compare two different array of objects and filter out the item which is having same ID from one arra-CodePudding

I am trying to compare and filter out the two different array of objects I need to check against all the elements in both the array of objects and need to filter out the item which is having the same ID, meaning I need to filter the identical Item from only one array another array should have that item. I have tried using filter some and find methods all are giving me the same element(id) from both the array, but I need to filter out from my 2nd array only.

// Example :

// Lets's say I am having following array of objects:

const a = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
  { id: '123456', description: 'PrdThree', version: '2', categories: '--' },
];

const b = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
];

// I am trying to get something like below:

const res = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
  { id: '123456', description: 'PrdThree', version: '2', categories: '--' },
];

Basically I need to check for each and every element and I need to ignore the items which is same in the array b only I should get the items in array a , and also I don't want that item from const b to be removed, I need to check both the arrays and if it matches I need to go with only A at the same time items inside the B should not be removed or deleted.

Solution I tried below giving me the same item from both the list. Could any one help me out how can I achieve this? Thanks in advance`

const a = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
  { id: '123456', description: 'PrdThree', version: '2', categories: '--' },
];

const b = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
];

let result = a.filter(o => !b.some(v => v.id === o.id));

console.log(result);

CodePudding user response：

you can try this :

let arr = []

    let res = a.filter(o => !b.some(v => v.id === o.id))
    let res2 = a.filter(o => b.some(x => x.id == o.id))
    arr.push(...res2 , ...res)

CodePudding user response：

You could take a closure over a Set and filter an array of all items.

const
    a = [{ id: '1234', description: 'PrdOne', version: '3', categories: '--' }, { id: '12345', description: 'PrdTwo', version: '2', categories: '--' }, { id: '123456', description: 'PrdThree', version: '2', categories: '--' }],
    b = [{ id: '1234', description: 'PrdOne', version: '3', categories: '--' }, { id: '12345', description: 'PrdTwo', version: '2', categories: '--' }],
    result = [...a, ...b].filter(
        (s => ({ id }) => !s.has(id) && s.add(id))
        (new Set)
    );

console.log(result);

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

CodePudding user response：

    const a = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
  { id: '123456', description: 'PrdThree', version: '2', categories: '--' },
];

const b = [
  { id: '1234', description: 'PrdOne', version: '3', categories: '--' },
  { id: '12345', description: 'PrdTwo', version: '2', categories: '--' },
];
function some(el){
    const objectKeys=Object.keys(el);
    for(let elemnets of b){
        if(objectKeys.length!=Object.keys(elemnets).length)continue;
        let found=1;
        for(let key of objectKeys){
            if(el[key]!=elemnets[key]){
                found=0;
                break;
            }
        }
        if(found)return true;
    }
    return false
}
let result = a.filter((el)=>{
    return some(el);
});

console.log(result);

try this