I want to find how many "taskId" of below collectionOne is present in collectionTwo. Here, "taskId" in below collectionOne is embedded inside a array in a document.
Also, Let me know the different approaches to do this (if possible).
db.collectionOne.find({"sId":"DCNrnPeKFrBv" }).pretty()
{
"_id" : "sX8o7mJhebs",
"sId" : "DCNrnPeKFrBv",
"data" : [
{
"oId" : "7SycYQ",
"taskId" : 146108906
},
{
"oId" : "7SycYQ",
"taskId" : 14623846
},
{
"oId" : "fANQ",
"taskId" : 1461982
},
{
"oId" : "fAeNQ",
"taskId" : 131732
},
{
"oId" : "t6AF5yn",
"taskId" : 197681
}
]
}
> db.collectionTwo.find().pretty().limit(2)
{
"_id" : 146108906,
"oId" : "7SycYQ",
"name" : "ABC"
}
{
"_id" : 1461982,
"oId" : "fANQ",
"name" : "XYZ"
}
In collectionTwo "_id" is equivalent to "taskId" of collectionOne.
Using the $lookup
operator to do a left join on collectionTwo
, you can get the counts as follows:
db.collectionOne.aggregate([
{ $match: { sId: "DCNrnPeKFrBv" }},
{ $lookup: {
from: "collectionTwo",
localField: "data.taskId",
foreignField: "_id",
as: "tasksCount"
} },
{ $addFields: {
tasksCount: { $size: "$tasksCount" }
} }
])
or if using older MongoDB server versions (below 3.2):
db.collectionOne.aggregate([
{ $unwind: "$data" },
{ $lookup: {
from: "collectionTwo",
localField: "data.taskId",
foreignField: "_id",
as: "tasks"
} },
{ $unwind: "$tasks" },
{ $group: {
_id: "$tasks._id",
count: { "$sum": 1 }
} },
{ $group: {
_id: null,
tasksCount: { "$sum": "$count" }
} }
])
--EDIT--
An alternative is to get a list of all the distinct taskIds in collectionOne, and use that list as count query on collectionTwo e.g.
var taskIds = db.collectionOne.distinct("data.taskId");
var tasksCount = db.collectionTwo.count({ "_id": { "$in": taskIds } });
printjson(taskIds);
printjson(tasksCount);
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