按字符串属性值对对象数组进行排序

Question

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' }
];

如何根据last_nomJavaScript中的值对它们进行排序？

我知道sort(a,b),但这似乎只适用于字符串和数字.我是否需要为toString()对象添加方法？

Answer 1

编写自己的比较函数很容易:

function compare( a, b ) {
  if ( a.last_nom < b.last_nom ){
    return -1;
  }
  if ( a.last_nom > b.last_nom ){
    return 1;
  }
  return 0;
}

objs.sort( compare );

或内联(c/o Marco Demaio):

objs.sort((a,b) => (a.last_nom > b.last_nom) ? 1 : ((b.last_nom > a.last_nom) ? -1 : 0)); 

Answer 2

您还可以创建一个动态排序函数,按照您传递的值对对象进行排序:

function dynamicSort(property) {
    var sortOrder = 1;
    if(property[0] === "-") {
        sortOrder = -1;
        property = property.substr(1);
    }
    return function (a,b) {
        /* next line works with strings and numbers, 
         * and you may want to customize it to your needs
         */
        var result = (a[property] < b[property]) ? -1 : (a[property] > b[property]) ? 1 : 0;
        return result * sortOrder;
    }
}

所以你可以有一个这样的对象数组:

var People = [
    {Name: "Name", Surname: "Surname"},
    {Name:"AAA", Surname:"ZZZ"},
    {Name: "Name", Surname: "AAA"}
];

......当你这样做时它会起作用:

People.sort(dynamicSort("Name"));
People.sort(dynamicSort("Surname"));
People.sort(dynamicSort("-Surname"));

实际上这已经回答了这个问题.下面的部分是因为许多人联系我,抱怨它不适用于多个参数.

多个参数

您可以使用以下函数生成具有多个排序参数的排序函数.

function dynamicSortMultiple() {
    /*
     * save the arguments object as it will be overwritten
     * note that arguments object is an array-like object
     * consisting of the names of the properties to sort by
     */
    var props = arguments;
    return function (obj1, obj2) {
        var i = 0, result = 0, numberOfProperties = props.length;
        /* try getting a different result from 0 (equal)
         * as long as we have extra properties to compare
         */
        while(result === 0 && i < numberOfProperties) {
            result = dynamicSort(props[i])(obj1, obj2);
            i++;
        }
        return result;
    }
}

这将使您能够做到这样的事情:

People.sort(dynamicSortMultiple("Name", "-Surname"));

将它添加到原型

(下面的实现灵感来自Mike R的回答)

我不建议更改本机对象原型,只是为了给出一个示例,以便您可以在自己的对象上实现它(对于支持它的环境,您也可以使用Object.defineProperty,如下一节所示,至少没有可枚举的负面副作用,如最后一部分所述)

原型实现将类似于以下(这是一个工作示例):

class MyArray extends Array {
    sortBy(...args) {
        return this.sort(dynamicSortMultiple.apply(null, args));
    }
}

将其添加到原型的"OK"方式

如果你的目标是IE v9.0,那么就像我之前提到的那样,使用像这样的Object.defineProperty(工作示例):

MyArray.from(People).sortBy("Name", "-Surname");

在绑定运算符到达之前,这可以是可接受的折衷方案.

所有这些原型乐趣都可以实现:

function dynamicSort(property) {
    var sortOrder = 1;
    if(property[0] === "-") {
        sortOrder = -1;
        property = property.substr(1);
    }
    return function (a,b) {
        /* next line works with strings and numbers, 
         * and you may want to customize it to your needs
         */
        var result = (a[property] < b[property]) ? -1 : (a[property] > b[property]) ? 1 : 0;
        return result * sortOrder;
    }
}

你应该读这个

如果你使用直接原型访问方法(Object.defineProperty很好)而其他代码没有检查hasOwnProperty,小猫会死!好吧,老实说,任何小猫都没有受到伤害,但可能事情会破裂,团队中的其他开发人员都会讨厌你:

看到最后一个"SortBy"？是啊.不酷.尽可能使用Object.defineProperty,否则单独保留Array.prototype.

Answer 3

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));

Answer 4

underscore.js

使用下划线,它小而且很棒......

sortBy_.sortBy(list,iterator,[context])返回列表的排序副本,按照通过迭代器运行每个值的结果按升序排列.迭代器也可以是要排序的属性的字符串名称(例如,长度).

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

var sortedObjs = _.sortBy( objs, 'first_nom' );

Answer 5

不要理解为什么人们会这么复杂:

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

对于更严格的引擎:

objs.sort(function(a, b){
  return a.last_nom == b.last_nom ? 0 : +(a.last_nom > b.last_nom) || -1;
});

交换运算符以按反向字母顺序排序.

