how to find array element using auto generated id in mongodb

Question

currently, I am struggling with how the MongoDB document system works. I want to fetch array elements with an auto-generated id but how to fetch that specific data that I don't know.

my current schema is

const ItemPricesSchema = new mongoose.Schema({
    _id : {
        type: String
    },
    ItemsPrices: {
        type: [{
            barcode : {
                type: String
            },
            itemName : {
                type: String
            },
            price : {
                type: String
            }
        }]
    }
});

current data is stored in this way

{
    "_id": "[email protected]",
    "ItemsPrices": [
        {
            "barcode": "345345",
            "itemName": "maggie",
            "price": "45",
            "_id": "620a971e11120abbde5f4c3a"
        },
        {
            "barcode": "356345",
            "itemName": "monster",
            "price": "70",
            "_id": "620a971e11120abbde5f4c3b"
        }
    ],
    "__v": 0
}

what I want to achieve is that I want to find array elements through ids

if I want a specific array element with id "620a971e11120abbde5f4c3b" what should I do?? I have tried $unwind , $in, $match...

the result should be like

{
    "_id": "[email protected]",
    "ItemsPrices": [
        {
            "barcode": "356345",
            "itemName": "monster",
            "price": "70",
            "_id": "620a971e11120abbde5f4c3b"
        }
    ],
    "__v": 0
}

what I tried is like this from the answer

router.get('/filter/:id', async (req, res) => {
    try {
        const item = await ItemPricesSchema.aggregate([
            {$project: {
                "ItemsPrices": {
                    $filter: {
                        input: "$ItemsPrices",
                        as: "item",
                        cond: {
                            $eq: [
                                "$$item._id",
                                "620a8dd1c88ae3eb88a8107a"
                               
                            ]
                        }

                    }
    
                }
    
            }
            }
        ])
        res.json(item);
        console.log(item);
        
    } catch (error) {
        res.status(500).json({message: error.message});
        
    }
})

and returns something like this (Empty arrays)

[
  {
    "_id": "[email protected]",
    "ItemsPrices": []
  },
  {
    "_id": "[email protected]",
    "ItemsPrices: []
  },
  {
    "_id": "[email protected]",
    "ItemsPrices": []
  },
  {
    "_id": "[email protected]",
    "ItemsPrices": []
  }
]

but If I search for price $$item.price


  cond: {
               $eq: [
                  "$$item.price",
                  "30"    
              ]
       }

it returns the perfect output

[
  {
    "_id": "[email protected]",
    "ItemsPrices": []
  },
  {
    "_id": "[email protected]",
    "ItemsPrices: []
  },
  {
    "_id": "[email protected]",
    "ItemsPrices": []
  },
  {
    "_id": "[email protected]",
    "ItemsPrices": [
        {
           "barcode":"234456345",
           "price":"30",
           "itemName":"monster",
           "_id":"620a8dd1c88ae3eb88a8107a"
        }
     ]
  }
]

@FekiHamza can you explain in detail, I don't understand how this query will find the second element of an array — Shashank Mistry
– Shashank Mistry, Commented Feb 14, 2022 at 19:08
I'm sorry, I misunderstood your question, does this help you ? stackoverflow.com/questions/43852393/… — Feki Hamza
– Feki Hamza, Commented Feb 14, 2022 at 19:11
@FekiHamza sorry mate, I am a beginner and this answer does not work for me, I tried but it just returns an empty array — Shashank Mistry
– Shashank Mistry, Commented Feb 14, 2022 at 19:21

cmgchess · Accepted Answer · 2022-02-15 09:29:41Z

3

You can do an aggregation with $project and apply $filter on the array part. In mongoose you can apply the aggregation query in a more or less similar way https://mongoosejs.com/docs/api/aggregate.html

db.collection.aggregate([
  {
    $project: {
      "ItemsPrices": {
        $filter: {
          input: "$ItemsPrices",
          as: "item",
          cond: {
            $eq: [
              "$$item._id",
              mongoose.Types.ObjectId("620a971e11120abbde5f4c3b")
            ]
          }
        }
      },
      "__v": 1  //when projecting 1 means in the final result this field appears
    }
  }
])

more examples

demo

edited Feb 15, 2022 at 9:29

answered Feb 14, 2022 at 19:35

cmgchess

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4 Comments

Shashank Mistry Over a year ago

Your approach works perfectly with all other strings in the array but not with _id. this is an auto-generated id and when we run a similar query it returns an empty array but if I search $$Item.price it works perfectly fine but if I write $$Item._id does not work. maybe because I am coding this in js or something Idk

Shashank Mistry Over a year ago

can you suggest the proper answer because I am new to mongo and don't know why this is the issue and how to solve this, you can see the main question that I changed above

cmgchess Over a year ago

can you also try mongoose.Types.ObjectId("620a971e11120abbde5f4c3b")

cmgchess Over a year ago

@ShashankMistry did it solve it so I can update the answer

J.F. · Accepted Answer · 2022-02-14 19:42:47Z

1

Option 1:

Use $filter in an aggregation query as explained by cmgchess

Option 2:

If you only want one object from array you can use $elemMatch like this:

db.collection.find({
  "ItemsPrices._id": "620a971e11120abbde5f4c3b"
},
{
  "ItemsPrices": {
    "$elemMatch": {
      "_id": "620a971e11120abbde5f4c3b"
    }
  }
})

Example here

But take care, using $elemMatch only the first element is returned. Check this other example where there are two objects with the desired _id but only returns one.

As said before, if you only one (or only exists one) maybe you can use find and $elemMatch to avoid a filter by the entire array. But if can be multiple values use $filter.

answered Feb 14, 2022 at 19:42

J.F.

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2 Comments

cmgchess Over a year ago

yours looks better since most probably there will be 1 record with a given _id i guess

Shashank Mistry Over a year ago

this is just an example in question, I have many documents, so this second approach is not working with multiple records.

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