Mongo field A greater than field B

Question

i\'m trying a simple query in Mongo which in MySQL would look like this.

select * from emails where bounceCount > sentCount;

So far I ha

Answer 1

db.so.find("this.bounceCount > this.sentCount") is what you are looking for.

Equivalent: db.so.find({"$where":"this.bounceCount > this.sentCount"})

Documentation: http://docs.mongodb.org/manual/reference/operator/where/

Shell output:

> db.so.insert({bounceCount:1, sentCount:2})
> db.so.insert({bounceCount:5, sentCount:3})
> db.so.insert({bounceCount:5, sentCount:4})
> db.so.insert({bounceCount:5, sentCount:7})
> db.so.insert({bounceCount:9, sentCount:7})

> db.so.find()
{ "_id" : ObjectId("516d7f30675a2a8d659d7594"), "bounceCount" : 1, "sentCount" : 2 }
{ "_id" : ObjectId("516d7f37675a2a8d659d7595"), "bounceCount" : 5, "sentCount" : 3 }
{ "_id" : ObjectId("516d7f3b675a2a8d659d7596"), "bounceCount" : 5, "sentCount" : 4 }
{ "_id" : ObjectId("516d7f3d675a2a8d659d7597"), "bounceCount" : 5, "sentCount" : 7 }
{ "_id" : ObjectId("516d7f40675a2a8d659d7598"), "bounceCount" : 9, "sentCount" : 7 }

> db.so.find({"bounceCount":5})
{ "_id" : ObjectId("516d7f37675a2a8d659d7595"), "bounceCount" : 5, "sentCount" : 3 }
{ "_id" : ObjectId("516d7f3b675a2a8d659d7596"), "bounceCount" : 5, "sentCount" : 4 }
{ "_id" : ObjectId("516d7f3d675a2a8d659d7597"), "bounceCount" : 5, "sentCount" : 7 }

> db.so.find("this.bounceCount > this.sentCount")
{ "_id" : ObjectId("516d7f37675a2a8d659d7595"), "bounceCount" : 5, "sentCount" : 3 }
{ "_id" : ObjectId("516d7f3b675a2a8d659d7596"), "bounceCount" : 5, "sentCount" : 4 }
{ "_id" : ObjectId("516d7f40675a2a8d659d7598"), "bounceCount" : 9, "sentCount" : 7 }