Answer 6

如果你有重复的姓氏,你可以按名字排序 -

obj.sort(function(a,b){
  if(a.last_nom< b.last_nom) return -1;
  if(a.last_nom >b.last_nom) return 1;
  if(a.first_nom< b.first_nom) return -1;
  if(a.first_nom >b.first_nom) return 1;
  return 0;
});

Answer 7

使用原型继承简单快速地解决此问题:

Array.prototype.sortBy = function(p) {
  return this.slice(0).sort(function(a,b) {
    return (a[p] > b[p]) ? 1 : (a[p] < b[p]) ? -1 : 0;
  });
}

示例/用法

objs = [{age:44,name:'vinay'},{age:24,name:'deepak'},{age:74,name:'suresh'}];

objs.sortBy('age');
// Returns
// [{"age":24,"name":"deepak"},{"age":44,"name":"vinay"},{"age":74,"name":"suresh"}]

objs.sortBy('name');
// Returns
// [{"age":24,"name":"deepak"},{"age":74,"name":"suresh"},{"age":44,"name":"vinay"}]

更新:不再修改原始数组.

Answer 8

截至2018年,有一个更短,更优雅的解决方案.只是用.Array.prototype.sort().

例:

var items = [
  { name: 'Edward', value: 21 },
  { name: 'Sharpe', value: 37 },
  { name: 'And', value: 45 },
  { name: 'The', value: -12 },
  { name: 'Magnetic', value: 13 },
  { name: 'Zeros', value: 37 }
];

// sort by value
items.sort(function (a, b) {
  return a.value - b.value;
});

Answer 9

使用JavaScriptsort方法

可以修改该sort方法以使用比较函数对数字数组、字符串甚至对象等任何内容进行排序。

比较函数作为可选参数传递给排序方法。

该比较函数接受 2 个参数，通常称为a和b。基于这两个参数，您可以修改排序方法以按您想要的方式工作。

1. 如果比较函数返回小于 0，则该sort()方法将a排序为比b更低的索引。简单地说，a 将出现在 b 之前。
2. 如果比较函数返回等于 0，则该sort()方法将元素位置保持原样。
3. 如果比较函数返回大于 0，则该sort()方法按比b更大的索引对a进行排序。只是a会出现在b之后。

使用上述概念应用于您的对象，其中a将是您的对象属性。

var objs = [
  { first_nom: 'Lazslo', last_nom: 'Jamf' },
  { first_nom: 'Pig', last_nom: 'Bodine' },
  { first_nom: 'Pirate', last_nom: 'Prentice' }
];
function compare(a, b) {
  if (a.last_nom > b.last_nom) return 1;
  if (a.last_nom < b.last_nom) return -1;
  return 0;
}
objs.sort(compare);
console.log(objs)
// for better look use console.table(objs)

Answer 10

您也可以使用自定义toString()方法(由默认比较函数调用)创建对象类型,而不是使用自定义比较函数:

function Person(firstName, lastName) {
    this.firtName = firstName;
    this.lastName = lastName;
}

Person.prototype.toString = function() {
    return this.lastName + ', ' + this.firstName;
}

var persons = [ new Person('Lazslo', 'Jamf'), ...]
persons.sort();

Answer 11

您可以使用

最简单的方法:Lodash

(https://lodash.com/docs/4.17.10#orderBy)

此方法类似于_.sortBy,但它允许指定迭代的排序顺序以进行排序.如果未指定订单,则所有值都按升序排序.否则,为降序指定"desc"的顺序,或为相应值的升序排序指定"asc".

参数

collection(Array | Object):要迭代的集合.[iteratees = [_.identity]](Array [] | Function [] | Object [] | string []):迭代次数排序依据.[orders](string []):迭代的排序顺序.

返回

(Array):返回新排序的数组.

var _ = require('lodash');
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"}
    ];

_.orderBy(homes, ['city', 'state', 'zip'], ['asc', 'desc', 'asc']);

Answer 12

尝试这个，

UPTO ES5

//Ascending Sort
items.sort(function (a, b) {
   return a.value - b.value;
});


//Descending Sort
items.sort(function (a, b) {
   return b.value - a.value;
});


IN ES6 & above:

// Ascending sort
items.sort((a, b) => a.value - b.value);

// Descending Sort
 items.sort((a, b) => b.value - a.value);

Answer 13

Lodash.js(超集Underscore.js)

最好不要为每一个简单的逻辑片段添加一个框架,但依靠经过良好测试的实用程序框架,加快开发速度并减少写入的错误数量并不是一件容易的事.

