mongoose aggregate using $exists in $cond

Question

I want to $project if a field exists, but not it's value, using mongoose model aggregate query. If it was possible using $exists in $cond, it would have looked something like this:

$project: {
    b: {
        $cond: {
            if    : {$exists: ['$b', true]},
            then  : true,
            else  : false
        }
    }
}

But, I have to use a boolean expression in the $cond operator. In the MongoDB shell, I can do something similar with:

{$eq: ['$b', undefined]}

and it yields the expected results, but with mongoose model aggregate for some reason, it always results with true.

for example, if I have the following documents:

{
    "a" : 1,
    "b" : 2
},
{
    "a" : 1
}

I need the following results:

{
    "b": true
},
{
    "b": false
}

How can I do something like that with mongoose?

Answer 1

$exists not supported in aggregate query of MongoDB. So in aggregate query instead of $exists can use $ifNull.

syntax:

{ $ifNull: [ <expression>, <replacement-expression-if-null> ] }

for more

Updated:

to get b value as true or false can try this query

db.test.aggregate([
    {
        $project: {
            b: { 
                $cond: [
                    {$ifNull: ['$b', false]}, // if
                    true, // then
                    false // else
                ]
            }
        }
    }
])

Explanation:

b = $cond: [ 'if condition satisfied', 'then true', 'else false' ];

where condition = {$ifNull: ['$b', false]} Here if $b not exist then condition = false otherwise condition = true.

so if condition = true then return then result that means b = true
if condition = false then return else result means b = false