How to push objects from array in object to another array?

Question

I get in response an array with deposits, and I need to get depositId, but multicurrency deposits have an array of several deposits. (First array) As a result, I need to have an array with deposits, without linkedDeposits. (On the example of the second array) I just can’t figure out how to do this

[ {
    depositId: 1111, 
    depositName: 'Test',
    depositCur: 'RUB'
}
,
{
    depositName: 'Test',
    linkedDeposits: [ {
        depositName: 'Test',
        depositId: 2222, 
        sepositName: 'Test RUB', 
        depositCur: 'RUB'
    }
    ,
    {
        depositName: 'Test',
        depositId: 3333, 
        sepositName: 'Test USD', 
        depositCur: 'USD'
    }
    ,
    ]
}
]

[ {
    depositId: 1111, 
    depositName: 'Test',
    depositCur: 'RUB'
}
,
{
   depositName: 'Test',
   depositId: 2222, 
   sepositName: 'Test RUB', 
   depositCur: 'RUB'
}
,
{
   depositName: 'Test',
   depositId: 3333, 
   sepositName: 'Test USD', 
   depositCur: 'USD'
}
]

The original will always be in such structure or there can be more element ? — Code Maniac
– Code Maniac, Commented Oct 7, 2019 at 12:36

Bilal Siddiqui · Accepted Answer · 2019-10-07 12:45:44Z

4

You could just check if depositId exists then add item else add linkDeposits array into final array:

var data = [ {
    depositId: 1111, 
    depositName: 'Test', 
    depositCur: 'RUB'
}, {
    depositName: 'Test',
    linkedDeposits: [ {
        depositName: 'Test',
        depositId: 2222, 
        sepositName: 'Test RUB', 
        depositCur: 'RUB'
    },{
        depositName: 'Test',
        depositId: 3333, 
        sepositName: 'Test USD', 
        depositCur: 'USD'
    }]
}];

var result = [];
data.forEach(function(item){
    if (item.depositId)
       result.push(item);
    else
       result = result.concat(item.linkedDeposits);
});

console.log(result);

answered Oct 7, 2019 at 12:45

Bilal Siddiqui

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Comments

Ulysse BN · Accepted Answer · 2019-10-07 12:59:09Z

1

One more solution that I find elegant since it doesn't mutate original array, nor any array:

// Simplified data to focus on actual answer.
const data = 
  [ { id: 1111 }
  , { linkedDeposits: [ { id: 2222 }, { id: 3333 } ] }
  ]

const newData = data.flatMap(deposit =>
  deposit.linkedDeposits || deposit
)

console.log(newData)

See array.flatMap() for more details about this method.

Beware, this function is part of ES2019 ecosystem and might not be available for all users. See browsers compatibility.

edited Oct 7, 2019 at 12:59

answered Oct 7, 2019 at 12:50

Ulysse BN

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8 Comments

user4676340 Over a year ago

Could you use the original data ?

Ulysse BN Over a year ago

I'd rather focus on the answer part. However, this would work with original data as well. I hope this is not why you downvoted :/

user4676340 Over a year ago

I actually wanted to upvote for the use of flatMap, sorry about that ! But, still, could you use the original data ?

Andreas Over a year ago

Just for the "one-liner"-fraction: deposit.linkedDeposits ? deposit.linkedDeposits : deposit can be reduced to deposit.linkedDeposits || deposit

user4676340 Over a year ago

@Andreas As per SOF guidelines, it's better to include relevant information into the answer (in case the website is down for some reason). An anyway, it's always nicer to not have to open a link and just see it in the answer !

|

score 1 · Accepted Answer · 2019-10-07 14:44:46Z

1

here is my take on your issue.