Lodash生成非常干净的代码并提升了更多功能的编程风格,从而减少了错误.在一瞥中,很明显代码的意图是什么.

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_.property简写user.age来指定应匹配的属性的路径.我们将通过嵌套的age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']);

想要扭转吗？没问题.使用_.reverse.

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

想要使用Chaining来结合两者吗？

const sortedObjs = _.chain(users).sortBy('user.age').reverse().value();

Answer 14

这里有很多好的答案,但我想指出,它们可以非常简单地扩展,以实现更复杂的排序.您唯一要做的就是使用OR运算符链接比较函数,如下所示:

objs.sort((a,b)=> fn1(a,b) || fn2(a,b) || fn3(a,b) )

其中fn1,, fn2...是返回[-1,0,1]的排序函数.这导致"按fn1排序","按fn2排序",这几乎等于SQL中的ORDER BY.

此解决方案基于||运算符的行为,该行为计算为可转换为true的第一个计算表达式.

最简单的形式只有一个内联函数,如下所示:

// ORDER BY last_nom
objs.sort((a,b)=> a.last_nom.localeCompare(b.last_nom) )

有两个步骤 last_nom,first_nom排序顺序将如下所示:

// ORDER_BY last_nom, first_nom
objs.sort((a,b)=> a.last_nom.localeCompare(b.last_nom) || 
                  a.first_nom.localeCompare(b.first_nom)  )

通用的比较函数可能是这样的:

// ORDER BY <n>
let cmp = (a,b,n)=>a[n].localeCompare(b[n])

此函数可以扩展为支持数字字段,大小写敏感,任意数据类型等.

您可以通过排序优先级链接它们来使用它:

// ORDER_BY last_nom, first_nom
objs.sort((a,b)=> cmp(a,b, "last_nom") || cmp(a,b, "first_nom") )
// ORDER_BY last_nom, first_nom DESC
objs.sort((a,b)=> cmp(a,b, "last_nom") || -cmp(a,b, "first_nom") )
// ORDER_BY last_nom DESC, first_nom DESC
objs.sort((a,b)=> -cmp(a,b, "last_nom") || -cmp(a,b, "first_nom") )

这里的要点是具有功能方法的纯JavaScript可以在没有外部库或复杂代码的情况下走很长的路.它也非常有效,因为不必进行字符串解析

Answer 15

用法示例:

objs.sort(sortBy('last_nom'));

脚本:

/**
 * @description
 * Returns a function which will sort an
 * array of objects by the given key.
 *
 * @param  {String}  key
 * @param  {Boolean} reverse
 * @return {Function}
 */
const sortBy = (key, reverse) => {

  // Move smaller items towards the front
  // or back of the array depending on if
  // we want to sort the array in reverse
  // order or not.
  const moveSmaller = reverse ? 1 : -1;

  // Move larger items towards the front
  // or back of the array depending on if
  // we want to sort the array in reverse
  // order or not.
  const moveLarger = reverse ? -1 : 1;

  /**
   * @param  {*} a
   * @param  {*} b
   * @return {Number}
   */
  return (a, b) => {
    if (a[key] < b[key]) {
      return moveSmaller;
    }
    if (a[key] > b[key]) {
      return moveLarger;
    }
    return 0;
  };
};

Answer 16

我没有看到这种特殊方法的建议,所以这里是一个简洁的比较方法我喜欢使用它适用于string和number:

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

const sortBy = fn => (a, b) => {
  const fa = fn(a)
  const fb = fn(b)
  return -(fa < fb) || +(fa > fb)
}
const getLastName = o => o.last_nom
const sortByLastName = sortBy(getLastName)

objs.sort(sortByLastName)
console.log(objs.map(getLastName))

以下是对以下内容的解释sortBy():

sortBy()接受一个fn从对象中选择要用作比较的值,并返回一个可以直接传递给的函数Array.prototype.sort().在这个例子中,我们将使用o.last_nom作为比较的值,因此每当我们通过Array.prototype.sort()诸如此类接收两个对象时

{ first_nom: 'Lazslo', last_nom: 'Jamf' }

和

{ first_nom: 'Pig', last_nom: 'Bodine' }

我们用

(a, b) => {
  const fa = fn(a)
  const fb = fn(b)
  return -(fa < fb) || +(fa > fb)
}

比较它们.

