mongoose aggregate using $exists in $cond

You could use two $project statement for this case and make use of the $ifNull operator (Just wont work when some_field is set as false in which case you can change the inner false to something more suitable)

[
{
        $project: {
        test: {
            $ifNull: [
                '$some_field',
                false
            ]
        }
    }
},
{
    $project: {
        test: {
            $cond: {
                if    : {$eq: ['$test', false]},
                then  : false,
                else  : true
            }
        }
    }
}

])

$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