = _.reverse(_.sortBy(users, ['user.age']));

想用链条将两者结合起来吗?

const { chain } = require('lodash');
const sortedObjs = chain(users).sortBy('user.age').reverse().value();

或者你什么时候更喜欢流动而不是链条?

const { flow, reverse, sortBy } = require('lodash/fp');
const sortedObjs = flow([sortBy('user.age'), reverse])(users);

在ES6/ES2015或更高版本中,您可以这样做:

objs.sort((a, b) => a.last_nom.localeCompare(b.last_nom));

ES6/ES2015之前

objs.sort(function(a, b) {
    return a.last_nom.localeCompare(b.last_nom)
});

给定原始示例:

var objs = [ 
    { first_nom: 'Lazslo', last_nom: 'Jamf'     },
    { first_nom: 'Pig',    last_nom: 'Bodine'   },
    { first_nom: 'Pirate', last_nom: 'Prentice' }
];

按多个字段排序:

objs.sort(function(left, right) {
    var last_nom_order = left.last_nom.localeCompare(right.last_nom);
    var first_nom_order = left.first_nom.localeCompare(right.first_nom);
    return last_nom_order || first_nom_order;
});

笔记

a.localeCompare(b)是普遍支持的,如果a<b,a==b,a>b,则分别返回-1,0,1。||在最后一行中,last_nom优先于first_nom。减法适用于数字字段:var age_order=left.age-right.age;取反顺序,返回-last_nom_order | |-first_nom_oorder | |-age_order;