记住这一点fn = o => o.last_nom,我们可以将比较功能扩展到等效功能

(a, b) => {
  const fa = a.last_nom
  const fb = b.last_nom
  return -(fa < fb) || +(fa > fb)
}

逻辑OR ||运算符具有短路功能,这在此非常有用.由于它的工作原理,上述功能的主体意味着

if (fa < fb) return -1
return +(fa > fb)

所以,如果fa < fb我们返回-1,否则如果fa > fb那时我们返回+1,但如果fa == fb,那么fa < fb并且fa > fb是假的,所以它返回false.

作为一个额外的奖励,这里是没有箭头功能的ECMAScript 5.1中的等价物,遗憾的是它并不那么简洁:

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

var sortBy = function (fn) {
  return function (a, b) {
    var fa = fn(a)
    var fb = fn(b)
    return -(fa < fb) || +(fa > fb)
  }
}

var getLastName = function (o) { return o.last_nom }
var sortByLastName = sortBy(getLastName)

objs.sort(sortByLastName)
console.log(objs.map(getLastName))

Answer 17

我有一段适合我的代码:

arr.sort((a, b) => a.name > b.name)

更新:不工作总是,所以它是不正确的:(

Answer 18

区分大小写

arr.sort((a, b) => a.name > b.name ? 1 : -1);

不区分大小写

arr.sort((a, b) => a.name.toLowerCase() > b.name.toLowerCase() ? 1 : -1);

笔记

如果顺序没有变化（在相同字符串的情况下），则条件>将失败并-1返回。但是如果字符串相同则返回 1 或 -1 将导致正确的输出

另一种选择是使用>=运算符而不是>

Answer 19

当您需要自然字符串排序（即）Intl.Collator时，针对特定情况对对象进行排序。["1","2","10","11","111"]

const files = [
 {name: "1.mp3", size: 123},
 {name: "10.mp3", size: 456},
 {name: "100.mp3", size: 789},
 {name: "11.mp3", size: 123},
 {name: "111.mp3", size: 456},
 {name: "2.mp3", size: 789},
];

const naturalCollator = new Intl.Collator(undefined, {numeric: true, sensitivity: 'base'});

files.sort((a, b) => naturalCollator.compare(a.name, b.name));

console.log(files);

注意： 的undefined构造函数参数Intl.Collator代表区域设置，可以是显式的 ISO 639-1 语言代码，例如en，或者是系统默认区域设置undefined。

浏览器支持Intl.Collator

Answer 20

我知道这个问题太旧了,但我没有看到任何类似于我的实现.
此版本基于Schwartzian变换习语.

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);
}

以下是如何使用它的示例:

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'));

Answer 21

排序(更多)复杂的对象数组

由于您可能会遇到更复杂的数据结构,例如此数组,我会扩展解决方案.

TL; DR

更可插拔的版本基于@ ege-Özcan的非常可爱的答案.

问题

我遇到了下面的内容,无法改变它.我也不想暂时压扁物体.我也不想使用下划线/ lodash,主要是出于性能原因以及自己实现它的乐趣.

var People = [
   {Name: {name: "Name", surname: "Surname"}, Middlename: "JJ"},
   {Name: {name: "AAA", surname: "ZZZ"}, Middlename:"Abrams"},
   {Name: {name: "Name", surname: "AAA"}, Middlename: "Wars"}
];

目标

我们的目标是主要通过对它进行排序People.Name.name,其次通过People.Name.surname

障碍

现在,在基本解决方案中使用括号表示法来计算要动态排序的属性.但是,在这里,我们还必须动态构造括号表示法,因为你会期望一些像People['Name.name']工作一样 - 但事实并非如此.

People['Name']['name']另一方面,简单地说,它是静态的,只允许你下降到第n级.

解

这里的主要补充是走下对象树并确定最后一个叶子的值,你必须指定,以及任何中间叶子.

var People = [
   {Name: {name: "Name", surname: "Surname"}, Middlename: "JJ"},
   {Name: {name: "AAA", surname: "ZZZ"}, Middlename:"Abrams"},
   {Name: {name: "Name", surname: "AAA"}, Middlename: "Wars"}
];

People.sort(dynamicMultiSort(['Name','name'], ['Name', '-surname']));
// Results in...
// [ { Name: { name: 'AAA', surname: 'ZZZ' }, Middlename: 'Abrams' },
//   { Name: { name: 'Name', surname: 'Surname' }, Middlename: 'JJ' },
//   { Name: { name: 'Name', surname: 'AAA' }, Middlename: 'Wars' } ]

