MongoDB : querying documents with two equal fields, $match and $eq

Question

What is the best way to return all documents in a collection if I want document.a == document.b?

I've tried

db.collection.aggregate([ { $match: { $eq: [ '$a', '$b' ] } }]) 

But it returns with no errors or results, because I assume it is literally matching strings "$a" and "$b". Is there a different way to specify that these are fields?

db.collection.aggregate([    { $project: {      eq: { $cond: [ { $eq: [ '$a', '$b' ] }, 1, 0 ] }  } }, { $match: { eq: 1 } }]) 

The above works, but requires the additional step of querying again with whatever documents it found or projecting all possible fields.

Is there a better way for achieving this query?

Answer 1

If I understood your question right you want those documents that have same values in field1 and field2.

For this try

db.coll.find({$where: function() { return this.field1 == this.field2 } } ); 

or more compact

db.coll.find({ $where : "this.field1 == this.field2" } );