How to Simulate subquery in MongoDB query condition - mysql

Question

Let's suppose that I have a product logs collection, all changes are being done on my products will be recorded in this collection ie :
+------------------------------+
| productId - status - comment |
| 1 0 .... |
| 2 0 .... |
| 1 1 .... |
| 2 1 .... |
| 1 2 .... |
| 3 0 .... |
+------------------------------+
I want to get all products which their status is 1 but hasn't became 2. In SQL the query would look something like :
select productId from productLog as PL1
where
status = 1
and productId not in (
select productId from productLog as PL2 where
PL1.productId = PL2.productId and PL2.status = 2
)
group by productId
I'm using native PHP MongoDB driver.

Answer 1

Well since the logic here on the subquery join is simply that exactly the same key matches the other then:
Setup
db.status.insert([
{ "productId": 1, "status": 0 },
{ "productId": 2, "status": 0 },
{ "productId": 1, "status": 1 },
{ "productId": 2, "status": 1 },
{ "productId": 1, "status": 2 },
{ "productId": 3, "status": 0 }
])
Then use .aggregate():
db.status.aggregate([
{ "$match": {
"status": { "$ne": 2 }
}},
{ "$group": {
"_id": "$productId"
}}
])
Or using map reduce (with a DBRef):
db.status.mapReduce(
function() {
if ( this.productId.$oid == 2 ) {
emit( this.prouctId.$oid, null )
}
},
function(key,values) {
return null;
},
{ "out": { "inline": 1 } }
);
But again the SQL here was as simple as:
select productId
from productLog
where status <> 2
group by productId
Without the superfluous join on exactly the same key value

Answer 2

This mongo query above doesn't meet the requirements in question,
the result of the mongo-query includes documents with productId=1,
however the result of the SQL in question doesn't. Because in sample data: there exists 1 record with status=2, and productId of that document is 1.
So, assuming db.productLog.insert executed as stated above, you can use the code below to get the results:
//First: subquery for filtering records having status=2:
var productsWithStatus2 = db.productLog .find({"status":2}).map(function(rec) { return rec.productId; });
//Second:final query to get productIds which there not exists having status=2 with same productId :
db.productLog.aggregate([ {"$match":{productId:{$nin:productsWithStatus2}}},{"$group": {"_id": "$productId"}}]) ;
//Alternative for Second final query:
//db.productLog.distinct("productId",{productId:{$nin:productsWithStatus2}});
//Alternative for Second final query,get results with product and status detail:
//db.productLog.find({productId:{$nin:productsWithStatus2}});