How to change value of object which is inside an array using JavaScript or jQuery?

Question

The code below comes from jQuery UI Autocomplete:

var projects = [
    {
        value: "jquery",
        label: "jQuery",
        desc: "the write less, do more, JavaScript library",
        icon: "jquery_32x32.png"
    },
    {
        value: "jquery-ui",
        label: "jQuery UI",
        desc: "the official user interface library for jQuery",
        icon: "jqueryui_32x32.png"
    },
    {
        value: "sizzlejs",
        label: "Sizzle JS",
        desc: "a pure-JavaScript CSS selector engine",
        icon: "sizzlejs_32x32.png"
    }
];

For example, I want to change the desc value of jquery-ui. How can I do that?

Additionally, is there a faster way to get the data? I mean give the object a name to fetch its data, just like the object inside an array? So it would be something like jquery-ui.jquery-ui.desc = ....

You would have to transform the array into a Javascript object in order to use the syntax projects["jquery-ui"].desc. Would that be worth the effort only to get nicer syntax? — Frédéric Hamidi
– Frédéric Hamidi, Commented Jan 14, 2011 at 10:26
I've updated my solution with your latest question. And you can use the "projects.jquery-ui.desc" notation. — Aston
– Aston, Commented Jan 14, 2011 at 10:28
** ↑ aston means you can use that notation if the object structure is changed as per his answer below. (Not with the OP's existing example structure.) — ashleedawg
– ashleedawg, Commented Oct 27, 2020 at 22:04
For new people, just use .find() it's a method for the array and is very useful in this case. See abe kur's answer on this. — Tigerrrrr
– Tigerrrrr, Commented Mar 2, 2021 at 3:22

General Grievance · Accepted Answer · 2024-02-14 17:07:30Z

587

It is quite simple

Find the index of the object using findIndex method.
Store the index in variable.
Do a simple update like this: yourArray[indexThatyouFind]

//Initailize array of objects.
let myArray = [
  {id: 0, name: "Jhon"},
  {id: 1, name: "Sara"},
  {id: 2, name: "Domnic"},
  {id: 3, name: "Bravo"}
],
    
//Find index of specific object using findIndex method.    
objIndex = myArray.findIndex(obj => obj.id == 1);

//Log object to Console.
console.log("Before update: ", myArray[objIndex])

//Update object's name property.
myArray[objIndex].name = "Laila"

//Log object to console again.
console.log("After update: ", myArray[objIndex])

edited Feb 14, 2024 at 17:07

General Grievance

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answered Jan 30, 2017 at 14:43

Umair Ahmed

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4 Comments

Bimal Grg Over a year ago

it will mutate myArray.

Umair Ahmed Over a year ago

Yes, But if you do not want to mutate. [...myArray.slice(0, objIndex), Object.assign({}, myArray[objIndex], myArray.slice(objIndex + 1))]

Craig Over a year ago

@UmairAhmed Shouldnt the above code be [...myArray.slice(0, objIndex), Object.assign({}, myArray[objIndex], ...myArray.slice(objIndex + 1))] ? I think ur missing the second ellipses.

trane77 Over a year ago

In my case it's perfect, I was looking for a way to get an object into an object array, then change one of its properties. I need the object to mutate because it's watched by a Vue.js component

Aston · Accepted Answer · 2014-02-13 10:26:51Z

206

You have to search in the array like:

function changeDesc( value, desc ) {
   for (var i in projects) {
     if (projects[i].value == value) {
        projects[i].desc = desc;
        break; //Stop this loop, we found it!
     }
   }
}

and use it like

var projects = [ ... ];
changeDesc ( 'jquery-ui', 'new description' );

UPDATE:

To get it faster:

var projects = {
   jqueryUi : {
      value:  'lol1',
      desc:   'lol2'
   }
};

projects.jqueryUi.desc = 'new string';

(In according to Frédéric's comment you shouldn't use hyphen in the object key, or you should use "jquery-ui" and projects["jquery-ui"] notation.)

edited Feb 13, 2014 at 10:26

answered Jan 14, 2011 at 10:05

Aston

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4 Comments

qinHaiXiang Over a year ago

Is there much faster way to get the data? I mean give the object a name to fetch its data.Just like the object inside array.So,can I used in that way : jquery-ui.jquery-ui.desc = ....

