按键排序 JavaScript object

Question

I need to sort JavaScript objects by key.

Hence the following:

{ 'b' : 'asdsad', 'c' : 'masdas', 'a' : 'dsfdsfsdf' }

Would become:

{ 'a' : 'dsfdsfsdf', 'b' : 'asdsad', 'c' : 'masdas' }

Answer 1

The other answers to this question are outdated, never matched implementation reality, and have officially become incorrect now that the ES6 / ES2015 spec has been published.

---

See the section on property iteration order in Exploring ES6 by Axel Rauschmayer :

All methods that iterate over property keys do so in the same order:

1. First all Array indices, sorted numerically.
2. Then all string keys (that are not indices), in the order in which they were created.
3. Then all symbols, in the order in which they were created.

So yes, JavaScript objects are in fact ordered, and the order of their keys/properties can be changed.

Here's how you can sort an object by its keys/properties, alphabetically:

 const unordered = { 'b': 'foo', 'c': 'bar', 'a': 'baz' }; console.log(JSON.stringify(unordered)); // → '{"b":"foo","c":"bar","a":"baz"}' const ordered = Object.keys(unordered).sort().reduce( (obj, key) => { obj[key] = unordered[key]; return obj; }, {} ); console.log(JSON.stringify(ordered)); // → '{"a":"baz","b":"foo","c":"bar"}'

Use var instead of const for compatibility with ES5 engines.

Answer 2

JavaScript objects ¹ are not ordered. It is meaningless to try to "sort" them. If you want to iterate over an object's properties, you can sort the keys and then retrieve the associated values:

 var myObj = { 'b': 'asdsadfd', 'c': 'masdasaf', 'a': 'dsfdsfsdf' }, keys = [], k, i, len; for (k in myObj) { if (myObj.hasOwnProperty(k)) { keys.push(k); } } keys.sort(); len = keys.length; for (i = 0; i < len; i++) { k = keys[i]; console.log(k + ':' + myObj[k]); }

---

Alternate implementation using Object.keys fanciness:

 var myObj = { 'b': 'asdsadfd', 'c': 'masdasaf', 'a': 'dsfdsfsdf' }, keys = Object.keys(myObj), i, len = keys.length; keys.sort(); for (i = 0; i < len; i++) { k = keys[i]; console.log(k + ':' + myObj[k]); }

---

^{1 Not to be pedantic, but there's no such thing as a JSON object .}

Answer 3

A lot of people have mention that "objects cannot be sorted" , but after that they are giving you a solution which works. Paradox, isn't it?

No one mention why those solutions are working. They are, because in most of the browser's implementations values in objects are stored in the order in which they were added. That's why if you create new object from sorted list of keys it's returning an expected result.

And I think that we could add one more solution – ES5 functional way:

function sortObject(obj) {
    return Object.keys(obj).sort().reduce(function (result, key) {
        result[key] = obj[key];
        return result;
    }, {});
}

ES2015 version of above (formatted to "one-liner"):

const sortObject = o => Object.keys(o).sort().reduce((r, k) => (r[k] = o[k], r), {})

Short explanation of above examples (as asked in comments):

Object.keys is giving us a list of keys in provided object ( obj or o ), then we're sorting those using default sorting algorithm, next .reduce is used to convert that array back into an object, but this time with all of the keys sorted.

Answer 4

Guys I'm figuratively shocked! Sure all answers are somewhat old, but no one did even mention the stability in sorting! So bear with me I'll try my best to answer the question itself and go into details here. So I'm going to apologize now it will be a lot to read.

Since it is 2018 I will only use ES6, the Polyfills are all available at the MDN docs, which I will link at the given part.

