How to get diff between tow queries in mongodb?

Question

I have collection

[{ 
    "_id" : "5c9a69ef42c82b0197a2ffb8", 
    "key" : "a",  
    "language" : "en", 
    "version" : "0.1", 
},
{ 
    "_id" : "5c9a69ef42c82b0197a2ffb8", 
    "key" : "b",  
    "language" : "en", 
    "version" : "0.1", 
},
{ 
    "_id" : "5c9a69ef42c82b0197a2ffb8", 
    "key" : "b",  
    "language" : "en", 
    "version" : "0.2", 
},
{ 
    "_id" : "5c9a69ef42c82b0197a2ffb8", 
    "key" : "c",  
    "language" : "en", 
    "version" : "0.2", 
}]

so I have 2 queries that getting version 0.1 and version 0.2, And I need to find the difference between them?

I need to get all of translation 0.2 that does not exist in translation 0.1.

Answer 1

Here is a solution to get all translations key that has a 0.2 version but not a 0.1 version.

• First I $sort the translations by versions.
• Then with $group , I get all the versions for a given key .
• $match remove all the key that have a version 0.1
• $replaceRoot keep only the last version for a key

db.collection.aggregate([
  {
    "$sort": {
      "version": 1
    }
  },
  {
    $group: {
      "_id": "$key",
      "items": {
        "$push": "$$ROOT"
      }
    }
  },
  {
    "$match": {
      "items.version": {
        "$ne": "0.1"
      }
    }
  },
  {
    "$replaceRoot": {
      "newRoot": {
        "$last": "$items"
      }
    }
  }
])

try it here

---

If you remove the $match , you can also retrieve the last version available for each translation.

• key a => version 0.1
• key b => version 0.2
• key c => version 0.2

db.collection.aggregate([
  {
    "$sort": {
      "version": 1
    }
  },
  {
    "$group": {
      "_id": "$key",
      "items": {
        "$push": "$$ROOT"
      }
    }
  },
  {
    "$replaceRoot": {
      "newRoot": {
        "$last": "$items"
      }
    }
  }
])

try it here