How to group query with multiple $cond? - mongodb

Question

I want to query like below, but this contains only one $cond.
How to query with two $cond?
collection.aggregate(
{
$match : {
'_id' : {$in:ids}
}
},
{
$group: {
_id: '$someField',
...
count: {$sum: { $cond: [ { $eq: [ "$otherField", false] } , 1, 0 ] }}
}
},
function(err, result){
...
}
);

Answer 1

You want to use a compound expression inside {$cond:[]} - something like:
collection.aggregate(
{
$match : {
'_id' : {$in:ids}
}
},
{
$group: {
_id: '$someField',
...
count: {$sum: { $cond: [ {$and : [ { $eq: [ "$otherField", false] },
{ $eq: [ "$anotherField","value"] }
] },
1,
0 ] }}
}
},
function(err, result){
...
}
);
The $and operator is documented here: http://docs.mongodb.org/manual/reference/operator/aggregation/#boolean-operators

Answer 2

you can add multiple $cond and multiple criterias inside $cond like this
`
collection.aggregate(
[
{
"$match": {
//matching criteria
}
},
{
"$project": {
"service": {
"$cond": {
"if": {
"$eq": [
"$foo",
"bar"
]
},
"then": "return string1",
"else": {
"$cond": {
"if": {
"$eq": [
"$foo",
"bar"
]
},
"then": "return string2",
"else": {
"$cond": {
"if": {
"$or": [
{
"$eq": [
"$foo",
"bar1"
]
},
{
"$eq": [
"$foo",
"bar2"
]
}
]
},
"then": "return string3",
"else": "$foo"
}
}
}
}
}
},
"_id": 0
}
}
]
)
`