---

Answer to the question:

If your keys are only numbers then you can safely use Object.keys() together with Array.prototype.reduce() to return the sorted object:

// Only numbers to show it will be sorted.
const testObj = {
  '2000': 'Articel1',
  '4000': 'Articel2',
  '1000': 'Articel3',
  '3000': 'Articel4',
};

// I'll explain what reduces does after the answer.
console.log(Object.keys(testObj).reduce((accumulator, currentValue) => {
  accumulator[currentValue] = testObj[currentValue];
  return accumulator;
}, {}));

/**
 * expected output:
 * {
 * '1000': 'Articel3',
 * '2000': 'Articel1',
 * '3000': 'Articel4',
 * '4000': 'Articel2' 
 *  } 
 */

// if needed here is the one liner:
console.log(Object.keys(testObj).reduce((a, c) => (a[c] = testObj[c], a), {}));

However if you are working with strings I highly recommend chaining Array.prototype.sort() into all of this:

// String example
const testObj = {
  'a1d78eg8fdg387fg38': 'Articel1',
  'z12989dh89h31d9h39': 'Articel2',
  'f1203391dhj32189h2': 'Articel3',
  'b10939hd83f9032003': 'Articel4',
};
// Chained sort into all of this.
console.log(Object.keys(testObj).sort().reduce((accumulator, currentValue) => {
  accumulator[currentValue] = testObj[currentValue];
  return accumulator;
}, {}));

/**
 * expected output:   
 * { 
 * a1d78eg8fdg387fg38: 'Articel1',
 * b10939hd83f9032003: 'Articel4',
 * f1203391dhj32189h2: 'Articel3',
 * z12989dh89h31d9h39: 'Articel2' 
 * }
 */

// again the one liner:
console.log(Object.keys(testObj).sort().reduce((a, c) => (a[c] = testObj[c], a), {}));

If someone is wondering what reduce does:

// Will return Keys of object as an array (sorted if only numbers or single strings like a,b,c).
Object.keys(testObj)

// Chaining reduce to the returned array from Object.keys().
// Array.prototype.reduce() takes one callback 
// (and another param look at the last line) and passes 4 arguments to it: 
// accumulator, currentValue, currentIndex and array
.reduce((accumulator, currentValue) => {

  // setting the accumulator (sorted new object) with the actual property from old (unsorted) object.
  accumulator[currentValue] = testObj[currentValue];

  // returning the newly sorted object for the next element in array.
  return accumulator;

  // the empty object {} ist the initial value for  Array.prototype.reduce().
}, {});

If needed here is the explanation for the one liner:

Object.keys(testObj).reduce(

  // Arrow function as callback parameter.
  (a, c) => 

  // parenthesis return! so we can safe the return and write only (..., a);
  (a[c] = testObj[c], a)

  // initial value for reduce.
  ,{}
);

• Docs for reduce: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce
• Why use parenthesis on JavaScript return statements: http://jamesknelson.com/javascript-return-parenthesis/

---

Why Sorting is a bit complicated:

In short Object.keys() will return an array with the same order as we get with a normal loop:

const object1 = {
  a: 'somestring',
  b: 42,
  c: false
};

console.log(Object.keys(object1));
// expected output: Array ["a", "b", "c"]

Object.keys() returns an array whose elements are strings corresponding to the enumerable properties found directly upon object. The ordering of the properties is the same as that given by looping over the properties of the object manually.

• https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Object/keys

Sidenote - you can use Object.keys() on arrays as well, keep in mind the index will be returned:

// simple array
const arr = ['a', 'b', 'c'];
console.log(Object.keys(arr)); // console: ['0', '1', '2']

But it is not as easy as shown by those examples, real world objects may contain numbers and alphabetical characters or even symbols (please don't do it).

Here is an example with all of them in one object:

// This is just to show what happens, please don't use symbols in keys.
const testObj = {
  '1asc': '4444',
  1000: 'a',
  b: '1231',
  '#01010101010': 'asd',
  2: 'c'
};

console.log(Object.keys(testObj));
// output: [ '2', '1000', '1asc', 'b', '#01010101010' ]

Now if we use Array.prototype.sort() on the array above the output changes:

console.log(Object.keys(testObj).sort());
// output: [ '#01010101010', '1000', '1asc', '2', 'b' ]

Here is a quote from the docs:

The sort() method sorts the elements of an array in place and returns the array. The sort is not necessarily stable. The default sort order is according to string Unicode code points.

