Filter and find documents which have more than one element in an array of objects which have same key value pair - mongodb

AneeshJose
June 2, 2022
144 views
0 votes
2 Answers

A sample collection

[
  {
    "_id":"id1",
    "usersArray":[
      {
        "name":"user1Name",
        "type":1
      },
      {
        "name":"user2Name",
        "type":1
      },
      {
        "name":"user3Name",
        "type":2
      },
    ]
  },
  {
    "_id":"id2",
    "usersArray":[
      {
        "name":"user4Name",
        "type":1
      },
      {
        "name":"user5Name",
        "type":3
      },
      {
        "name":"user6Name",
        "type":2
      },
    ]
  },
  {
    "_id":"id3",
    "usersArray":[
      {
        "name":"user7Name",
        "type":1
      },
      {
        "name":"user8Name",
        "type":1
      },
      {
        "name":"user9Name",
        "type":2
      },
    ]
  },
]

What I need is a filter that will return the documents with _id id1 and id3 since they have more than 1 objects in usersArray which have key-value pairs "type": 1. It would be great if you could resolve this with just the filter.

Tags: filter mongodb

Answers

- YongShun
- June 2, 2022 at 2:32 pm
- 0 votes
0
1. $expr – Allow using aggregation operator.
  
  1.1. $gt – Filter the document with the result of 1.1.1 greater than 1.
  
  1.1.1. $size – Get the size of the result 1.1.1.1.
  
  1.1.1.1. $filter – Filter the document(s) with type: 1 in usersArray.
```
db.collection.find({
  $expr: {
    $gt: [
      {
        $size: {
          $filter: {
            input: "$usersArray",
            cond: {
              $eq: [
                "$$this.type",
                1
              ]
            }
          }
        }
      },
      1
    ]
  }
})
```
Sample Mongo Playground

In case usersArray is possibly null or not array, then you need to check the usersArray is an array, else assign an empty array via $cond and $isArray.
```
db.collection.find({
  $expr: {
    $gt: [
      {
        $size: {
          $filter: {
            input: {
              $cond: {
                if: {
                  $isArray: "$usersArray"
                },
                then: "$usersArray",
                else: []
              }
            },
            cond: {
              $eq: [
                "$$this.type",
                1
              ]
            }
          }
        }
      },
      1
    ]
  }
})
```
Sample Mongo Playground (With check array)
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- Takis
- June 2, 2022 at 3:51 pm
- 0 votes
0
Query
- make it set($union with empty), get the $size
- get the original $size
- if set size < original size => it contained duplicated type
*this is a way to do it not for type:1 duplicate only, this keeps documents that dont have duplicate same types, i am not sure what you need. (for example if type:2 is duplicated document wont pass the filter also)

Playmongo
```
aggregate(
[{"$match": 
   {"$expr": 
     {"$lt": 
       [{"$size": {"$setUnion": ["$usersArray.type", []]}},
        {"$size": "$usersArray.type"}]}}}])
```
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