Frédéric Hamidi Over a year ago

your update won't work because of the hyphen - in the object name. You would have to write "jquery-ui": {} and projects["jquery-ui"].desc respectively.

stackdave Over a year ago

look at the abe kur's answer, that is the right, this and anothers are long

Tigerrrrr Over a year ago

For new people, just use .find() it's a method for the array and is very useful in this case. See abe kur's answer on this.

Brett Hayes · Accepted Answer · 2019-02-22 07:07:55Z

201

The best solution, thanks to ES6.

This returns a new array with a replaced description for the object that contains a value equal to "jquery-ui".

const newProjects = projects.map(p =>
  p.value === 'jquery-ui'
    ? { ...p, desc: 'new description' }
    : p
);

edited Feb 22, 2019 at 7:07

Brett Hayes

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answered Aug 7, 2018 at 2:59

kintaro

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7 Comments

Maged Mohamed Over a year ago

@FrederikoCesar not in all cases, iterating over each object costs more than slicing the array and inject the new object using spread operator

Thanasis Over a year ago

what if you want to change the value on the fly? Without creating another var? Best way is the index method: const targetIndex = summerFruits.findIndex(f=>f.id===3);

SgtPepperAut Over a year ago

this is great and short. could you write out how you would update two values at the same time from the? the shorthand notation is sure hard to understand for me. ? : are if else whats "..."

kintaro Over a year ago

@SgtPepperAut maybe this way: proj.map(p => ['jquery-ui', 'other-value'].includes(p.value) ? { ...p, desc: 'new-description' } : p )

Tigerrrrr Over a year ago

For new people, just use .find() it's a method for the array and is very useful in this case. See abe kur's answer on this.

|

OhhhThatVarun · Accepted Answer · 2019-10-20 18:19:19Z

108

Using map is the best solution without using extra libraries.(using ES6)

const state = [
{
    userId: 1,
    id: 100,
    title: "delectus aut autem",
    completed: false
},
{
    userId: 1,
    id: 101,
    title: "quis ut nam facilis et officia qui",
    completed: false
},
{
    userId: 1,
    id: 102,
    title: "fugiat veniam minus",
    completed: false
},
{
    userId: 1,
    id: 103,
    title: "et porro tempora",
    completed: true
}]

const newState = state.map(obj =>
    obj.id === "101" ? { ...obj, completed: true } : obj
);

edited Oct 20, 2019 at 18:19

OhhhThatVarun

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answered Sep 3, 2019 at 20:44

Ankit Kumar Rajpoot

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2 Comments

Noam Gal Over a year ago

this solution was provided already: stackoverflow.com/a/51718190/6543875

JeremyFc Over a year ago

This solution only works when ternary operator is used. Isn't is possible to use this method without ternary operator like --> const newState = state.map(obj => [{ ...obj, completed: true }] ); ?

Konrad · Accepted Answer · 2022-07-13 18:40:22Z

ES6 way, without mutating original data.

var projects = [
{
    value: "jquery",
    label: "jQuery",
    desc: "the write less, do more, JavaScript library",
    icon: "jquery_32x32.png"
},
{
    value: "jquery-ui",
    label: "jQuery UI",
    desc: "the official user interface library for jQuery",
    icon: "jqueryui_32x32.png"
}];

//find the index of object from array that you want to update
const objIndex = projects.findIndex(obj => obj.value === 'jquery-ui');

// Make sure to avoid incorrect replacement
// When specific item is not found
if (objIndex === -1) {
  return;
}

// make new object of updated object.   
const updatedObj = { ...projects[objIndex], desc: 'updated desc value'};

// make final new array of objects by combining updated object.
const updatedProjects = [
  ...projects.slice(0, objIndex),
  updatedObj,
  ...projects.slice(objIndex + 1),
];

console.log("original data=", projects);
console.log("updated data=", updatedProjects);

Frédéric Hamidi · Accepted Answer · 2011-01-14 10:07:18Z

37

You can use $.each() to iterate over the array and locate the object you're interested in:

$.each(projects, function() {
    if (this.value == "jquery-ui") {
        this.desc = "Your new description";
    }
});

answered Jan 14, 2011 at 10:07

Frédéric Hamidi

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Comments

Fred · Accepted Answer · 2020-03-18 04:12:33Z

24

given the following data, we want to replace berries in the summerFruits list with watermelon.

const summerFruits = [
{id:1,name:'apple'}, 
{id:2, name:'orange'}, 
{id:3, name: 'berries'}];

const fruit = {id:3, name: 'watermelon'};

Two ways you can do this.