The time and space complexity of the sort cannot be guaranteed as it is implementation dependent.

• https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort

You have to make sure that one of them returns the desired output for you. In reallife examples people tend to mix up things expecially if you use different information inputs like APIs and Databases together.

---

So what's the big deal?

Well there are two articles which every programmer should understand:

In-place algorithm :

In computer science, an in-place algorithm is an algorithm which transforms input using no auxiliary data structure. However a small amount of extra storage space is allowed for auxiliary variables. The input is usually overwritten by the output as the algorithm executes. In-place algorithm updates input sequence only through replacement or swapping of elements. An algorithm which is not in-place is sometimes called not-in-place or out-of-place.

So basically our old array will be overwritten! This is important if you want to keep the old array for other reasons. So keep this in mind.

Sorting algorithm

Stable sort algorithms sort identical elements in the same order that they appear in the input. When sorting some kinds of data, only part of the data is examined when determining the sort order. For example, in the card sorting example to the right, the cards are being sorted by their rank, and their suit is being ignored. This allows the possibility of multiple different correctly sorted versions of the original list. Stable sorting algorithms choose one of these, according to the following rule: if two items compare as equal, like the two 5 cards, then their relative order will be preserved, so that if one came before the other in the input, it will also come before the other in the output.

An example of stable sort on playing cards. When the cards are sorted by rank with a stable sort, the two 5s must remain in the same order in the sorted output that they were originally in. When they are sorted with a non-stable sort, the 5s may end up in the opposite order in the sorted output.

This shows that the sorting is right but it changed. So in the real world even if the sorting is correct we have to make sure that we get what we expect! This is super important keep this in mind as well. For more JavaScript examples look into the Array.prototype.sort() - docs: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort

Answer 5

现在是 2019 年，我们有 2019 年的方法来解决这个问题:)

Object.fromEntries(Object.entries({b: 3, a:8, c:1}).sort())

Answer 6

ES6 - here is the 1 liner

 var data = { zIndex:99, name:'sravan', age:25, position:'architect', amount:'100k', manager:'mammu' }; console.log(Object.entries(data).sort().reduce( (o,[k,v]) => (o[k]=v,o), {} ));

Answer 7

This works for me

/**
 * Return an Object sorted by it's Key
 */
var sortObjectByKey = function(obj){
    var keys = [];
    var sorted_obj = {};

    for(var key in obj){
        if(obj.hasOwnProperty(key)){
            keys.push(key);
        }
    }

    // sort keys
    keys.sort();

    // create new array based on Sorted Keys
    jQuery.each(keys, function(i, key){
        sorted_obj[key] = obj[key];
    });

    return sorted_obj;
};

Answer 8

This is an old question, but taking the cue from Mathias Bynens' answer, I've made a short version to sort the current object, without much overhead.

    Object.keys(unordered).sort().forEach(function(key) {
        var value = unordered[key];
        delete unordered[key];
        unordered[key] = value;
    });

after the code execution, the "unordered" object itself will have the keys alphabetically sorted.

Answer 9

Using lodash this will work:

some_map = { 'b' : 'asdsad', 'c' : 'masdas', 'a' : 'dsfdsfsdf' }

// perform a function in order of ascending key
_(some_map).keys().sort().each(function (key) {
  var value = some_map[key];
  // do something
});

// or alternatively to build a sorted list
sorted_list = _(some_map).keys().sort().map(function (key) {
  var value = some_map[key];
  // return something that shall become an item in the sorted list
}).value();

Just food for thought.

Answer 10

Suppose it could be useful in VisualStudio debugger which shows unordered object properties.

(function(s) {
    var t = {};

    Object.keys(s).sort().forEach(function(k) {
        t[k] = s[k]
    });

    return t
})({
    b: 2,
    a: 1,
    c: 3
});

The same as inline version:

(function(s){var t={};Object.keys(s).sort().forEach(function(k){t[k]=s[k]});return t})({b:2,a:1,c:3})

Answer 11

I am actually very surprised that over 30 answers were given, and yet none gave a full deep solution for this problem. Some had shallow solution, while others had deep but faulty (it'll crash if undefined, function or symbol will be in the json).