// same logic as above, but strong deviation for dynamic properties 
function dynamicSort(properties) {
  var sortOrder = 1;
  // determine sort order by checking sign of last element of array
  if(properties[properties.length - 1][0] === "-") {
    sortOrder = -1;
    // Chop off sign
    properties[properties.length - 1] = properties[properties.length - 1].substr(1);
  }
  return function (a,b) {
    propertyOfA = recurseObjProp(a, properties)
    propertyOfB = recurseObjProp(b, properties)
    var result = (propertyOfA < propertyOfB) ? -1 : (propertyOfA > propertyOfB) ? 1 : 0;
    return result * sortOrder;
  };
}

/**
 * Takes an object and recurses down the tree to a target leaf and returns it value
 * @param  {Object} root - Object to be traversed.
 * @param  {Array} leafs - Array of downwards traversal. To access the value: {parent:{ child: 'value'}} -> ['parent','child']
 * @param  {Number} index - Must not be set, since it is implicit.
 * @return {String|Number}       The property, which is to be compared by sort.
 */
function recurseObjProp(root, leafs, index) {
  index ? index : index = 0
  var upper = root
  // walk down one level
  lower = upper[leafs[index]]
  // Check if last leaf has been hit by having gone one step too far.
  // If so, return result from last step.
  if (!lower) {
    return upper
  }
  // Else: recurse!
  index++
  // HINT: Bug was here, for not explicitly returning function
  // https://stackoverflow.com/a/17528613/3580261
  return recurseObjProp(lower, leafs, index)
}

/**
 * Multi-sort your array by a set of properties
 * @param {...Array} Arrays to access values in the form of: {parent:{ child: 'value'}} -> ['parent','child']
 * @return {Number} Number - number for sort algorithm
 */
function dynamicMultiSort() {
  var args = Array.prototype.slice.call(arguments); // slight deviation to base

  return function (a, b) {
    var i = 0, result = 0, numberOfProperties = args.length;
    // REVIEW: slightly verbose; maybe no way around because of `.sort`-'s nature
    // Consider: `.forEach()`
    while(result === 0 && i < numberOfProperties) {
      result = dynamicSort(args[i])(a, b);
      i++;
    }
    return result;
  }
}

例

关于JSBin的工作示例

Answer 22

还有一个选择:

var someArray = [...];

function generateSortFn(prop, reverse) {
    return function (a, b) {
        if (a[prop] < b[prop]) return reverse ? 1 : -1;
        if (a[prop] > b[prop]) return reverse ? -1 : 1;
        return 0;
    };
}

someArray.sort(generateSortFn('name', true));

默认情况下按升序排序.

Answer 23

一个简单的方法:

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

看到'.toLowerCase()'在比较字符串时防止错误是必要的.

Answer 24

这是我对此的看法：

该order参数是可选的，升序默认为“ASC”。

适用于重音字符，并且不区分大小写。

注意：它排序并返回ORIGINAL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;
}

片段

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;
}

Answer 25

EgeÖzcan代码的附加desc参数

function dynamicSort(property, desc) {
    if (desc) {
        return function (a, b) {
            return (a[property] > b[property]) ? -1 : (a[property] < b[property]) ? 1 : 0;
        }   
    }
    return function (a, b) {
        return (a[property] < b[property]) ? -1 : (a[property] > b[property]) ? 1 : 0;
    }
}

Answer 26

将Ege的动态解决方案与Vinay的想法相结合,您将获得一个非常强大的解决方案:

Array.prototype.sortBy = function() {
    function _sortByAttr(attr) {
        var sortOrder = 1;
        if (attr[0] == "-") {
            sortOrder = -1;
            attr = attr.substr(1);
        }
        return function(a, b) {
            var result = (a[attr] < b[attr]) ? -1 : (a[attr] > b[attr]) ? 1 : 0;
            return result * sortOrder;
        }
    }
    function _getSortFunc() {
        if (arguments.length == 0) {
            throw "Zero length arguments not allowed for Array.sortBy()";
        }
        var args = arguments;
        return function(a, b) {
            for (var result = 0, i = 0; result == 0 && i < args.length; i++) {
                result = _sortByAttr(args[i])(a, b);
            }
            return result;
        }
    }
    return this.sort(_getSortFunc.apply(null, arguments));
}

用法:

// Utility for printing objects
Array.prototype.print = function(title) {
    console.log("************************************************************************");
    console.log("**** "+title);
    console.log("************************************************************************");
    for (var i = 0; i < this.length; i++) {
        console.log("Name: "+this[i].FirstName, this[i].LastName, "Age: "+this[i].Age);
    }
}

