CODEWITHSUNDEEP
javascriptarrayssortingobjectjavascript-objects
DayIsGreen
DayIsGreen

Reputation: 265

How to reconstruct array of objects based on duplicate keys?

I'm getting an array object that looks like this. My goal is to combine or merge them based on duplicate keys.

documents: [
    {
        image: 'sample image 1',
        id_side: 'Front',
        type: 'Passport'
    },
    {
        image: 'sample image 2',
        id_side: 'Back',
        type: 'Passport'
    },
    {
        image: 'sample image 3',
        id_side: 'Back',
        type: 'License'
    }
]

How can I arrange it to look like this?

documents: [
    {
        documentType: 'Passport',
        requiredDocs: [
            {
                image: 'sample image 1',
                id_side: 'Front',
                type: 'Passport'
            },
            {
                image: 'sample image 2',
                id_side: 'Back',
                type: 'Passport'
            }
        ]
    },
    {
        documentType: 'License',
        requiredDocs: [
            {
                image: 'sample image 3',
                id_side: 'Back',
                type: 'License'
            }
        ]
    }
]

I have found a similar question but I can't seem to make it work in my case. See the similar question in this link: How to merge/combine array of objects based on duplicate keys?

Upvotes: 0

Views: 266

Answers (4)

Arjun Singh
Arjun Singh

Reputation: 312

First you can use an Object {} to store as key value pair such that unique items will be treated as key:

const data = [{
  image: 'sample image 1',
  id_side: 'Front',
  type: 'Passport'
}, {
  image: 'sample image 2',
  id_side: 'Back',
  type: 'Passport'
}, {
  image: 'sample image 3',
  id_side: 'Back',
  type: 'License'
}];
const Obj = {}, arr = []
data.forEach(element => {
  if (Obj[element.type]) Obj[element.type].push(element)
  else Obj[element.type] = [element]
})
for (const eachType in Obj) {
  arr.push({ documentType: eachType, requiredDocs: Obj[eachType] })
}
console.log(arr)

Upvotes: 0

Abbas Shaikh
Abbas Shaikh

Reputation: 324

You can achieve your output with this

const documents = [
  {
    image: "sample image 1",
    id_side: "Front",
    type: "Passport"
  },
  {
    image: "sample image 2",
    id_side: "Back",
    type: "Passport"
  },
  {
    image: "sample image 3",
    id_side: "Back",
    type: "License"
  }
];

let output = {};
  documents.forEach((docs) => {
    if (output[docs.type]) {
      output[docs.type].requiredDocs.push(docs);
    } else {
      output[docs.type] = {
        documentType: docs.type,
        requiredDocs: [docs]
      };
    }
  });

  console.log(Object.values(output));

Upvotes: 2

Jack Bashford
Jack Bashford

Reputation: 44125

This assumes that the key that you want to deal with is type. Build an object mapping from the type to all the relevant objects, and then map it to the desired format.

const obj={documents:[{image:'sample image 1',id_side:'Front',type:'Passport'},{image:'sample image 2',id_side:'Back',type:'Passport'},{image:'sample image 3',id_side:'Back',type:'License'}]};

let types = obj.documents.reduce((a, { type, ...r }) => {
  a[type] = a[type] || [];
  a[type].push({ type, ...r });
  return a;
}, {});

let result = Object.entries(types).map(([ documentType, requiredDocs ]) => ({ documentType, requiredDocs }));

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: auto; }

Upvotes: 0

Robby Cornelissen
Robby Cornelissen

Reputation: 97162

Use reduce() to create an accumulation object, and then map() the entries of that object to your desired output format:

const data = [{
  image: 'sample image 1',
  id_side: 'Front',
  type: 'Passport'
}, {
  image: 'sample image 2',
  id_side: 'Back',
  type: 'Passport'
}, {
  image: 'sample image 3',
  id_side: 'Back',
  type: 'License'
}];

const result = Object.entries(data.reduce((a, v) => ({
  ...a,
  [v.type]: [...a[v.type] || [], v]
}), {})).map(([documentType, requiredDocs]) => ({
  documentType,
  requiredDocs
}));

console.log(result);

Upvotes: 0

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