Here is the full solution:

function sortObject(unordered, sortArrays = false) {
  if (!unordered || typeof unordered !== 'object') {
    return unordered;
  }

  if (Array.isArray(unordered)) {
    const newArr = unordered.map((item) => sortObject(item, sortArrays));
    if (sortArrays) {
      newArr.sort();
    }
    return newArr;
  }

  const ordered = {};
  Object.keys(unordered)
    .sort()
    .forEach((key) => {
      ordered[key] = sortObject(unordered[key], sortArrays);
    });
  return ordered;
}

const json = {
  b: 5,
  a: [2, 1],
  d: {
    b: undefined,
    a: null,
    c: false,
    d: true,
    g: '1',
    f: [],
    h: {},
    i: 1n,
    j: () => {},
    k: Symbol('a')
  },
  c: [
    {
      b: 1,
      a: 1
    }
  ]
};
console.log(sortObject(json, true));

Answer 12

Underscore version :

function order(unordered)
{
return _.object(_.sortBy(_.pairs(unordered),function(o){return o[0]}));
}

If you don't trust your browser for keeping the order of the keys, I strongly suggest to rely on a ordered array of key-value paired arrays.

_.sortBy(_.pairs(c),function(o){return o[0]})

Answer 13

Maybe a bit more elegant form:

 /** * Sorts a key-value object by key, maintaining key to data correlations. * @param {Object} src key-value object * @returns {Object} */ var ksort = function ( src ) { var keys = Object.keys( src ), target = {}; keys.sort(); keys.forEach(function ( key ) { target[ key ] = src[ key ]; }); return target; }; // Usage console.log(ksort({ a:1, c:3, b:2 }));

PS and the same with ES6+ syntax:

function ksort( src ) {
  const keys = Object.keys( src );
  keys.sort();
  return keys.reduce(( target, key ) => {
        target[ key ] = src[ key ];
        return target;
  }, {});
};

Answer 14

function sortObjectKeys(obj){
    return Object.keys(obj).sort().reduce((acc,key)=>{
        acc[key]=obj[key];
        return acc;
    },{});
}

sortObjectKeys({
    telephone: '069911234124',
    name: 'Lola',
    access: true,
});

Answer 15

Here is a one line solution (not the most efficient but when it comes to thin objects like in your example I'd rather use native JS functions then messing up with sloppy loops)

 const unordered = { 'b' : 'asdsad', 'c' : 'masdas', 'a' : 'dsfdsfsdf' } const ordered = Object.fromEntries(Object.entries(unordered).sort()) console.log(ordered); // a->b->c

Answer 16

// if keys are char/string
const sortObject = (obj) => Object.fromEntries(Object.entries(obj).sort( ));
let obj = { c: 3, a: 1 };
obj = sortObject(obj)

// if keys are numbers
const sortObject = (obj) => Object.fromEntries(Object.entries(obj).sort( (a,b)=>a-b ));
let obj = { 3: 'c', 1: 'a' };
obj = sortObject(obj)

Answer 17

recursive sort, for nested object and arrays

function sortObjectKeys(obj){
    return Object.keys(obj).sort().reduce((acc,key)=>{
        if (Array.isArray(obj[key])){
            acc[key]=obj[key].map(sortObjectKeys);
        }
        if (typeof obj[key] === 'object'){
            acc[key]=sortObjectKeys(obj[key]);
        }
        else{
            acc[key]=obj[key];
        }
        return acc;
    },{});
}

// test it
sortObjectKeys({
    telephone: '069911234124',
    name: 'Lola',
    access: true,
    cars: [
        {name: 'Family', brand: 'Volvo', cc:1600},
        {
            name: 'City', brand: 'VW', cc:1200, 
            interior: {
                wheel: 'plastic',
                radio: 'blaupunkt'
            }
        },
        {
            cc:2600, name: 'Killer', brand: 'Plymouth',
            interior: {
                wheel: 'wooden',
                radio: 'earache!'
            }
        },
    ]
});

Answer 18

Here is a clean lodash-based version that works with nested objects

/**
 * Sort of the keys of an object alphabetically
 */
const sortKeys = function(obj) {
  if(_.isArray(obj)) {
    return obj.map(sortKeys);
  }
  if(_.isObject(obj)) {
    return _.fromPairs(_.keys(obj).sort().map(key => [key, sortKeys(obj[key])]));
  }
  return obj;
};

It would be even cleaner if lodash had a toObject() method ...

