add key to nested array with condition

Question

I have a simple datastructure in mongodb:

{
  _id: ObjectID,
 name: 'Name',
 birthday: '25.05.2001'
 items: [
          {
             _id: ObjectID,
             name: 'ItemName',
             info: 'ItemInfo',
           },
           {
             _id: ObjectID,
             name: 'ItemName',
             info: 'ItemInfo',
           }
        ]
}

Now i want a query, that takes a ObjectID (_id) of an item as criteria and gives me back the object with all items in the array AND projects a new field "selected" with value true or false into a field in the result of each array item:

I tried that with this query:

 { $unwind: '$items' },
 {
    $project: {
       selected: {
          $cond: { if: { 'items._id': itemObjectID }, then: true, else: false },
      },
    },
  },

but MongoDB gives me back an error:

MongoError: FieldPath field names may not contain '.'.

Have no clue why its not working, any help or ideas? Thank you very much!

Ashh · Accepted Answer · 2018-06-12 06:13:28Z

1

What you are missing here is $eq aggregation operator which checks the condition for the equality.

You can try below aggregation here if you want to check for ObjectId then you need to put mongoose.Types.ObjectId(_id)

db.collection.aggregate([
  { "$unwind": "$items" },
  { "$addFields": {
    "items.selected": {
      "$eq": [
        1111,
        "$items._id"
      ]
    }
  }},
  { "$group": {
    "_id": "$_id",
    "name": { "$first": "$name" },
    "items": {
      "$push": {
        "_id": "$items._id",
        "selected": "$items.selected"
      }
    }
  }}
])

Will give following output

[
  {
    "_id": ObjectId("5a934e000102030405000000"),
    "items": [
      {
        "_id": 1111,
        "selected": true
      },
      {
        "_id": 2222,
        "selected": false
      }
    ],
    "name": "Name"
  }
]

You can check it here

edited Jun 12, 2018 at 6:13

answered Jun 11, 2018 at 17:33

Ashh

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1 Comment

Vizari Over a year ago

Thanks! I found my solution with your help. I posted it as an answer to my own question because of the large code part...thx

Vizari · Accepted Answer · 2018-06-11 20:37:53Z

0

@Ashish: Thank you very much for your help! Your answer helped me to build the right query for me:

  db.collection.aggregate([
  {
    $unwind: "$items"
  },
  {
    $project: {
      "items.name": 0,
      "birthday": 0
    }
  },
  {
    "$addFields": {
      "items.selected": {
        "$eq": [
          1111,
          "$items._id"
        ]
      }
    }
  },
  {
    $group: {
      _id: "$_id",
      "name": {
        "$first": "$name"
      },
      items: {
        $push: "$items"
      }
    }
  },
  {
    $match: {
      "items._id": {
        $eq: 1111
      }
    }
  },
])

and leads to a result that looks like:

[
  {
    "_id": ObjectId("5a934e000102030405000000"),
    "items": [
      {
        "_id": 1111,
        "selected": true
      },
      {
        "_id": 2222,
        "selected": false
      }
    ],
    "name": "Name"
  }
]

answered Jun 11, 2018 at 20:37

Vizari

357 bronze badges

2 Comments

Ashh Over a year ago

HI @Vizari, Actually what you are doing wrong here is $match and $project... The $match should be used as soon as possible and the $project should be used at the end of the aggregation... And also you don't need to use $project here... You can see my updated answer...

Vizari Over a year ago

Hi @Ashish, thanks for your reply. If i dont use the $project i have to add each field of the root document that i want to have in my result with "name": { "$first": "$name" } and $push: "$items"for my nested object?. And if i put the '$match' at the beginning of my aggregation i have only the nested object that matches that criteria in my result, but i need all of them because they have to be marked as "selected: false"

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