What is the best way to transfer an item from one object to another within the same array using javascript?

Question

What is the best way to transfer an item from one object to another within the same array using javascript?

I'm dealing with an array structure like the following:

[
        {
            "ID": 227886,
            "post_author": "54"
        },
        {
            "ID": 227545,
            "post_author": "18"
        },
        {
            "ID": 229317,
            "post_author": "22"
        },
        {
            "img": "bang.jpg",
            "id": 229317
        },
        {
            "img": "other.png",
            "id": 227886
        },
        {
            "img": "name.jpg",
            "id": 227545
        }
]

My goal is to merge the objects with an "img" key into the objects whose "ID" matches the "img" object's "id" value.

The desired output should be:

[
        {
            "ID": 227886,
            "post_author": "54",
            "img": "other.png",
            "id": 227886
        },
        {
            "ID": 227545,
            "post_author": "18",
            "img": "name.jpg",
            "id": 227545
        },
        {
            "ID": 229317,
            "post_author": "22",
            "img": "bang.jpg",
            "id": 229317
        }  
]

Is there a way to merge objects within the same array based on matching "id" and "ID" values?

javascript arrays

Answer 1

Answer №1

To organize the data based on either ID or id, you can utilize the reduce function and then convert the resulting object into an array using Object.values.

const arr = [ { "ID": 227886, "post_author": "54" }, { "ID": 227545, "post_author": "18" }, { "ID": 229317, "post_author": "22" }, { "img": "bang.jpg", "id": 229317 }, { "img": "other.png", "id": 227886 }, { "img": "name.jpg", "id": 227545 }];

const result = Object.values(arr.reduce((a, o)=>{
    if(o.ID){
        a[o.ID] = {...(a[o.ID] || {}),...o};
    } else {
        a[o.id] = {...(a[o.id] || {}),...o};
    };
    return a;
},{}));
console.log(result);

Answer 2

To organize the data based on either ID or id, you can utilize the reduce function and then convert the resulting object into an array using Object.values.

const arr = [ { "ID": 227886, "post_author": "54" }, { "ID": 227545, "post_author": "18" }, { "ID": 229317, "post_author": "22" }, { "img": "bang.jpg", "id": 229317 }, { "img": "other.png", "id": 227886 }, { "img": "name.jpg", "id": 227545 }];

const result = Object.values(arr.reduce((a, o)=>{
    if(o.ID){
        a[o.ID] = {...(a[o.ID] || {}),...o};
    } else {
        a[o.id] = {...(a[o.id] || {}),...o};
    };
    return a;
},{}));
console.log(result);

Answer 3

Answer №2

To create two separate arrays from the initial array - one containing objects with the post author and the other without:

const postAuthorList = initialArray.filter(obj => obj.hasOwnProperty('post_author'));
const nonPostAuthorList = initialArray.filter(obj => !obj.hasOwnProperty('post_author'));

Next, iterate through the postAuthorList to find the corresponding object from the nonPostAuthorList and merge them using Object.assign():

postAuthorList.forEach(postAuthor => Object.assign(postAuthor, nonPostAuthorList.find(obj => obj.id === postAuthor.ID)))

After running this code, the postAuthorList will now have the combined data. The specifics of the code may vary, but these are the general steps to follow.

Answer 4

To create two separate arrays from the initial array - one containing objects with the post author and the other without:

const postAuthorList = initialArray.filter(obj => obj.hasOwnProperty('post_author'));
const nonPostAuthorList = initialArray.filter(obj => !obj.hasOwnProperty('post_author'));

Next, iterate through the postAuthorList to find the corresponding object from the nonPostAuthorList and merge them using Object.assign():

postAuthorList.forEach(postAuthor => Object.assign(postAuthor, nonPostAuthorList.find(obj => obj.id === postAuthor.ID)))

After running this code, the postAuthorList will now have the combined data. The specifics of the code may vary, but these are the general steps to follow.

Answer 5

Answer №3

To achieve this result, you can utilize the power of Object.assign for the elements in the b array and the map() function of the a array. Here's how you can do it:

let c = a.map((row, index) => Object.assign({}, row, b[index]));

Here is the complete code snippet:

let a = [
    {
        ID: 227886,
        post_author: "69",
    },
    {
        ID: 227545,
        post_author: "22",
    },
    {
        ID: 229317,
        post_author: "18",
    },
    {
        img: "moon.jpg",
        id: 229317,
    },
    {
        img: "sun.png",
        id: 227886,
    },
    {
        img: "star.jpg",
        id: 227545,
    },
];

let b = [
    {
        img: "sun.png",
        id: 227886,
    },
    {
        img: "star.jpg",
        id: 227545,
    },
    {
        img: "moon.jpg",
        id: 229317,
    },
];

let c = a.map((row, index) => Object.assign({}, row, b[index]));

console.log(c);

Answer 6

To achieve this result, you can utilize the power of Object.assign for the elements in the b array and the map() function of the a array. Here's how you can do it:

let c = a.map((row, index) => Object.assign({}, row, b[index]));

Here is the complete code snippet:

let a = [
    {
        ID: 227886,
        post_author: "69",
    },
    {
        ID: 227545,
        post_author: "22",
    },
    {
        ID: 229317,
        post_author: "18",
    },
    {
        img: "moon.jpg",
        id: 229317,
    },
    {
        img: "sun.png",
        id: 227886,
    },
    {
        img: "star.jpg",
        id: 227545,
    },
];

let b = [
    {
        img: "sun.png",
        id: 227886,
    },
    {
        img: "star.jpg",
        id: 227545,
    },
    {
        img: "moon.jpg",
        id: 229317,
    },
];

let c = a.map((row, index) => Object.assign({}, row, b[index]));

console.log(c);

What is the best way to transfer an item from one object to another within the same array using javascript?

Answer №1

Answer №2

Answer №3

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