Answer 19

Object.keys(unordered).sort().reduce(
    (acc,curr) => ({...acc, [curr]:unordered[curr]})
    , {}
)

Answer 20

As already mentioned, objects are unordered.

However...

You may find this idiom useful:

var o = { 'b' : 'asdsad', 'c' : 'masdas', 'a' : 'dsfdsfsdf' };

var kv = [];

for (var k in o) {
  kv.push([k, o[k]]);
}

kv.sort()

You can then iterate through kv and do whatever you wish.

> kv.sort()
[ [ 'a', 'dsfdsfsdf' ],
  [ 'b', 'asdsad' ],
  [ 'c', 'masdas' ] ]

Answer 21

There's a great project by @sindresorhus called sort-keys that works awesome.

You can check its source code here:

https://github.com/sindresorhus/sort-keys

Or you can use it with npm:

$ npm install --save sort-keys

Here are also code examples from his readme

const sortKeys = require('sort-keys');

sortKeys({c: 0, a: 0, b: 0});
//=> {a: 0, b: 0, c: 0}

sortKeys({b: {b: 0, a: 0}, a: 0}, {deep: true});
//=> {a: 0, b: {a: 0, b: 0}}

sortKeys({c: 0, a: 0, b: 0}, {
    compare: (a, b) => -a.localeCompare(b)
});
//=> {c: 0, b: 0, a: 0}

Answer 22

Use this code if you have nested objects or if you have nested array obj.

var sortObjectByKey = function(obj){
    var keys = [];
    var sorted_obj = {};
    for(var key in obj){
        if(obj.hasOwnProperty(key)){
            keys.push(key);
        }
    }
    // sort keys
    keys.sort();

    // create new array based on Sorted Keys
    jQuery.each(keys, function(i, key){
        var val = obj[key];
        if(val instanceof Array){
            //do for loop;
            var arr = [];
            jQuery.each(val,function(){
                arr.push(sortObjectByKey(this));
            }); 
            val = arr;

        }else if(val instanceof Object){
            val = sortObjectByKey(val)
        }
        sorted_obj[key] = val;
    });
    return sorted_obj;
};

Answer 23

Just use lodash to unzip map and sortBy first value of pair and zip again it will return sorted key.

If you want sortby value change pair index to 1 instead of 0

var o = { 'b' : 'asdsad', 'c' : 'masdas', 'a' : 'dsfdsfsdf' };
console.log(_(o).toPairs().sortBy(0).fromPairs().value())

Answer 24

Sorts keys recursively while preserving references.

function sortKeys(o){
    if(o && o.constructor === Array)
        o.forEach(i=>sortKeys(i));
    else if(o && o.constructor === Object)
        Object.entries(o).sort((a,b)=>a[0]>b[0]?1:-1).forEach(e=>{
            sortKeys(e[1]);
            delete o[e[0]];
            o[e[0]] = e[1];
        });
}

Example:

let x = {d:3, c:{g:20, a:[3,2,{s:200, a:100}]}, a:1};
let y = x.c;
let z = x.c.a[2];
sortKeys(x);
console.log(x); // {a: 1, c: {a: [3, 2, {a: 1, s: 2}], g: 2}, d: 3}
console.log(y); // {a: [3, 2, {a: 100, s: 200}}, g: 20}
console.log(z); // {a: 100, s: 200}

Answer 25

This is a lightweight solution to everything I need for JSON sorting.

