What I'm trying to do in Scala 2.11 and akka is have one case class but two different validations based on which route is being hit.
For example, let's consider the case class below
case class User(_id: String, name: String, age: Int, address: String)
Now while the /create route is hit, I don't need _id but I need all the other fields.
But while /update route is hit, I need the _id and the fields that are to be updated (which could be one or all three)
Only declaring Option doesn't serve the purpose because then my /create route goes for a toss.
Even extending case classes doesn't really work seamlessly (there's too much code duplicity).
I would love if something like this was possible
case class User(_id: String, name: String, age: Int, address: String)
case class SaveUser() extends User {
require(name.nonEmpty)
require(age.nonEmpty)
require(address.nonEmpty)
}
case class UpdateUser() extends User {
require(_id.nonEmpty)
}
Is there an elegant solution to this? Or do I have to create two identical case classes?
My suggestion would be to encode different case classes for different requirements, but if you insist you must share code between these two cases a possible solution would be to parameterize the case class
case class User[Id[_], Param[_]](_id: Id[String], name: Param[String], age: Param[Int], address: Param[String])
Then you define an alias for the Identity type constructor and your two uses of the case class
type Identity[T] = T
type SaveUser = User[Option, Identity]
type UpdateUser = User[Identity, Option]
Related
Say I have a case class that has a type parameter T:
case class ProdutOption[T](name: String, value:T)
And this case class is contained in another case class like:
case class Product[T](id: String, options: Set[ProductOption[T]])
What is the correct way to model this?
I want my options parameter to be able to have different types of T... I don't want all items in the set to be forced to a specific type.
Is this scenario possible to model using type parameters like this?
Everything depends on whether your set of possible option types is closed (i.e. known statically) or open (unknown until runtime) and how exactly do you want to process these options.
If you can enumerate all possible options, use an ADT (algebraic data type):
sealed trait ProductOption[+T] {
def name: String
def value: T
}
object ProjectOption {
case class SomeNumericOption(value: Int) extends ProductOption[Int] {
def name: String = "numeric"
}
case class SomeTextualOption(value: String) extends ProductOption[String] {
def name: String = "textual"
}
}
case class Product(id: String, option: Set[ProductOption[_]])
You can then process these options in a type-safe way using pattern matching.
If your option hierarchy is not known then you should introduce some method for processing this option into the base trait, which will no longer be sealed:
trait ProductOption {
def name: String
def processMe(): Whatever = ...
}
case class Product(id: String, option: Set[ProductOption])
What exactly would processMe do depends on your use case.
This is a classical example of so called expression problem.
It's a little unclear what exactly you're looking for, but it may be useful to consider covariance for ProductOption:
case class ProductOption[+T](name: String, value: T)
This says that if S is a supertype of T, ProductOption[T] is a subtype of ProductOption[S].
Specifically this then means that a Product[Any] allows options to be any kind of ProductOption:
Product("my favorite product", Set(ProductOption[String]("color", "blue"), ProductOption[ShirtSize]("size", ShirtSize.TwoXL)) // type is Product[AnyRef]
During some simple scala coding exercise I ran into ideological problem of case classes without parameters and constructor parameters duplication.
It all started with the following two quite simple classes:
trait Namespace
case class Reply[T](namespace: Namespace, correlation: String, data: Try[T])
abstract class Request(val namespace: Namespace, val id: String = UUID.randomUUID().toString) {
def success[T](data: T) = Reply(namespace, id, Try(data))
def failure(msg: String) = Reply(namespace, id, Failure(new RuntimeException(msg)))
}
Now let's assume i have an entity Post and I want to add All class as a command to query all records of type Post. In my current set up it would be easier to actually write the following:
case class All extends Request(Posts)
However in this case I get compiler warning that case classes without parameters are deprecated. So one might suggest to rewrite it into the following:
case object All extends Request(Posts)
However in this case object All will be instantiated only once along with its id field which would like to avoid having unique id for each request.
Could you please suggest a better way of representing All command so that it would not be required to duplicate constructor arguments?
Thanks in advance?
The actual warning is that
case classes without a parameter list are not allowed; use either case
objects or case classes with an explicit `()' as a parameter list.
So give this class an empty parameter list, just as suggested by the compiler:
case class All() extends Requests(Posts)
Don't use a case class or case object, just use a companion apply instead. You don't really want an object here anyway, if you need a unique ID for every request.
class All extends Requests(Posts)
object All {
def apply(): All = new All()
}
getPosts(All())
please take a look at the following code:
scala> sealed abstract class Person(val name: String)
defined class Person
scala> case class Student(id: Int, name: String) extends Person(name)
<console>:8: error: overriding value name in class Person of type String;
value name needs `override' modifier
case class Student(id: Int, name: String) extends Person(name)
^
This might be a trivial question, but after searching the web for quite some time, I wasn't able to figure out how to simply pass the string that Student's constructor will be provided as name to the Person's constructor. I don't want to override anything. What am I doing wrong?
Thank you very much in advance!
All constructor parameters of a case class are vals. That's the whole point. Roughly speaking, it is what gives you the ability to enjoy the benefits cases classes provide compared to regular classes: copying, extraction, pattern matching, etc.
If you want Student to be a case class, you should override name. Theoretically, you can avoid overriding it by giving the val a different name: case class Student(id: Int, studentName: String) extends Person(studentName) - this works, but just doesn't make very much sense - you end up having two different member vals whose values are always identical.
Alternatively, if there is an actual reason why you don't want to override name (I can't imagine what one could possibly be, but if ...), then Student should not be a case class: class Student(val id: Int, name: String) extends Person(name).
