How do I display only part of the array after findOne in MongoDB?

Question

Restaurants is a collection and has objects like below:

{
    _id: new ObjectId("61723c7378b6d3a5a02d908e"),
    name: 'The Blue Hotel',
    location: 'Noon city, New York',
    phoneNumber: '122-536-7890',
    website: 'http://www.bluehotel.com',
    priceRange: '$$$',
    cuisines: [ 'Mexican', 'Italian' ],
    overallRating: 0,
    serviceOptions: { dineIn: true, takeOut: true, delivery: true },
    reviews: [
      {
        _id: new ObjectId("61736a0f65b9931b9e428789"),
        title: 'asd',
        reviewer: 'khoh',
        rating: 3,
        dateOfReview: '5/12/2002',
        review: 'hey'
      },
        _id: new ObjectId("61736a0f65b9931b9e428790"),
        title: 'dom',
        reviewer: 'firuu',
        rating: 4,
        dateOfReview: '25/1/2002',
        review: ' bruh'
      }
    ]
  }

I am using the below code to find this object based on the review id provided

async get(reviewId) {

    const restaurantsCollection = await restaurants();
    reviewId = ObjectId(reviewId)
    const r = await restaurantsCollection.findOne({reviews: {$elemMatch: {_id: reviewId}}})
    return r

This returns the whole object from the restaurant collection, what do I do if I want only the review displayed whose id is provided in get(reviewID)

Output:

{
  _id: new ObjectId("61736a0f65b9931b9e428790"),
   title: 'dom',
   reviewer: 'firuu',
   rating: 4,
   dateOfReview: '25/1/2002',
   review: ' bruh'
}

Pramisha C P · Accepted Answer · 2021-10-23 20:21:03Z

3

With a projection, specify the fields to return

The following returns only the review whose id is provided in get(reviewID)

async get(reviewId) {
  const restaurantsCollection = await restaurants();
  reviewId = ObjectId(reviewId)

  const r = await restaurantsCollection.findOne(
    { reviews: { $elemMatch: { _id: reviewId } } },
    { "reviews.$": 1 }
  )

  return r
}

Test Here

You can also use find instead of fineOne

edited Oct 23, 2021 at 20:21

answered Oct 23, 2021 at 19:55

Pramisha C P

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5 Comments

Swayam Shah Over a year ago

It returns all the reviews, I want to display only the matched review

Pramisha C P Over a year ago

Hi, I have updated the code. Please check @swombhai

Swayam Shah Over a year ago

I still get the entire object when I return r, not sure why

Takis Over a year ago

the findOne method of node js as far as i know takes the projection as option,so it should replaced by {"projection" : { "reviews.$": true }} test it maybe

Swayam Shah Over a year ago

Is there a way to represent the _id with a string instead of new ObjectID? Can toString() be added to findOne function?

Takis · Accepted Answer · 2021-10-23 20:42:15Z

1

Query

replace the ObjectId("61736a0f65b9931b9e428789") with reviewId
this will return the reviews that match the _id in an array
if you want to get the first only, in case there is always max 1 you can replace the last project with {"$project": {"_id": 0, "review": {"$arrayElemAt": ["$reviews", 0]}}}

*not sure if this is what you need

Test code here

aggregate(
[{"$match": {"reviews._id": ObjectId("61736a0f65b9931b9e428789")}}
 {"$set": 
   {"reviews": 
     {"$filter": 
       {"input": "$reviews",
        "cond": 
        {"$eq": ["$$this._id", ObjectId("61736a0f65b9931b9e428789")]}}}}},
  {"$project": {"_id": 0, "reviews": 1}}])

edited Oct 23, 2021 at 20:42

answered Oct 23, 2021 at 19:53

Takis

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3 Comments

Swayam Shah Over a year ago

Do I have to put toArray() anywhere? I replaced the code but didn't get the output

Takis Over a year ago

yes aggreagte will return a cursor , use result.toArray()

Takis Over a year ago

the @Pramisha answer looks more like your code, i think use the other answer, this is aggregate solution, here you can do it with find only.

Web Dev · Accepted Answer · 2021-10-23 19:44:45Z

0

This might not be the correct answer of your question but you can try something like this.

const r = await restaurantsCollection.findOne({reviews: {$elemMatch: {_id: reviewId}}})?.reviews.find(review => review._id.equals(reviewId))

answered Oct 23, 2021 at 19:44

Web Dev

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