First approach:

//create a copy of summer fruits.
const summerFruitsCopy = [...summerFruits];

//find index of item to be replaced
const targetIndex = summerFruits.findIndex(f=>f.id===3); 

//replace the object with a new one.
summerFruitsCopy[targetIndex] = fruit;

Second approach: using map, and spread:

const summerFruitsCopy = summerFruits.map(fruitItem => 
fruitItem .id === fruit.id ? 
    {...summerFruits, ...fruit} : fruitItem );

summerFruitsCopy list will now return an array with updated object.

edited Mar 18, 2020 at 4:12

Fred

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answered Oct 17, 2019 at 10:56

meol

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1 Comment

Thanasis Over a year ago

The first method is best. No need to move to another var and then back. On the fly method. I up voted your solution.

abe kur · Accepted Answer · 2017-07-19 15:11:34Z

23

you can use .find so in your example

   var projects = [
            {
                value: "jquery",
                label: "jQuery",
                desc: "the write less, do more, JavaScript library",
                icon: "jquery_32x32.png"
            },
            {
                value: "jquery-ui",
                label: "jQuery UI",
                desc: "the official user interface library for jQuery",
                icon: "jqueryui_32x32.png"
            },
            {
                value: "sizzlejs",
                label: "Sizzle JS",
                desc: "a pure-JavaScript CSS selector engine",
                icon: "sizzlejs_32x32.png"
            }
        ];

let project = projects.find((p) => {
    return p.value === 'jquery-ui';
});

project.desc = 'your value'

answered Jul 19, 2017 at 15:11

abe kur

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1 Comment

Crayons Over a year ago

While this functionally does the same thing, findIndex performs significantly better than find. If your source array is large, I recommend findIndex instead. Refer to the answer with the most upvotes.

anson · Accepted Answer · 2021-12-09 09:36:01Z

23

It's easily can be accomplished with underscore/lodash library:

  _.chain(projects)
   .find({value:"jquery-ui"})
   .merge({desc: "new desc"}).value();

Docs:
https://lodash.com/docs#find
https://lodash.com/docs#merge

edited Dec 9, 2021 at 9:36

anson

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answered Mar 15, 2016 at 11:31

user2765479

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3 Comments

Mika Over a year ago

What if 'jquery-ui' is not found by the find function?

AndrewBenjamin Over a year ago

Property 'find' does not exist on type 'LoDashExplicitArrayWrapper'

parwatcodes Over a year ago

The result of such sequences must be unwrapped with _#value. lodash.com/docs/4.17.4#chain .value()

Stephen Quan · Accepted Answer · 2021-09-13 22:43:44Z

This is another answer involving find. This relies on the fact that find:

iterates through every object in the array UNTIL a match is found
each object is provided to you and is MODIFIABLE

Here's the critical Javascript snippet:

projects.find( function (p) {
    if (p.value !== 'jquery-ui') return false;
    p.desc = 'your value';
    return true;
} );

Here's an alternate version of the same Javascript:

projects.find( function (p) {
    if (p.value === 'jquery-ui') {
        p.desc = 'your value';
        return true;
    }
    return false;
} );

Here's an even shorter (and somewhat more evil version):

projects.find( p => p.value === 'jquery-ui' && ( p.desc = 'your value', true ) );

Here's a full working version:

  let projects = [
            {
                value: "jquery",
                label: "jQuery",
                desc: "the write less, do more, JavaScript library",
                icon: "jquery_32x32.png"
            },
            {
                value: "jquery-ui",
                label: "jQuery UI",
                desc: "the official user interface library for jQuery",
                icon: "jqueryui_32x32.png"
            },
            {
                value: "sizzlejs",
                label: "Sizzle JS",
                desc: "a pure-JavaScript CSS selector engine",
                icon: "sizzlejs_32x32.png"
            }
        ];

projects.find( p => p.value === 'jquery-ui' && ( p.desc = 'your value', true ) );

console.log( JSON.stringify( projects, undefined, 2 ) );

elasticrash · Accepted Answer · 2011-01-14 10:04:53Z

19

you need to know the index of the object you are changing. then its pretty simple

projects[1].desc= "new string";

answered Jan 14, 2011 at 10:04

elasticrash

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Comments

Mohammadreza · Accepted Answer · 2021-06-24 16:17:08Z

19

const users = [
  { name: "Alex", age: 25 },
  { name: "John", age: 32 },
];

const newUsers = users.map((user) => ({
  ...user,
  age: user.age + 5, // just for example
}));