I have a very specific scenario, in which I have some different abstract classes the have child case classes that can have different parameters, for example:
abstract class ball() {}
case class football(_name: String, _shape: String) extends ball
case class basketball(_name: String, _size: Int) extends ball
and a different abstract class:
abstract class food() {}
case class vegetarian(_name: String, calories: Int) extends food
case class meat(_name: String, proteinCount: Int) extends food
Now, the problem I'm facing is that I need to somehow extract the name of all of those without knowing what class it is, I just know that ALWAYS, EACH CLASS has a parameters named _name.
Supposing we have an object of any of above classes, I'm trying to do it like this:
object.getClass.getDeclaredField("_name").get(this)
But I'm getting the error:
can not access a member of class package.food with modifiers "private"
I tried putting val and var before parameters in class but it doesnt help. I also tried doing "setAccessible(true)" in a line before get(this), which also doesn't help.
The obvious clean solution would be to have a least a common trait to all these classes:
trait HasName {
def _name: String
}
and then you can safely do obj.asInstanceOf[HasName]._name. Better yet if you manage to keep around the static information that obj is a HasName, in which case obj._name suffices.
If you can't do any of that, reflection is the way to go. You can do it pretty easily using a structural type, in this case:
obj.asInstanceOf[{ def _name: String }]._name
Note that this will be slower than the above HasName solution, and completely unchecked at compile time.
I have an application based on Squeryl. I define my models as case classes, mostly since I find convenient to have copy methods.
I have two models that are strictly related. The fields are the same, many operations are in common, and they are to be stored in the same DB table. But there is some behaviour that only makes sense in one of the two cases, or that makes sense in both cases but is different.
Until now I only have used a single case class, with a flag that distinguishes the type of the model, and all methods that differ based on the type of the model start with an if. This is annoying and not quite type safe.
What I would like to do is factor the common behaviour and fields in an ancestor case class and have the two actual models inherit from it. But, as far as I understand, inheriting from case classes is frowned upon in Scala, and is even prohibited if the subclass is itself a case class (not my case).
What are the problems and pitfalls I should be aware in inheriting from a case class? Does it make sense in my case to do so?
My preferred way of avoiding case class inheritance without code duplication is somewhat obvious: create a common (abstract) base class:
abstract class Person {
def name: String
def age: Int
// address and other properties
// methods (ideally only accessors since it is a case class)
}
case class Employer(val name: String, val age: Int, val taxno: Int)
extends Person
case class Employee(val name: String, val age: Int, val salary: Int)
extends Person
If you want to be more fine-grained, group the properties into individual traits:
trait Identifiable { def name: String }
trait Locatable { def address: String }
// trait Ages { def age: Int }
case class Employer(val name: String, val address: String, val taxno: Int)
extends Identifiable
with Locatable
case class Employee(val name: String, val address: String, val salary: Int)
extends Identifiable
with Locatable
Since this is an interesting topic to many, let me shed some light here.
You could go with the following approach:
// You can mark it as 'sealed'. Explained later.
sealed trait Person {
def name: String
}
case class Employee(
override val name: String,
salary: Int
) extends Person
case class Tourist(
override val name: String,
bored: Boolean
) extends Person
Yes, you have to duplicate the fields. If you don't, it simply would not be possible to implement correct equality among other problems.
However, you don't need to duplicate methods/functions.
If the duplication of a few properties is that much of an importance to you, then use regular classes, but remember that they don't fit FP well.
Alternatively, you could use composition instead of inheritance:
case class Employee(
person: Person,
salary: Int
)
// In code:
val employee = ...
println(employee.person.name)
Composition is a valid and a sound strategy that you should consider as well.
And in case you wonder what a sealed trait means — it is something that can be extended only in the same file. That is, the two case classes above have to be in the same file. This allows for exhaustive compiler checks:
val x = Employee(name = "Jack", salary = 50000)
x match {
case Employee(name) => println(s"I'm $name!")
}
Gives an error:
warning: match is not exhaustive!
missing combination Tourist
Which is really useful. Now you won't forget to deal with the other types of Persons (people). This is essentially what the Option class in Scala does.
If that does not matter to you, then you could make it non-sealed and throw the case classes into their own files. And perhaps go with composition.
case classes are perfect for value objects, i.e. objects that don't change any properties and can be compared with equals.
But implementing equals in the presence of inheritance is rather complicated. Consider a two classes:
class Point(x : Int, y : Int)
and
class ColoredPoint( x : Int, y : Int, c : Color) extends Point
So according to the definition the ColorPoint(1,4,red) should be equal to the Point(1,4) they are the same Point after all. So ColorPoint(1,4,blue) should also be equal to Point(1,4), right? But of course ColorPoint(1,4,red) should not equal ColorPoint(1,4,blue), because they have different colors. There you go, one basic property of the equality relation is broken.
update
You can use inheritance from traits solving lots of problems as described in another answer. An even more flexible alternative is often to use type classes. See What are type classes in Scala useful for? or http://www.youtube.com/watch?v=sVMES4RZF-8
In these situations I tend to use composition instead of inheritance i.e.
sealed trait IVehicle // tagging trait
case class Vehicle(color: String) extends IVehicle
case class Car(vehicle: Vehicle, doors: Int) extends IVehicle
val vehicle: IVehicle = ...
vehicle match {
case Car(Vehicle(color), doors) => println(s"$color car with $doors doors")
case Vehicle(color) => println(s"$color vehicle")
}
Obviously you can use a more sophisticated hierarchy and matches but hopefully this gives you an idea. The key is to take advantage of the nested extractors that case classes provide