function sortObj(obj) {
    if (typeof obj !== "object" || obj === null)
        return obj;

    if (Array.isArray(obj))
        return obj.map((e) => sortObj(e)).sort();

    return Object.keys(obj).sort().reduce((sorted, k) => {
        sorted[k] = sortObj(obj[k]);
        return sorted;
    }, {});
}

Answer 26

Solution:

function getSortedObject(object) {
  var sortedObject = {};

  var keys = Object.keys(object);
  keys.sort();

  for (var i = 0, size = keys.length; i < size; i++) {
    key = keys[i];
    value = object[key];
    sortedObject[key] = value;
  }

  return sortedObject;
}

// Test run
getSortedObject({d: 4, a: 1, b: 2, c: 3});

Explanation:

Many JavaScript runtimes store values inside an object in the order in which they are added.

To sort the properties of an object by their keys you can make use of the Object.keys function which will return an array of keys. The array of keys can then be sorted by the Array.prototype.sort() method which sorts the elements of an array in place (no need to assign them to a new variable).

Once the keys are sorted you can start using them one-by-one to access the contents of the old object to fill a new object (which is now sorted).

Below is an example of the procedure (you can test it in your targeted browsers):

 /** * Returns a copy of an object, which is ordered by the keys of the original object. * * @param {Object} object - The original object. * @returns {Object} Copy of the original object sorted by keys. */ function getSortedObject(object) { // New object which will be returned with sorted keys var sortedObject = {}; // Get array of keys from the old/current object var keys = Object.keys(object); // Sort keys (in place) keys.sort(); // Use sorted keys to copy values from old object to the new one for (var i = 0, size = keys.length; i < size; i++) { key = keys[i]; value = object[key]; sortedObject[key] = value; } // Return the new object return sortedObject; } /** * Test run */ var unsortedObject = { d: 4, a: 1, b: 2, c: 3 }; var sortedObject = getSortedObject(unsortedObject); for (var key in sortedObject) { var text = "Key: " + key + ", Value: " + sortedObject[key]; var paragraph = document.createElement('p'); paragraph.textContent = text; document.body.appendChild(paragraph); }

Note: Object.keys is an ECMAScript 5.1 method but here is a polyfill for older browsers:

if (!Object.keys) {
  Object.keys = function (object) {
    var key = [];
    var property = undefined;
    for (property in object) {
      if (Object.prototype.hasOwnProperty.call(object, property)) {
        key.push(property);
      }
    }
    return key;
  };
}

Answer 27

I transfered some Java enums to javascript objects.

These objects returned correct arrays for me. if object keys are mixed type (string, int, char), there is a problem.

