mongodb join multiple collections [duplicate] - mongodb

Question

This question already has answers here:
Multiple join conditions using the $lookup operator
(4 answers)
Closed 5 years ago.
I'd like to "join" 3 Collections in MongoDB by using mongochef. The collections are "Order", "Employee" and "City". I tried to use temp collections, but it is not effective.
Now I am using var = a for the first "join".
If I 'd like to show "a", there are displayed 20 results only.
Do you have an idea or another solution?
var a = db.Order.aggregate([
{
$lookup:
{
from: "City",
localField: "City Key",
foreignField: "City Key",
as: "lsg"
}
},
{
$unwind: "$lsg"
},
{
$project:
{
"_id":1,
"Salesperson Key":1,
"City": "$lsg.City"
}
}
])
a;
var b = db.Employee.aggregate([
{
$lookup:
{
from: "a",
localField: "Employee Key",
foreignField: "Salesperson Key",
as: "lsg2"
}
},
{
$unwind: "$lsg2"
},
{
$project:
{
"_id":1,
"Employee":1
}
}
])
Thanks in advance for replies.

Answer 1

you can put multiple $lookup stages, so you could use a query like this (could't test it but should work)
But you should avoid multiple joins, keep in mind that MongoDB is not a relational database...
db.Order.aggregate([
{
$lookup:{
from:"City",
localField:"City Key",
foreignField:"City Key",
as:"lsg"
}
},
{
$unwind:"$lsg"
},
{
$lookup:{
from:"Employee",
localField:"Salesperson Key",
foreignField:"Employee Key",
as:"lsg2"
}
},
{
$unwind:"$lsg2"
},
{
$project:{
"_id":1,
"Employee":1,
"Salesperson Key":1,
"City":"$lsg.City"
}
}
]);