// Setup sample data
var arrObj = [
    {FirstName: "Zach", LastName: "Emergency", Age: 35},
    {FirstName: "Nancy", LastName: "Nurse", Age: 27},
    {FirstName: "Ethel", LastName: "Emergency", Age: 42},
    {FirstName: "Nina", LastName: "Nurse", Age: 48},
    {FirstName: "Anthony", LastName: "Emergency", Age: 44},
    {FirstName: "Nina", LastName: "Nurse", Age: 32},
    {FirstName: "Ed", LastName: "Emergency", Age: 28},
    {FirstName: "Peter", LastName: "Physician", Age: 58},
    {FirstName: "Al", LastName: "Emergency", Age: 51},
    {FirstName: "Ruth", LastName: "Registration", Age: 62},
    {FirstName: "Ed", LastName: "Emergency", Age: 38},
    {FirstName: "Tammy", LastName: "Triage", Age: 29},
    {FirstName: "Alan", LastName: "Emergency", Age: 60},
    {FirstName: "Nina", LastName: "Nurse", Age: 54}
];

//Unit Tests
arrObj.sortBy("LastName").print("LastName Ascending");
arrObj.sortBy("-LastName").print("LastName Descending");
arrObj.sortBy("LastName", "FirstName", "-Age").print("LastName Ascending, FirstName Ascending, Age Descending");
arrObj.sortBy("-FirstName", "Age").print("FirstName Descending, Age Ascending");
arrObj.sortBy("-Age").print("Age Descending");

Answer 27

一个简单的函数,用于按属性对对象数组进行排序

function sortArray(array, property, direction) {
    direction = direction || 1;
    array.sort(function compare(a, b) {
        let comparison = 0;
        if (a[property] > b[property]) {
            comparison = 1 * direction;
        } else if (a[property] < b[property]) {
            comparison = -1 * direction;
        }
        return comparison;
    });
    return array; // Chainable
}

用法:

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

sortArray(objs, "last_nom"); // Asc
sortArray(objs, "last_nom", -1); // Desc

Answer 28

你为什么不写短代码？

objs.sort((a, b) => a.last_nom > b.last_nom && 1 || -1)

Answer 29

根据您的示例,您需要按两个字段(姓氏,名字)排序,而不是一个.您可以使用Alasql库在一行中进行此排序:

var res = alasql('SELECT * FROM ? ORDER BY last_nom, first_nom',[objs]);

在jsFiddle上试试这个例子.

Answer 30

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

Answer 31

您可以使用可重用的排序功能。

Array.prototype.order = function (prop, methods = {}) {
    if (prop?.constructor == Object) {
        methods = prop;
        prop = null;
    }
    const [orderType_a, orderType_b] = methods.reverse ? [1, -1] : [-1, 1];

    const $ = x => prop
        ? methods.insensitive
            ? String(x[prop]).toLowerCase()
            : x[prop]
        : methods.insensitive
            ? String(x).toLowerCase()
            : x;

    const fn = (a, b) => $(a) < $(b) ? orderType_a : $(b) < $(a) ? orderType_b : 0;
    return this.sort(fn);
};

它可用于对array 中的Array和Object进行排序
。

let items = [{ x: "Z" }, 3, "1", "0", 2, { x: "a" }, { x: 0 }];
items
    .order("x", { insensitive: 1 })
    // [ { x: 0 }, { x: 'a' }, 3, '1', '0', 2, { x: 'Z' } ]
    .order({ reverse: 1 })
    // [ { x: 0 }, { x: 'a' }, 3, 2, { x: 'Z' }, '1', '0' ]
    .sort(x => typeof x == "string" || typeof x == "number" ? -1 : 0)
    // [ '0', '1', 2, 3, { x: 0 }, { x: 'a' }, { x: 'Z' } ]

第一个（可选） > 对数组中包含的对象进行排序。
二是方法 > { reverse: any, insensitive: any }

Answer 32

您可能需要将它们转换为小写以防止混淆.

objs.sort(function (a,b) {

var nameA=a.last_nom.toLowerCase(), nameB=b.last_nom.toLowerCase()

if (nameA < nameB)
  return -1;
if (nameA > nameB)
  return 1;
return 0;  //no sorting

})

Answer 33

function compare(propName) {
    return function(a,b) {
        if (a[propName] < b[propName])
            return -1;
        if (a[propName] > b[propName])
            return 1;
        return 0;
    };
}

objs.sort(compare("last_nom"));

Answer 34

鉴于原始示例:

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)被普遍支持,并返回1,0,1如果a<b,a==b,a>b分别.
• ||在最后一行last_nom优先于first_nom.
• 减法适用于数字字段: var age_order = left.age - right.age;
• 否定逆序, return -last_nom_order || -first_nom_order || -age_order;

