How to uppercase Javascript object keys?

Question

Anyone know a good way to turn this?:

var obj = [{key1: value1,key2: value2},{key3: value3,key4: value4}];

into:

var obj = [{Key1: value1,Key2: value2},{Key3: value3,Key4: value4}];

Any reason you'd want to do this? Any code which is using the old keys will break, because keys are case sensitive. — Marc B, Sep 13 '11 at 21:19
Are you looking for a regex or a way to do it while the code is running? — beatgammit, Sep 13 '11 at 22:02
@Marc B, I very well might choose to do this if, for instance, I'm making jqgrids and the column names have a capitalized first letter. Then say I'm getting json back from an existing API and the keys actually have to match case so the jqgrid will function properly. In this case, I would want to guarantee that the keys I'm getting from the json have an uppercase first letter as well and ergo, match. — Brandon Minton, Sep 14 '11 at 15:17
@tjameson, given the frequency of which I have to perform this operation, speed is not a bottleneck so I would say whatever function works including a regex. — Brandon Minton, Sep 14 '11 at 15:19

Joe · Accepted Answer · 2021-06-01T14:35:06.173

21

Loop through delete and replace:

var obj = [{key1: 1,key2: 1},{key3: 1,key4: 1}];
for(var i = 0; i<obj.length;i++) {

    var a = obj[i];
    for (var key in a) {
        if (a.hasOwnProperty(key)) {
          a[key.charAt(0).toUpperCase() + key.substring(1)] = a[key];
          delete a[key];
          
        }
    }
    obj[i] = a;

}

edited Jun 01 '21 at 14:35

answered Sep 13 '11 at 21:25

Joe

80,724
18
127
145

+1, but note if the starting object contains both a lowercase and initial-caps version of a particular keyname then the latter will be overwritten. I.e., `{key1:1, Key1:2}` will end up as `{Key1:1}`. (Also, the `obj[i] = a;` after the `for` is redundant isn't it?) – nnnnnn Sep 13 '11 at 23:56
The only difference is that I used slice instead of substring. Otherwise, this is excellent. – Brandon Minton Sep 14 '11 at 16:12
1

You can delete the original property (a[key]) after you set the new property, and then you won't need the temp variable. – Programmer May 31 '21 at 08:51
Object.keys(a).forEach((key) => { a[key.charAt(0).toUpperCase() + key.substring(1)] = a[key]; delete a[key]; }); – A. Horst Feb 03 '22 at 12:25

score 9 · Answer 2 · answered Mar 03 '20 at 20:32

9

As of 2019 you can use Object.fromEntries:

let populations = {london: 8.9, beijing: 21.54, mumbai: 18.41};  // March 2020

let entries = Object.entries(populations);
let capsEntries = entries.map((entry) => [entry[0][0].toUpperCase() + entry[0].slice(1), entry[1]]);
let capsPopulations = Object.fromEntries(capsEntries);

console.log(capsPopulations);

answered Mar 03 '20 at 20:32

Jeremiah England

662
9
15

4

You can also use destructuring here: `entries .map (([k, v]) => [k [0] .toUpperCase () + k .slice (1), v])`. – Scott Sauyet Mar 03 '20 at 20:38

score 3 · Answer 3 · edited Oct 07 '21 at 18:51

3

Another approach (more clean)

import * as _ from 'lodash';


function capitalizeObjectKeys(obj) {
  return _.transform(obj, (result, val, key) => {
    result[key.charAt(0).toUpperCase() + key.slice(1)] = val;
  });
}

https://stackblitz.com/edit/js-hsqutg?embed=1&file=index.js

edited Oct 07 '21 at 18:51

Nathan Arthur

8,287
7
55
80

answered Jun 18 '21 at 13:53

Sorin Veștemean

1,772
19
17

score 1 · Answer 4 · answered Jan 15 '22 at 08:13

In my case this code worked well. Both in uppercase and lowercase with map:

to uppercase :