 var Helper = { isEmpty: function (obj) { return !obj || obj === null || obj === undefined || Array.isArray(obj) && obj.length === 0; }, isObject: function (obj) { return (typeof obj === 'object'); }, sortObjectKeys: function (object) { return Object.keys(object) .sort(function (a, b) { c = a - b; return c }); }, containsItem: function (arr, item) { if (arr && Array.isArray(arr)) { return arr.indexOf(item) > -1; } else { return arr === item; } }, pushArray: function (arr1, arr2) { if (arr1 && arr2 && Array.isArray(arr1)) { arr1.push.apply(arr1, Array.isArray(arr2) ? arr2 : [arr2]); } } }; function TypeHelper() { var _types = arguments[0], _defTypeIndex = 0, _currentType, _value; if (arguments.length == 2) { _defTypeIndex = arguments[1]; } Object.defineProperties(this, { Key: { get: function () { return _currentType; }, set: function (val) { _currentType.setType(val, true); }, enumerable: true }, Value: { get: function () { return _types[_currentType]; }, set: function (val) { _value.setType(val, false); }, enumerable: true } }); this.getAsList = function (keys) { var list = []; Helper.sortObjectKeys(_types).forEach(function (key, idx, array) { if (key && _types[key]) { if (!Helper.isEmpty(keys) && Helper.containsItem(keys, key) || Helper.isEmpty(keys)) { var json = {}; json.Key = key; json.Value = _types[key]; Helper.pushArray(list, json); } } }); return list; }; this.setType = function (value, isKey) { if (!Helper.isEmpty(value)) { Object.keys(_types).forEach(function (key, idx, array) { if (Helper.isObject(value)) { if (value && value.Key == key) { _currentType = key; } } else if (isKey) { if (value && value.toString() == key.toString()) { _currentType = key; } } else if (value && value.toString() == _types[key]) { _currentType = key; } }); } else { this.setDefaultType(); } return isKey ? _types[_currentType] : _currentType; }; this.setTypeByIndex = function (index) { var keys = Helper.sortObjectKeys(_types); for (var i = 0; i < keys.length; i++) { if (index === i) { _currentType = keys[index]; break; } } }; this.setDefaultType = function () { this.setTypeByIndex(_defTypeIndex); }; this.setDefaultType(); } var TypeA = { "-1": "Any", "2": "2L", "100": "100L", "200": "200L", "1000": "1000L" }; var TypeB = { "U": "Any", "W": "1L", "V": "2L", "A": "100L", "Z": "200L", "K": "1000L" }; console.log('keys of TypeA', Helper.sortObjectKeys(TypeA));//keys of TypeA ["-1", "2", "100", "200", "1000"] console.log('keys of TypeB', Helper.sortObjectKeys(TypeB));//keys of TypeB ["U", "W", "V", "A", "Z", "K"] var objectTypeA = new TypeHelper(TypeA), objectTypeB = new TypeHelper(TypeB); console.log('list of objectA = ', objectTypeA.getAsList()); console.log('list of objectB = ', objectTypeB.getAsList());

Types:

var TypeA = {
    "-1": "Any",
    "2": "2L",
    "100": "100L",
    "200": "200L",
    "1000": "1000L"
};

var TypeB = {
    "U": "Any",
    "W": "1L",
    "V": "2L",
    "A": "100L",
    "Z": "200L",
    "K": "1000L"
};


Sorted Keys(output):

Key list of TypeA -> ["-1", "2", "100", "200", "1000"]

Key list of TypeB -> ["U", "W", "V", "A", "Z", "K"]

Answer 28

The one line:

Object.entries(unordered)
  .sort(([keyA], [keyB]) => keyA > keyB)
  .reduce((obj, [key,value]) => Object.assign(obj, {[key]: value}), {})

Answer 29

Simple and readable snippet, using lodash.

You need to put the key in quotes only when calling sortBy. It doesn't have to be in quotes in the data itself.

_.sortBy(myObj, "key")

Also, your second parameter to map is wrong. It should be a function, but using pluck is easier.

_.map( _.sortBy(myObj, "key") , "value");

Answer 30

Pure JavaScript answer to sort an Object. This is the only answer that I know of that will handle negative numbers. This function is for sorting numerical Objects.

Input obj = {1000: {}, -1200: {}, 10000: {}, 200: {}};

function osort(obj) {
var keys = Object.keys(obj);
var len = keys.length;
var rObj = [];
var rK = [];
var t = Object.keys(obj).length;
while(t > rK.length) {
    var l = null;
    for(var x in keys) {
        if(l && parseInt(keys[x]) < parseInt(l)) {
            l = keys[x];
            k = x;
        }
        if(!l) { // Find Lowest
            var l = keys[x];
            var k = x;
        }
    }
    delete keys[k];
    rK.push(l);
}

for (var i = 0; i < len; i++) {

    k = rK[i];
    rObj.push(obj[k]);
}
return rObj;
}

The output will be an object sorted by those numbers with new keys starting at 0.

Answer 31

Just to simplify it and make it more clear the answer from Matt Ball

 //your object var myObj = { b : 'asdsadfd', c : 'masdasaf', a : 'dsfdsfsdf' }; //fixed code var keys = []; for (var k in myObj) { if (myObj.hasOwnProperty(k)) { keys.push(k); } } keys.sort(); for (var i = 0; i < keys.length; i++) { k = keys[i]; alert(k + ':' + myObj[k]); }

Answer 32

Not sure if this answers the question, but this is what I needed.