// newUsers = [
// {name:"Alex" , age:30},
// {name:"John , age:37}
// ]

answered Jun 24, 2021 at 16:17

Mohammadreza

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1 Comment

lupodellasleppa Over a year ago

somehow missing the () after the arrow function raised an error for me so this one looks to be the most correct answer for ES6

tsadkan yitbarek · Accepted Answer · 2019-03-25 07:16:27Z

13

I think this way is better

const index = projects.findIndex(project => project.value==='jquery-ui');
projects[index].desc = "updated desc";

edited Mar 25, 2019 at 7:16

answered Feb 6, 2019 at 12:16

tsadkan yitbarek

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1 Comment

avalanche1 Over a year ago

in your findIndex you're assigning a value instead of comparing

mwangaben · Accepted Answer · 2021-11-10 02:52:46Z

9

let users = [
    {id: 1, name: 'Benedict'},
    {id: 2, name: 'Myles'},
    {id: 3, name: 'Happy'},
]

 users.map((user, index) => {
 if(user.id === 1){
  users[index] = {id: 1, name: 'Baba Benny'};    
 }
 
 return user
})


console.log(users)

What this code does is map over the object and then match the desired with if statement,

if(user.id === 1)

once there is match somewhere use its index to swap

 users[index] = {id: 1, name: 'Baba Benny'};

the object in the array and then return the modified array

edited Nov 10, 2021 at 2:52

answered Aug 17, 2021 at 11:04

mwangaben

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5 Comments

Sivaram Rasathurai Over a year ago

instead of writing the code, please explain what is happening there

Jeremy Caney Over a year ago

There’s 29 answers already. The accepted answer has 183 upvotes. Why do we need another answer? What is this answer offering that’s not already covered by the previous 29 answers?

mwangaben Over a year ago

@JeremyCaney This is because there so many ways to skin a cat, also by provide readers with verities of code flow they can pick one or mix and match the approaches

sayinmehmet47 Over a year ago

@mwangaben return value will be user, not users

mwangaben Over a year ago

@sayinmehmet47 thanks for pointing it out i have already fixed

MD SHAYON · Accepted Answer · 2021-09-18 11:29:28Z

6

Change value with conditions using for each loop

projects.forEach((p,index)=>{
    if(index === 1){
       p.value = "Updated jquery-ui"
    }
})

answered Sep 18, 2021 at 11:29

MD SHAYON

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Comments

Leo Lanese · Accepted Answer · 2017-09-07 15:11:37Z

4

// using higher-order functions to avoiding mutation
var projects = [
            {
                value: "jquery",
                label: "jQuery",
                desc: "the write less, do more, JavaScript library",
                icon: "jquery_32x32.png"
            },
            {
                value: "jquery-ui",
                label: "jQuery UI",
                desc: "the official user interface library for jQuery",
                icon: "jqueryui_32x32.png"
            },
            {
                value: "sizzlejs",
                label: "Sizzle JS",
                desc: "a pure-JavaScript CSS selector engine",
                icon: "sizzlejs_32x32.png"
            }
        ];

// using higher-order functions to avoiding mutation
index = projects.findIndex(x => x.value === 'jquery-ui');
[... projects.slice(0,index), {'x': 'xxxx'}, ...projects.slice(index + 1, projects.length)];

answered Sep 7, 2017 at 15:11

Leo Lanese

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2 Comments

lazzy_ms Over a year ago

this ... before projects are necessary?

rmcsharry Over a year ago

@lazzy_ms the ... is known as the spread operator. google it :)

gandharv garg · Accepted Answer · 2019-07-23 10:30:58Z

4

You can use map function --

const answers = this.state.answers.map(answer => {
  if(answer.id === id) return { id: id, value: e.target.value }
  return answer
})

this.setState({ answers: answers })

answered Jul 23, 2019 at 10:30

gandharv garg

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Comments

keroles Monsef · Accepted Answer · 2020-07-10 01:15:11Z

4

try using forEach(item,index) helper

var projects = [
    {
        value: "jquery",
        label: "jQuery",
        desc: "the write less, do more, JavaScript library",
        icon: "jquery_32x32.png"
    },
    {
        value: "jquery-ui",
        label: "jQuery UI",
        desc: "the official user interface library for jQuery",
        icon: "jqueryui_32x32.png"
    },
    {
        value: "sizzlejs",
        label: "Sizzle JS",
        desc: "a pure-JavaScript CSS selector engine",
        icon: "sizzlejs_32x32.png"
    }
];

