Compare a mongo diff on two collections

Try the following in the shell, it will iterate each item within a collection and try to match each document based on ID.

Say we have 2 collections db.col1 and db.col2:

> db.col1.find()
{ "_id" : 1, "item" : 1 }
{ "_id" : 2, "item" : 2 }
{ "_id" : 3, "item" : 3 }
{ "_id" : 4, "item" : 4 }

> db.col2.find()
{ "_id" : 1, "item" : 1 }
{ "_id" : 2, "item" : 2 }
{ "_id" : 3, "item" : 3 }
{ "_id" : 4, "item" : 4 }

We can then create a javascript function to compare 2 collections

function compareCollection(col1, col2){
    if(col1.count() !== col2.count()){
        return false;
    }

    var same = true;

    var compared = col1.find().forEach(function(doc1){
        var doc2 = col2.findOne({_id: doc1._id});

        same = same && JSON.stringify(doc1)==JSON.stringify(doc2);
    });

    return same;
}

Then call is like the following:

> compareCollection(db.col1, db.col2)
true

If we then have a 3rd collections db.col3

> db.col3.find()
{ "_id" : 1, "item" : 1 }

And compare this one

> compareCollection(db.col1, db.col3)
false

we’ll get the expected result.

If we also have a 4th collection which has matching documents but diffrent data db.col4

> db.col4.find()
{ "_id" : 1, "item" : 10 }
{ "_id" : 2, "item" : 2 }
{ "_id" : 3, "item" : 3 }
{ "_id" : 4, "item" : 4 }

This will also return false

> compareCollection(db.col1, db.col4)
false