mongodb - How to exclude from search results documents with fields which are not present in query?

Question

I have two documents:

1. { p1:"a", p2:"b" }
2. { p1:"a", p2:"b", p3:"c" }

What I should to do with query: { p1:"a", p2:"b" } to find only first document? So I want find only documents with fields what I specified. If document has more fields (than query) it should not be presented in search results.

Answer 1

I must admit I know of no normal querying method by which to solve this problem. There is only one way I know of and that is to use MongoDBs object comparison. To do this you would change your structure to be something along the lines of:

{
    ps: [a,b]
}

or:

{
    ps: {p1:a,p2:b}
}

And then you would query like:

db.col.find({ p: [a,b] })

or:

db.col.find({ p: {p1:a, p2:b} })

There is one immedate problem with this though. It is key order dependant which means that if your a and b are actually the other way around in another document it won't match. So you will need to make sure you care about order when saving if you do this.

Hope it helps,