let search_to_change = 'jquery'

projects.forEach((item,index)=>{
   if(item.value == search_to_change )
      projects[index].desc = 'your description ' 
})

answered Jul 10, 2020 at 1:15

keroles Monsef

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Comments

Deathstalker · Accepted Answer · 2020-11-10 22:49:20Z

Here is a nice neat clear answer. I wasn't 100% sure this would work but it seems to be fine. Please let me know if a lib is required for this, but I don't think one is. Also if this doesn't work in x browser please let me know. I tried this in Chrome IE11 and Edge they all seemed to work fine.

    var Students = [
        { ID: 1, FName: "Ajay", LName: "Test1", Age: 20},
        { ID: 2, FName: "Jack", LName: "Test2", Age: 21},
        { ID: 3, FName: "John", LName: "Test3", age: 22},
        { ID: 4, FName: "Steve", LName: "Test4", Age: 22}
    ]

    Students.forEach(function (Student) {
        if (Student.LName == 'Test1') {
            Student.LName = 'Smith'
        }
        if (Student.LName == 'Test2') {
            Student.LName = 'Black'
        }
    });

    Students.forEach(function (Student) {
        document.write(Student.FName + " " + Student.LName + "<BR>");
    });

Output should be as follows

Ajay Smith

Jack Black

John Test3

Steve Test4

David_ · Accepted Answer · 2021-02-23 04:58:07Z

Assuming you wanted to run a bit more complicated codes during the modification, you might reach for an if-else statement over the ternary operator approach

// original 'projects' array;
var projects = [
    {
        value: "jquery",
        label: "jQuery",
        desc: "the write less, do more, JavaScript library",
        icon: "jquery_32x32.png"
    },
    {
        value: "jquery-ui",
        label: "jQuery UI",
        desc: "the official user interface library for jQuery",
        icon: "jqueryui_32x32.png"
    },
    {
        value: "sizzlejs",
        label: "Sizzle JS",
        desc: "a pure-JavaScript CSS selector engine",
        icon: "sizzlejs_32x32.png"
    }
];

// modify original 'projects' array, and save modified array into 'projects' variable
projects = projects.map(project => {
// When there's an object where key 'value' has value 'jquery-ui'
    if (project.value == 'jquery-ui') {

// do stuff and set a new value for where object's key is 'value'
        project.value = 'updated value';

// do more stuff and also set a new value for where the object's key is 'label', etc.
        project.label = 'updated label';

// now return modified object
        return project;
    } else {
// just return object as is
        return project;
    }
});

// log modified 'projects' array
console.log(projects);

KARTHIKEYAN.A · Accepted Answer · 2022-11-25 18:00:25Z

2

We can change in the following way

const oldArray = [{username: gopal, age: 20}, {username: gopi, age: 21}]
const obj = {username: gopal, age: 25}
const result = oldArray.map(d => d.username === 'gopi' ? d.age = obj.age : d)

answered Nov 25, 2022 at 18:00

KARTHIKEYAN.A

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Comments

Mahima Agrawal · Accepted Answer · 2017-07-27 04:43:32Z

1

We can also use Array's map function to modify object of an array using Javascript.

function changeDesc(value, desc){
   projects.map((project) => project.value == value ? project.desc = desc : null)
}

changeDesc('jquery', 'new description')

answered Jul 27, 2017 at 4:43

Mahima Agrawal

1,4351 gold badge14 silver badges13 bronze badges

1 Comment

Fernando Caride Over a year ago

this will return [null, object with the updated value, null]

