Reputation: 3924
This is my code.
db.article.aggregate(
{ $project : {
title : 1 ,
stats : {
pv : "$name",
foo : "$foo",
dpv : 1
}
}}
);
In that name,foo are there in collection. That dpv is not, but i need to add that one.
OUTPUT CAME
{
PV:"shan",
foo:"sd"
}
I need that dpv also in the ouput with null value.
Is it possible ? How to achieve this ? I am new to mongoDb. (using node.js)
Upvotes: 0
Views: 1144
Reputation: 59
Asya is correct in that you need to specify a literal 1
, and while the solution works a more correct solution would be:
dpv: {$literal: 1}
Upvotes: 3
Reputation: 42352
The syntax fieldname:1
means "pass through this field as is.
You want to have a literal 1
value - the simplest way is to create an expression that will return 1. I suggest: dpv:{$add:[1]}
Upvotes: 0