Answer 35

这是一个简单的问题,不知道为什么人们有这么复杂的解决方案.
一个简单的排序函数(基于快速排序算法):

function sortObjectsArray(objectsArray, sortKey)
        {
            // Quick Sort:
            var retVal;

            if (1 < objectsArray.length)
            {
                var pivotIndex = Math.floor((objectsArray.length - 1) / 2);  // middle index
                var pivotItem = objectsArray[pivotIndex];                    // value in the middle index
                var less = [], more = [];

                objectsArray.splice(pivotIndex, 1);                          // remove the item in the pivot position
                objectsArray.forEach(function(value, index, array)
                {
                    value[sortKey] <= pivotItem[sortKey] ?                   // compare the 'sortKey' proiperty
                        less.push(value) :
                        more.push(value) ;
                });

                retVal = sortObjectsArray(less, sortKey).concat([pivotItem], sortObjectsArray(more, sortKey));
            }
            else
            {
                retVal = objectsArray;
            }

            return retVal;
        }

使用示例:

var myArr = 
        [
            { val: 'x', idx: 3 },
            { val: 'y', idx: 2 },
            { val: 'z', idx: 5 },
        ];
myArr = sortObjectsArray(myArr, 'idx');

Answer 36

使用Ramda,

npm install ramda

import R from 'ramda'
var objs = [ 
    { first_nom: 'Lazslo', last_nom: 'Jamf'     },
    { first_nom: 'Pig',    last_nom: 'Bodine'   },
    { first_nom: 'Pirate', last_nom: 'Prentice' }
];
var ascendingSortedObjs = R.sortBy(R.prop('last_nom'), objs)
var descendingSortedObjs = R.reverse(ascendingSortedObjs)

Answer 37

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

Answer 38

使用 xPrototype 的sortBy：

var o = [
  { Name: 'Lazslo', LastName: 'Jamf'     },
  { Name: 'Pig',    LastName: 'Bodine'   },
  { Name: 'Pirate', LastName: 'Prentice' },
  { Name: 'Pag',    LastName: 'Bodine'   }
];


// Original
o.each(function (a, b) { console.log(a, b); });
/*
 0 Object {Name: "Lazslo", LastName: "Jamf"}
 1 Object {Name: "Pig", LastName: "Bodine"}
 2 Object {Name: "Pirate", LastName: "Prentice"}
 3 Object {Name: "Pag", LastName: "Bodine"}
*/


// Sort By LastName ASC, Name ASC
o.sortBy('LastName', 'Name').each(function(a, b) { console.log(a, b); });
/*
 0 Object {Name: "Pag", LastName: "Bodine"}
 1 Object {Name: "Pig", LastName: "Bodine"}
 2 Object {Name: "Lazslo", LastName: "Jamf"}
 3 Object {Name: "Pirate", LastName: "Prentice"}
*/


// Sort by LastName ASC and Name ASC
o.sortBy('LastName'.asc, 'Name'.asc).each(function(a, b) { console.log(a, b); });
/*
 0 Object {Name: "Pag", LastName: "Bodine"}
 1 Object {Name: "Pig", LastName: "Bodine"}
 2 Object {Name: "Lazslo", LastName: "Jamf"}
 3 Object {Name: "Pirate", LastName: "Prentice"}
*/


// Sort by LastName DESC and Name DESC
o.sortBy('LastName'.desc, 'Name'.desc).each(function(a, b) { console.log(a, b); });
/*
 0 Object {Name: "Pirate", LastName: "Prentice"}
 1 Object {Name: "Lazslo", LastName: "Jamf"}
 2 Object {Name: "Pig", LastName: "Bodine"}
 3 Object {Name: "Pag", LastName: "Bodine"}
*/


// Sort by LastName DESC and Name ASC
o.sortBy('LastName'.desc, 'Name'.asc).each(function(a, b) { console.log(a, b); });
/*
 0 Object {Name: "Pirate", LastName: "Prentice"}
 1 Object {Name: "Lazslo", LastName: "Jamf"}
 2 Object {Name: "Pag", LastName: "Bodine"}
 3 Object {Name: "Pig", LastName: "Bodine"}
*/

Answer 39

我只是增强了EgeÖzcan的动态排序能力，以深入研究内部物体。如果数据如下所示：

obj = [
    {
        a: { a: 1, b: 2, c: 3 },
        b: { a: 4, b: 5, c: 6 }
    },
    { 
        a: { a: 3, b: 2, c: 1 },
        b: { a: 6, b: 5, c: 4 }
}];

