Mongodb:Right way to collect data from two collections? - mongodb

Question

I have two collections: one is items and the second one is user_item_history. I want to fetch items with their status. Status of each item is stored in user_item_history, and other details of the item are in the items collection. we have to filter data for particular user and category of item. so user_id and category is in user_item_history collection.
user_item_history:
{
"_id" : NumberLong(25424),
"_class" : "com.samepinch.domain.registration.UserItemHistory",
"user_id" : NumberLong(25416),
"item_id" : NumberLong(26220),
"catagoryPreference" : "BOTH",
"preference" : 0.6546536707079772,
"catagory" : "FOOD",
"status" : 1,
"createdDate" : ISODate("2015-09-02T07:50:36.760Z"),
"updatedDate" : ISODate("2015-09-02T07:55:24.105Z")
}
items:
{
"_id" : NumberLong(26220),
"_class" : "com.samepinch.domain.item.Item",
"itemName" : "Shoes",
"categoryName" : "SHOPPING",
"attributes" : [
"WESTERN",
"CASUAL",
"ELEGANT",
"LATEST"
],
"isAccessed" : false,
"imageUrl" : "0bd2838e-9349-432a-a200-6e6b659e853eitemcompressed.jpg",
"catagoryPreference" : "FEMALE",
"startDate" : ISODate("2015-11-26T18:30:00Z"),
"endDate" : ISODate("2015-11-27T18:30:00Z"),
"location" : {
"coordinates" : [
77.24149558372778,
28.56973445677584
],
"type" : "Point",
"radius" : 2000
},
"createdDate" : ISODate("2015-11-16T10:49:11.858Z"),
"updatedDate" : ISODate("2015-11-16T10:49:11.858Z")
}
As the final result, I would like to have documents of this format:
{
item_id:26220,
status:1,
imageUrl: "0bd2838e-9349-432a-a200-6e6b659e853eitemcompressed.jpg"
}

Answer 1

Update to MongoDB 3.2 and you'll be able to use the $lookup aggregation stage, which works similarly to SQL joins.
One-to-many relationship
If there are many corresponding user_item_history documents for each items document, you can get a list of item statuses as an array.
Query
db.items.aggregate([
{
$lookup:
{
from: "user_item_history",
localField: "_id",
foreignField: "item_id",
as: "item_history"
}
},
{
$project:
{
item_id: 1,
status: "$item_history.status",
imageUrl: 1
}
}])
Example Output
{
"_id" : NumberLong(26220),
"imageUrl" : "0bd2838e-9349-432a-a200-6e6b659e853eitemcompressed.jpg",
"status" : [ 1 ]
},
{
"_id" : NumberLong(26233),
"imageUrl" : "0bd2838e-9349-432a-a200-6e6b659e853eitemcompressed.jpg",
"status" : [ 1, 2 ]
}
One-to-one relationship
If there's only one corresponding history document for every item, you can use the following approach to get the exact format you requested:
Query
db.items.aggregate([
{
$lookup:
{
from: "user_item_history",
localField: "_id",
foreignField: "item_id",
as: "item_history"
}
},
{
$unwind: "$item_history"
},
{
$project:
{
item_id: 1,
status: "$item_history.status",
imageUrl: 1
}
}])
Example Output
{
"_id" : NumberLong(26220),
"imageUrl" : "0bd2838e-9349-432a-a200-6e6b659e853eitemcompressed.jpg",
"status" : 1
}
Please bear in mind that with every additional aggregation pipeline stage you add, the performance deteriorates. So you may prefer the one-to-many query even if you have a one-to-one relationship.
Applying filtering
In your edit, you added this:
we have to filter data for particular user and category of item. so user_id and category is in user_item_history collection
To filter your results, you should add a $match step to your query:
db.items.aggregate([
{
$lookup:
{
from: "user_item_history",
localField: "_id",
foreignField: "item_id",
as: "item_history"
}
},
{
$unwind: "$item_history"
},
{
$match:
{
"item_history.user_id": NumberLong(25416),
"item_history.catagory": "FOOD"
}
},
{
$project:
{
item_id: 1,
status: "$item_history.status",
imageUrl: 1
}
}])
Please note that "category" is misspelled as "catagory" in your example data, so I also had to misspell it in the query above.