Start by reducnig your array, then remove the deposits that still contains inner deposits. That leaves you with the expected result.

const original = [{
  depositId: 1111,
  depositName: 'Test',
  depositCur: 'RUB'
}, {
  depositName: 'Test',
  linkedDeposits: [{
    depositName: 'Test',
    depositId: 2222,
    sepositName: 'Test RUB',
    depositCur: 'RUB'
  }, {
    depositName: 'Test',
    depositId: 3333,
    sepositName: 'Test USD',
    depositCur: 'USD'
  }]
}];

const modified = original.reduce(
  // Maryannah's solution (less compact)
  // (p, n) => (Array.prototype.concat.apply([], [...p, ...(n.linkedDeposits || []), n])),
  // CodeManiac's solution
  (p, n) => [...p, ...(n.linkedDeposits || []), n],
  []
).filter(deposit => !deposit.linkedDeposits);

console.log(modified);

(The code is minified to its maximum though.)

edited Oct 7, 2019 at 14:44

answered Oct 7, 2019 at 12:51

user4676340

3 Comments

Code Maniac Over a year ago

can be minified further i guess, (p, n) => [...p, ...(n.linkedDeposits || []), n],...

user4676340 Over a year ago

@CodeManiac not sure it flattens the array by doing so. Care to make a snippet out of it ?

Code Maniac Over a year ago

JS bin demo

cccn · Accepted Answer · 2019-10-07 12:45:28Z

0

You can use reduce method, like this:

const data = [{
	depositId: 1111,
	depositName: 'Test',
	depositCur: 'RUB'
}, {
	depositName: 'Test',
	linkedDeposits: [{
		depositName: 'Test',
		depositId: 2222,
		sepositName: 'Test RUB',
		depositCur: 'RUB'
	},
	{
		depositName: 'Test',
		depositId: 3333,
		sepositName: 'Test USD',
		depositCur: 'USD'
	}]
}]

const result = data.reduce((acc, elem) => {
	!elem.hasOwnProperty('linkedDeposits') ? acc.push(elem) : elem.linkedDeposits.forEach(el => acc.push(el))
	return acc
}, [])

console.log(result)

answered Oct 7, 2019 at 12:45

cccn

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Comments

RamAnji · Accepted Answer · 2019-10-07 12:46:36Z

-1

let tempVal = [ {
    depositId: 1111, 
    depositName: Test, 
    depositCur: RUB
}
,
{
    depositName: Test,
    linkedDeposits: [ {
        depositName: Test,
        depositId: 2222, 
        sepositName: Test RUB, 
        depositCur: RUB
    }
    ,
    {
        depositName: Test,
        depositId: 3333, 
        sepositName: Test USD, 
        depositCur: USD
    }
    ,
    ]
}
]

let newArray = [];
for(let i = 0; tempVal.length > i; i++){
if(tempVal[i].depositId != null && tempVal[i].depositId != undefined){
  this.newArray({depositId: tempVal[i].depositId, depositName: tempVal[i].depositName, depositCur: tempVal[i].depositCur})
}
if(tempVal[i].linkedDeposits.length > 0){
  for(let j = 0; tempVal[i].linkedDeposits.length > j;j++){
  if(tempVal[i].linkedDeposits[j].depositId != null && tempVal[i].linkedDeposits[j].depositId != undefined){
  this.newArray({depositId: tempVal[i].linkedDeposits[j].depositId, depositName: tempVal[i].linkedDeposits[j].depositName, depositCur: tempVal[i].linkedDeposits[j].depositCur})
   }
  }
}
}

answered Oct 7, 2019 at 12:46

RamAnji

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1 Comment

Gibor Over a year ago

you're doing alot of unecessary calculations and loops without using ANY js arrays method... its like writing it in C -.-

Gibor · Accepted Answer · 2019-10-07 13:07:06Z

-1

Use the array.forEach() to loop on all deposites, if they have linkedDeposites array - concat it to result array. else- push the single deposite to result array.

let res = [];
sourceArr.forEach( deposite => {
  if( deposite.hasOwnProperty('linkedDeposites') )
     res = res.concat(deposite.linkedDeposites);
  else
    res.push(deposite);
});

NOTE: this will not work if any deposite inside "linkedDeposites" has its own "linkedDeposites". if that is the case comment down below and I'll rewrite my answer to take care of that as well.

edited Oct 7, 2019 at 13:07

answered Oct 7, 2019 at 12:46

Gibor

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2 Comments

Bilal Siddiqui Over a year ago

concat doesn't auto add data into any array of its arg but it returns an array with merged value. So it should be res = res.concat(deposite.linkedDeposites);

Gibor Over a year ago

@BilalSiddiqui you're right, tnx for the correction :)

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