Merge two array object : one has key value pair and another is just array

Question

I have two array objects one is ArrayX and other is ArrayY. ArrayX has user_id and store_ids[] and ArrayY has store_id and store_name I want to merge both the arrays according to the ArrayX's store_ids.

```
//First array
ArrayX = [
  {
    user_id: 'user 4',
    store_ids: [ 'store 2','store 4', 'store 1' ],
  },
  {
    user_id: 'user 6',
    store_ids: [ 'store 1', 'store 2' ],
  }
]


//second array
ArrayY = [
  {
    store_id: 'store 4',
    store_name: 'store D'
  },
  {
    store_id: 'store 2',
    store_name: 'store B'
  },
  {
    store_id: 'store 1',
    store_name: 'store A'
  },
  {
    store_id: 'store 3',
    store_name: 'store C'
  }
] 
```

and what i wanted is given below.

```
ArrayZ = [
  {
    user_id: 'user 4',
    store_ids: [ 'store 2','store 4', 'store 1' ],
    store_info : [
            {
                    store_id: 'store 2',
                    store_name: 'store B'
            },
            {
                    store_id: 'store 4',
                    store_name: 'store D'
            },
            {
                    store_id: 'store 1',
                    store_name: 'store A'
            }       
        ]
  },
  {
    user_id: 'user 6',
    store_ids: [ 'store 1', 'store 2' ],
    store_info: [
            {
                    store_id: 'store 1',
                    store_name: 'store A',
            },
            {
                store_id: 'store 2',
                store_name: 'store B',
            }
        ]
  }
]
```

I tried the map function but not getting the desired result that is mentioned above.

let ArrayZ = ArrayX.map((item, i) => Object.assign({}, item, ArrayY[i])) 

[
  {
    user_id: 'user 4',
    store_ids: [ 'store 2', 'store 4','store 1' ],
    store_id: 'store 4',
    store_name: 'store D',
  },
  {
    user_id: 'user 6',
    store_ids: [ 'store 1', 'store 2'],
    store_id: 'store 2',
    store_name: 'store B',
  },
Thi is what i am getting.

can anyone suggest something on this.

Stefan Dobre · Accepted Answer · 2020-09-24 09:19:22Z

1

This is how i would do it:

const arrayZ = [];

const sort = () =>{
    
    arrayX.forEach(element => {
        let store_info = [];
        arrayY.forEach(el =>{
            if(element.store_ids.includes(el.store_id)){
                store_info.push(el);
                console.log(store_info)
            }
        })
        element.store_info = store_info;
        arrayZ.push(element);
    })
} 

sort();

console.log(arrayZ);

You can even refactor the function a bit to take 2 arrays as arguments.... Like this you keep both arrayX and arrayY intact, if you need it...

answered Sep 24, 2020 at 9:19

Stefan Dobre

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Nina Scholz · Accepted Answer · 2020-09-24 13:12:32Z

1

You could take an object for all stores and then map new objects with the store names.

This approach requires only two loops.

const
    array1 = [{ user_id: 'user 4', store_ids: ['store 2', 'store 4', 'store 1'] }, { user_id: 'user 6',  store_ids: ['store 1', 'store 2'] }],
    array2 = [{ store_id: 'store 4', store_name: 'store D' }, { store_id: 'store 2', store_name: 'store B' }, { store_id: 'store 1', store_name: 'store A' }, { store_id: 'store 3', store_name: 'store C' }],
    stores = Object.fromEntries(array2.map(o => [o.store_id, o])),
    result = array1.map(o => ({ ...o, store_info: o.store_ids.map(id => stores[id]) }));

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Sep 24, 2020 at 13:12

answered Sep 24, 2020 at 8:10

Nina Scholz

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1 Comment

Vandana Maurya Over a year ago

Thanks for the effort mate, but this script is not giving the exact array object what I wanted. checkout the first suggestion.

kako-jun · Accepted Answer · 2020-09-24 08:19:03Z

0

I always write as follows. I think it's readable.

    const ArrayZ = ArrayX.map((x) => {
      const store_infos = [];
      for (const store_id of x.store_ids) {
        const store_info = ArrayY.find((y) => y.store_id === store_id);
        store_infos.push(store_info);
      }

      return {
        user_id: x.user_id,
        store_ids: x.store_ids,
        store_info: store_infos,
      };
    });

    console.log(JSON.stringify(ArrayZ, null, 2));

answered Sep 24, 2020 at 8:19

kako-jun

1831 silver badge9 bronze badges

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