Mongo Shell convert string to nuuid and compare - mongodb

Question

I would like to do the following:
db
.getCollection('COLLECTION')
.find({ $where : function() { return NUUID(this.Col1) != this.Col2 }})
Where Col1 contains NUUID strings and Col2 contains NUUIDs.
I would like rows where the NUUID value of Col1 (which is a string) is not equal to Col2.
I receive the error: ReferenceError: NUUID is not defined

Answer 1

Using https://github.com/mongodb/mongo-csharp-driver/blob/master/uuidhelpers.js , the solution is:
db
.getCollection('TestColl')
.find({$where: function(){
function CSUUID(uuid) {
var hex = uuid.replace(/[{}-]/g, ""); // remove extra characters
var a = hex.substr(6, 2) + hex.substr(4, 2) + hex.substr(2, 2) + hex.substr(0, 2);
var b = hex.substr(10, 2) + hex.substr(8, 2);
var c = hex.substr(14, 2) + hex.substr(12, 2);
var d = hex.substr(16, 16);
hex = a + b + c + d;
var base64 = HexToBase64(hex);
return new BinData(3, base64);
}
function HexToBase64(hex) {
var base64Digits = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
var base64 = "";
var group;
for (var i = 0; i < 30; i += 6) {
group = parseInt(hex.substr(i, 6), 16);
base64 += base64Digits[(group >> 18) & 0x3f];
base64 += base64Digits[(group >> 12) & 0x3f];
base64 += base64Digits[(group >> 6) & 0x3f];
base64 += base64Digits[group & 0x3f];
}
group = parseInt(hex.substr(30, 2), 16);
base64 += base64Digits[(group >> 2) & 0x3f];
base64 += base64Digits[(group << 4) & 0x3f];
base64 += "==";
return base64;
}
return CSUUID(this['Col1']).toString() != this['Col2'].toString()
}
})