Maps.iterate.sorted = function (o, callback) {
    var keys = Object.keys(o), sorted = keys.sort(), k; 
    if ( callback ) {
            var i = -1;
            while( ++i < sorted.length ) {
                    callback(k = sorted[i], o[k] );
            }
    }

    return sorted;
}

Called as :

Maps.iterate.sorted({c:1, b:2, a:100}, function(k, v) { ... } ) 

Answer 33

the best way to do it is

 const object =  Object.keys(o).sort().reduce((r, k) => (r[k] = o[k], r), {})

 //else if its in reverse just do 

 const object = Object.keys(0).reverse ()

You can first convert your almost-array-like object to a real array, and then use .reverse():

Object.assign([], {1:'banana', 2:'apple', 
3:'orange'}).reverse();
// [ "orange", "apple", "banana", <1 empty slot> ]

The empty slot at the end if cause because your first index is 1 instead of 0. You can remove the empty slot with .length-- or .pop().

Alternatively, if you want to borrow .reverse and call it on the same object, it must be a fully-array-like object. That is, it needs a length property:

Array.prototype.reverse.call({1:'banana', 2:'apple', 
3:'orange', length:4});
// {0:"orange", 1:"apple", 3:"banana", length:4}

Note it will return the same fully-array-like object object, so it won't be a real array. You can then use delete to remove the length property.

Answer 34

i liked this aswer above except it doesnt work:

/**
 * Return an Object sorted by it's Key
 */
var sortObjectByKey = function(obj){
    var keys = [];
    var sorted_obj = {};

    for(var key in obj){
        if(obj.hasOwnProperty(key)){
            keys.push(key);
        }
    }

    // sort keys
    keys.sort();

    // create new array based on Sorted Keys
    jQuery.each(keys, function(i, key){
        sorted_obj[key] = obj[key];
    });

    return sorted_obj;
};

looks like the sorted_obj there isnt guaranty of order, for that map seems to be a better fit, i ended up changing the object for a map, something like this:

/**
 * Return an Object sorted by it's Key
 */
var sortObjectByKey = function(obj){
    var keys = [];
    let sorted_map = new Map();

    for(var key in obj){
        if(obj.hasOwnProperty(key)){
            keys.push(key);
        }
    }

    // sort keys
    keys.sort();

    // create new map based on Sorted Keys
    keys.forEach(key => {
        sorted_map.set(key, obj[key]);
    });

    return sorted_map;
};

Answer 35

 const sortObjectByKeys = (object) => Object.fromEntries( Object.entries(object).sort(([k1], [k2]) => k1 < k2 ? -1 : 1) ) const object = {b: 'asdsad', c: 'masdas', a: 'dsfdsfsdf' } const orderedObject = sortObjectByKeys(object) console.log(orderedObject)

Answer 36

This question comes up when searching for how to sort an object in Typescript as well, so adding here what worked for me.

With typescript, I was having trouble generalizing the types with the reduce approach. Here's what I ended up doing:

function sortObject<T>(obj: T, compareFn?: ((a: unknown, b: unknown) => number)) {
    let keys: unknown[];

    if (compareFn) {
        keys = Object.keys(obj).sort(compareFn);
    } else {
        keys = Object.keys(obj).sort();
    }

    // Storing the data in temp object and deleting from the original
    const temp = {} as typeof obj;

    for (let i = 0; i < keys.length; ++i) {
        const typedKey = keys[i] as keyof typeof obj;
        temp[typedKey] = obj[typedKey];
        delete obj[typedKey];
    }

    // Restore the data in sorted order back into the original object
    for (let i = 0; i < keys.length; ++i) {
        const typedKey = keys[i] as keyof typeof obj;
        obj[typedKey] = temp[typedKey];
    }

    return obj;
}

Answer 37

This works as of Jan 2023

 var obj = { 'b':'cVal', 'a':'aVal', 'c':'bVal' } console.log(JSON.stringify(obj)) obj = Object.fromEntries(Object.entries(obj).sort()) console.log(JSON.stringify(obj));