Milianor · Accepted Answer · 2020-11-26 15:58:43Z

The power of javascript destructuring

const projects = [
  {
    value: 'jquery',
    label: 'jQuery',
    desc: 'the write less, do more, JavaScript library',
    icon: 'jquery_32x32.png',
    anotherObj: {
      value: 'jquery',
      label: 'jQuery',
      desc: 'the write less, do more, JavaScript library',
      icon: 'jquery_32x32.png',
    },
  },
  {
    value: 'jquery-ui',
    label: 'jQuery UI',
    desc: 'the official user interface library for jQuery',
    icon: 'jqueryui_32x32.png',
  },
  {
    value: 'sizzlejs',
    label: 'Sizzle JS',
    desc: 'a pure-JavaScript CSS selector engine',
    icon: 'sizzlejs_32x32.png',
  },
];

function createNewDate(date) {
  const newDate = [];
  date.map((obj, index) => {
    if (index === 0) {
      newDate.push({
        ...obj,
        value: 'Jquery??',
        label: 'Jquery is not that good',
        anotherObj: {
          ...obj.anotherObj,
          value: 'Javascript',
          label: 'Javascript',
          desc: 'Write more!!! do more!! with JavaScript',
          icon: 'javascript_4kx4k.4kimage',
        },
      });
    } else {
      newDate.push({
        ...obj,
      });
    }
  });

  return newDate;
}

console.log(createNewDate(projects));

user8909781 · Accepted Answer · 2017-11-21 07:05:06Z

0

Find the index first:

function getIndex(array, key, value) {
        var found = false;
        var i = 0;
        while (i<array.length && !found) {
          if (array[i][key]==value) {
            found = true;
            return i;
          }
          i++;
        }
      }

Then:

console.log(getIndex($scope.rides, "_id", id));

Then do what you want with this index, like:

$scope[returnedindex].someKey = "someValue";

Note: please do not use for, since for will check all the array documents, use while with a stopper, so it will stop once it is found, thus faster code.

answered Nov 21, 2017 at 7:05

user8909781

Comments

MiMFa · Accepted Answer · 2020-11-06 07:48:15Z

0

You can create your specific function like the below, then use that everywhere you need.

var each    = (arr, func) => 
                Array.from(
                    (function* (){
                        var i = 0;
                        for(var item of arr)
                            yield func(item, i++);
                    })()
                );

Enjoy..

edited Nov 6, 2020 at 7:48

user10153652

answered Oct 28, 2020 at 7:58

MiMFa

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1 Comment

Danish Over a year ago

Welcome to the community, if you explain what is going on and how it is working, then obviously everyone can Enjoy..

Aathi · Accepted Answer · 2020-11-23 23:25:11Z

0

upsert(array, item) { 
        const i = array.findIndex(_item => _item.id === item.id);
        if (i > -1) {
            let result = array.filter(obj => obj.id !== item.id);
            return [...result, item]
        }
        else {
            return [...array, item]
        };
    }

answered Nov 23, 2020 at 23:25

Aathi

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Comments

abhisekpaul · Accepted Answer · 2016-05-04 23:12:17Z

-1

to update multiple items with the matches use:

_.chain(projects).map(item => {
      item.desc = item.value === "jquery-ui" ? "new desc" : item.desc;
      return item;
    })

answered May 4, 2016 at 23:12

abhisekpaul

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Comments

Mohammed Kamran Azam · Accepted Answer · 2016-10-17 16:41:49Z

-1

Try this code. it uses jQuery grep function

array = $.grep(array, function (a) {
    if (a.Id == id) {
        a.Value= newValue;
    }
    return a;
});

answered Oct 17, 2016 at 16:41

Mohammed Kamran Azam

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Comments

Lorelorelore · Accepted Answer · 2018-12-14 10:27:26Z

-1

Here i am using angular js. In javascript you can use for loop to find.

    if($scope.bechval>0 &&$scope.bechval!=undefined)
    {

                angular.forEach($scope.model.benhmarkghamlest, function (val, key) {
                $scope.model.benhmarkghamlest[key].bechval = $scope.bechval;

            });
    }
    else {
        alert("Please sepecify Bechmark value");
    }

edited Dec 14, 2018 at 10:27

Lorelorelore

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answered Dec 14, 2018 at 9:27

Hari Lakkakula

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Comments

Aran Tamool · Accepted Answer · 2023-02-08 08:29:20Z

-1

The easiest way is to do this

    var projects = [
{
    
    
    value: "jquery",
    label: "jQuery",
    desc: "the write less, do more, JavaScript library",
    icon: "jquery_32x32.png"
},
{
    value: "jquery-ui",
    label: "jQuery UI",
    desc: "the official user interface library for jQuery",
    icon: "jqueryui_32x32.png"
}];

projects.find(data => data.value === "jquery").label ="xxxx"

console.log("------------>",projects)

answered Feb 8, 2023 at 8:29

Aran Tamool

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