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Merge 3 arrays by matching ID's

I have a small exercise, to merge three arrays into one array by matching ID's, I'd like to add the likes and comments array into the posts array how would I go about this?

const posts = 
  [ { _id: 1
    , message: 'this is post one'
    , likes:    [ { id: 1111 } , { id: 2222 } ] 
    , comments: [ { id: 3333 } , { id: 4444 } ] 
    } 
  ] 
const comments = 
  [ { id: 3333, message: 'this is comment 1' } 
  , { id: 4444, message: 'this is comment 2' } 
  ] 
const likes = 
  [ { id: 1111, user: 'Peter' } 
  , { id: 2222, user: 'John'  } 
  ] 

expected output:

newArray = 
  [ { _id: 1
    , message: 'this is post one'
    , likes: 
      [ { id: 1111, user: 'Peter'} 
      , { id: 2222, user: 'John'} ] 
    , comments: 
      [ { id: 3333, message: 'this is comment 1'} 
      , { id: 4444, message: 'this is comment 2'} 
  ] } ] 

Here is a similar approach I've used when working with two arrays:

const
  arr1 = [{ groupId: "Category", options: [{ name: "cookies", id: 1111 }, { name: "tv", id: 2222 }] }, { groupId: "Brand", options: [{ name: "random", id: 3333 }] }, { groupId: "Price", options: [{ name: "random2", id: 4444 }] }],
  arr2 = [{ id: 1111, url: "/test", other: 'other-prop' }, { id: 2222, url: "/test1" }, { id: 3333, url: "/test2" }],

  mergeById = (arr1, arr2) =>
    arr1.map(({ options, ...rest }) => (
      {
        ...rest,
        options: options.map(option => (
          {
            ...option,
            ...arr2.find(_option => _option.id === option.id)
          }
        ))
      }
    ));

const merged = mergeById(arr1, arr2);

console.log(JSON.stringify(merged, null, 2));
about 4 years ago · Juan Pablo Isaza
1 Respuestas
Responde la pregunta

0

const newPosts = posts.map(post => ({
    ...post,
    likes: likes.filter(like => post.likes.map(like => like.id).includes(like.id)),
    comments: comments.filter(comment => post.comments.map(comment => comment.id).includes(comment.id)),
}))

console.log(newPosts)
about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
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