Imagine i have the following models:
# MODEL A
schemaA = mongoose.Schema
_bId:
type: mongoose.Schema.Types.ObjectId
ref: "B"
# MODEL B
schemaB = mongoose.Schema
_cId:
type: mongoose.Schema.Types.ObjectId
ref: "C"
_dId:
type: mongoose.Schema.Types.ObjectId
ref: "D"
# MODEL C
schemaC = mongoose.Schema
_eId:
type: mongoose.Schema.Types.ObjectId
ref: "E"
Model D and E do not have any other object references and for the sake of convenience are therefore not listed anymore.
What is the best practice to populate model "A" with all references? Currently i resolve this task as follows (it is an instance-method because i need it quite often):
schemaA.methods =
populateAll: (cb) ->
@
.populate
path: "_bId"
model: "B"
populate:
path: "_cId"
model: "C"
populate:
path: "_eId"
model: "E"
, (error) =>
return cb error, @ if error?
D.findById @._bId._dId
.exec (error, d) =>
return cb error, @ if error?
@._bId._dId = d
return cb error, @
This is the only way i found to populate all references because populating multiple paths with then again multiple paths in different models is quite difficult. I already tried a solution like the one below but as one can imagine, it would only overwrite previous populations:
@
.populate
path: "_bId"
model: "B"
populate:
path: "_cId"
model: "C"
populate:
path: "_eId"
model: "E"
.populate
path: "_bId"
model: "B"
populate:
path: "_dId"
model: "D"
@
.populate
path: "_bId"
model: "B"
populate: [
{
path: "_cId"
model: "C"
populate:
path: "_eId"
model: "E"
}
{
path: "_dId"
}
]
, (error) =>
This solution works totally fine and i just found out about this.
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