如果您想对某个物业进行排序，我认为我的增强功能会非常有帮助。我向这样的对象添加新功能：

Object.defineProperty(Object.prototype, 'deepVal', {
    enumerable: false,
    writable: true,
    value: function (propertyChain) {
        var levels = propertyChain.split('.');
        parent = this;
        for (var i = 0; i < levels.length; i++) {
            if (!parent[levels[i]])
                return undefined;
            parent = parent[levels[i]];
        }
        return parent;
    }
});

并更改了_dynamicSort的返回函数：

return function (a,b) {
        var result = ((a.deepVal(property) > b.deepVal(property)) - (a.deepVal(property) < b.deepVal(property)));
        return result * sortOrder;
    }

现在你可以通过排序AA是这样的：

obj.sortBy('a.a');

见Commplete脚本的jsfiddle

Answer 40

对对象数组进行排序

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

// Property to sort by
var args = "last_nom";

// Function to sort the data by the given property
function sortByProperty(property) {
    return function (a, b) {
        var sortStatus = 0,
            aProp = a[property].toLowerCase(),
            bProp = b[property].toLowerCase();
        if (aProp < bProp) {
            sortStatus = -1;
        } else if (aProp > bProp) {
            sortStatus = 1;
        }
        return sortStatus;
    };
}

// Implementation
var sortedArray = booksArray.sort(sortByProperty(args));

console.log("sortedArray: " + JSON.stringify(sortedArray) );

控制台日志输出：

"sortedArray: 
[{"first_nom":"Pig","last_nom":"Bodine"},
{"first_nom":"Lazslo","last_nom":"Jamf"},
{"first_nom":"Pirate","last_nom":"Prentice"}]"

基于此来源改编：代码片段：如何按属性对 JSON 对象数组进行排序

Answer 41

方式1：

您可以使用Underscore.js。首先导入下划线。

 import * as _ from 'underscore';
 let SortedObjs = _.sortBy(objs, 'last_nom');

方法2：使用比较功能。

function compare(first, second) {
     if (first.last_nom < second.last_nom)
         return -1;
     if (first.last_nom > second.last_nom)
       return 1;
    return 0;
 }

objs.sort(compare);

Answer 42

这个对我有用。这里 它将保持undefined到最后。

 function sort(items, property, direction) {

    function compare(a, b) {
      if(!a[property] && !b[property]) {
        return 0;
      } else if(a[property] && !b[property]) {
        return -1;
      } else if(!a[property] && b[property]) {
        return 1;
      } else {
        const value1 = a[property].toString().toUpperCase(); // ignore upper and lowercase
        const value2 = b[property].toString().toUpperCase(); // ignore upper and lowercase
        if (value1 < value2) {
          return direction === 0 ? -1 : 1;
        } else if (value1 > value2) {
          return direction === 0 ? 1 : -1;
        } else {
          return 0;
        }
        
      }
    }
    
    return items.sort(compare);
   } 
   
   var items = [
  { name: 'Edward', value: 21 },
  { name: 'Sharpe', value: 37 },
  { name: 'And', value: 45 },
  { name: 'The', value: -12 },
  { name: undefined, value: -12 },
  { name: 'Magnetic', value: 13 },
  { name: 'Zeros', value: 37 }
];
   console.log('Ascending Order:- ');
   console.log(sort(items, 'name', 0));
   console.log('Decending Order:- ');
   console.log(sort(items, 'name', 1));
    
    

Answer 43

简单回答：

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

细节：

今天很简单，您可以将字符串与localeCompare. 正如 Mozilla Doc 所说：

该localeCompare()方法返回一个数字，指示引用字符串是before、 或after、 或is the same as the given string in sort order。

    //example1:
    console.log("aaa".localeCompare("aab")); //-1
    console.log("aaa".localeCompare("aaa")); //0
    console.log("aab".localeCompare("aaa")); //1

    //example2:
    const a = 'réservé'; // with accents, lowercase
    const b = 'RESERVE'; // no accents, uppercase

    console.log(a.localeCompare(b));
    // expected output: 1
    console.log(a.localeCompare(b, 'en', { sensitivity: 'base' }));
    // expected output: 0

有关更多详细信息，请参阅 Mozilla 文档localeCompare：

Answer 44

我已经在各种项目中使用了这个实用程序，并且效果很好。它也非常模块化：

• 传递要排序的键的名称
• 选择升序还是降序排序

sortArrayOfObjsByKeyUtil.js

// 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;
  });

sortArrayOfObjsByKeyUtil.test.js

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);
  });
});