const newDataUpperCase = data.map( function( item ){
        for(var key in item){
            var upper = key.toUpperCase();
            if( upper !== key ){ 
              item[ upper ] = item;
                delete item[key];
            }
        }
        return item;
    });

to lowercase :


const newDataUpperCase = data.map( function( item ){
        for(var key in item){
            var lower= key.toLowerCase();
            if( lower!== key ){ 
              item[ lower] = item;
                delete item[key];
            }
        }
        return item;
    });

score 1 · Answer 5 · answered Apr 29 '22 at 07:57

This will help you:

const griddata = [
  {
      "id_pk": 238,
      "acT_ID": 238,
      "desc": "Record 2",
      "parent": 0,
      "compdays": 5,
      "logical": "1",
      "quantity": "1",
      "lisT_ID": 75,
      "empL_ID": 1388,
      "default": 8,
      "level": "0",
      "sortorder": 2,
      "vfpRecNo": 0,
      "isDeleted": false,
      "clientID": 1,
      "empl_num": "1388",
      "duedate": "05/04/2022"
  }
]

const upperCaseKeys = (data) => {
    debugger;
    let gridData = [];
    if (data != null && data != undefined && data.length > 0) {
      let keys = Object.keys(data[0]);
      let upperCaseKey = [];

      for (let j = 0; j < data.length; j++) {
        upperCaseKey = [];
        for (let i = 0; i < keys.length; i++) {
          set(upperCaseKey, keys[i].toUpperCase(), data[j][keys[i]]);
        }
        upperCaseKey.push(...upperCaseKey);

        gridData.push(Object.assign({}, upperCaseKey));
      }
    }
    console.log("upperCaseKeys",gridData)
  }

const set = (obj, prop, value) => {
    obj[prop] = value;
}

upperCaseKeys(griddata)

Output:

upperCaseKeys [
  {
    ID_PK: 238,
    ACT_ID: 238,
    DESC: 'Record 2',
    PARENT: 0,
    COMPDAYS: 5,
    LOGICAL: '1',
    QUANTITY: '1',
    LIST_ID: 75,
    EMPL_ID: 1388,
    DEFAULT: 8,
    LEVEL: '0',
    SORTORDER: 2,
    VFPRECNO: 0,
    ISDELETED: false,
    CLIENTID: 1,
    EMPL_NUM: '1388',
    DUEDATE: '05/04/2022'
  }
]

score 0 · Answer 6 · answered Dec 17 '20 at 16:14

    const transform = (arrObj) => {
    let newObj=[];
    for(let obj of arrObj) {
      let temp=new Object();
      let keys = Object.keys(obj);
      let values = Object.values(obj);
      let i=0;
      keys.forEach(key => {
          key=key[0].toUpperCase() + key.slice(1);
          temp[`${key}`]=values[i++];
      });
      newObj.push(temp);
    }
    return newObj;
  };
  const arrObj = [{first: 'satya', last:'prakash'}, {college: 'AIT'}];
  console.log(transform(arrObj));

score 0 · Answer 7 · answered Feb 21 '21 at 06:33

If you need to capitalize just first letter you can go with Jeremiah's answer

or if you need to capitalize first letter of each word consider the following

let populations = {"london": 8.9, "beijing": 21.54, "mumbai": 18.41, "new york": 19.4};
let entries = Object.entries(populations);
// Capitalize first letter of each word 
let capsEntries = entries.map((entry) => [entry[0].replace(/(^\w{1})|(\s+\w{1})/g, letter => letter.toUpperCase()), entry[1]]);
let capsPopulations = Object.fromEntries(capsEntries);

console.log(capsPopulations)

Regex adopted from this answer

Regex Explanation:

(^\w{1}): match first char of string
|: or
(\s{1}\w{1}): match one char that came after one space
g: match all
match => match.toUpperCase(): replace with can take function, so; replace match with upper case match

score -5 · Answer 8 · answered Jun 23 '17 at 05:38

-5

use lodash's _.capitalize function

_.capitalize('firstName') => FirstName

answered Jun 23 '17 at 05:38

feeziman007

7

How to uppercase Javascript object keys?

8 Answers8

Linked