Extractor does not return exception - scala

We have the following example and it seems that extractors do not work while converting Json to case class.
import play.api.libs.json.Reads._
import play.api.libs.json._
import play.api.libs.json.Format.GenericFormat
val json: JsValue = Json.parse("""
{
"firstName" : "John",
"lastName" : "Doe"
}
""")
trait BasePublicForm {
def firstName: String
def lastName: String
}
case class CustomerPublicForm(firstName: String, lastName: String) extends BasePublicForm
case class LeadPublicForm(firstName: String, lastName: String ) extends BasePublicForm
object CustomerPublicForm {
implicit val writesPublicLeadFormRequest: Writes[CustomerPublicForm] = Json.writes[CustomerPublicForm]
implicit val readsPublicLeadFormRequest: Reads[CustomerPublicForm] = Json.reads[CustomerPublicForm]
def apply(firstName: String, lastName: String): CustomerPublicForm = {
if(firstName.equalsIgnoreCase("John")) {
throw new Exception("John Exception")
}
new CustomerPublicForm(firstName, lastName)
}
}
object LeadPublicForm {
def apply(firstName: String, lastName: String): LeadPublicForm = {
new LeadPublicForm(firstName, lastName)
}
}
val s = json.validate[CustomerPublicForm] match {
case JsSuccess(form, _) => {
form
// do something with place
}
case e: JsError => {
// error handling flow
throw new Exception("Error")
}
}
s
Link -> https://scastie.scala-lang.org/eZrHTOVkQvSUJmoAJMTXfQ
Any ideas why it does not return Exception as expected?

The code doesn't throw an Exception (which is probably not a good idea anyway), as the json input is valid and so .validate returns CustomerPublicForm(John,Doe)

The Json.reads[CustomerPublicForm] seems to be circumventing your apply method.
I'm not sure of the exact details of Play's macro Reads generation, but it's probably just going straight for your class constructor rather than the apply method. Since the apply method you wrote has the same signature as the existing constructor, you'd be better off if you moved the validation (exception throwing) logic into the constructor and just removing the custom apply.
Alternatively you could make a custom apply with a different signature (not that it makes sense in this case, but I'm speaking generally here), and then using non-macro code to implement the Reads for that class.

Related

DSL in scala using case classes

My use case has case classes something like
case class Address(name:String,pincode:String){
override def toString =name +"=" +pincode
}
case class Department(name:String){
override def toString =name
}
case class emp(address:Address,department:Department)
I want to create a DSL like below.Can anyone share the links about how to create a DSL and any suggestions to achieve the below.
emp.withAddress("abc","12222").withDepartment("HR")
Update:
Actual use case class may have more fields close to 20. I want to avoid redudancy of code
I created a DSL using reflection so that we don't need to add every field to it.
Disclamer: This DSL is extremely weakly typed and I did it just for fun. I don't really think this is a good approach in Scala.
scala> create an Employee where "homeAddress" is Address("a", "b") and "department" is Department("c") and that_s it
res0: Employee = Employee(a=b,null,c)
scala> create an Employee where "workAddress" is Address("w", "x") and "homeAddress" is Address("y", "z") and that_s it
res1: Employee = Employee(y=z,w=x,null)
scala> create a Customer where "address" is Address("a", "b") and "age" is 900 and that_s it
res0: Customer = Customer(a=b,900)
The last example is the equivalent of writing:
create.a(Customer).where("address").is(Address("a", "b")).and("age").is(900).and(that_s).it
A way of writing DSLs in Scala and avoid parentheses and the dot is by following this pattern:
object.method(parameter).method(parameter)...
Here is the source:
// DSL
object create {
def an(t: Employee.type) = new ModelDSL(Employee(null, null, null))
def a(t: Customer.type) = new ModelDSL(Customer(null, 0))
}
object that_s
class ModelDSL[T](model: T) {
def where(field: String): ValueDSL[ModelDSL2[T], Any] = new ValueDSL(value => {
val f = model.getClass.getDeclaredField(field)
f.setAccessible(true)
f.set(model, value)
new ModelDSL2[T](model)
})
def and(t: that_s.type) = new { def it = model }
}
class ModelDSL2[T](model: T) {
def and(field: String) = new ModelDSL(model).where(field)
def and(t: that_s.type) = new { def it = model }
}
class ValueDSL[T, V](callback: V => T) {
def is(value: V): T = callback(value)
}
// Models
case class Employee(homeAddress: Address, workAddress: Address, department: Department)
case class Customer(address: Address, age: Int)
case class Address(name: String, pincode: String) {
override def toString = name + "=" + pincode
}
case class Department(name: String) {
override def toString = name
}
I really don't think you need the builder pattern in Scala. Just give your case class reasonable defaults and use the copy method.
i.e.:
employee.copy(address = Address("abc","12222"),
department = Department("HR"))
You could also use an immutable builder:
case class EmployeeBuilder(address:Address = Address("", ""),department:Department = Department("")) {
def build = emp(address, department)
def withAddress(address: Address) = copy(address = address)
def withDepartment(department: Department) = copy(department = department)
}
object EmployeeBuilder {
def withAddress(address: Address) = EmployeeBuilder().copy(address = address)
def withDepartment(department: Department) = EmployeeBuilder().copy(department = department)
}
You could do
object emp {
def builder = new Builder(None, None)
case class Builder(address: Option[Address], department: Option[Department]) {
def withDepartment(name:String) = {
val dept = Department(name)
this.copy(department = Some(dept))
}
def withAddress(name:String, pincode:String) = {
val addr = Address(name, pincode)
this.copy(address = Some(addr))
}
def build = (address, department) match {
case (Some(a), Some(d)) => new emp(a, d)
case (None, _) => throw new IllegalStateException("Address not provided")
case _ => throw new IllegalStateException("Department not provided")
}
}
}
and use it as emp.builder.withAddress("abc","12222").withDepartment("HR").build().
You don't need optional fields, copy, or the builder pattern (exactly), if you are willing to have the build always take the arguments in a particular order:
case class emp(address:Address,department:Department, id: Long)
object emp {
def withAddress(name: String, pincode: String): WithDepartment =
new WithDepartment(Address(name, pincode))
final class WithDepartment(private val address: Address)
extends AnyVal {
def withDepartment(name: String): WithId =
new WithId(address, Department(name))
}
final class WithId(address: Address, department: Department) {
def withId(id: Long): emp = emp(address, department, id)
}
}
emp.withAddress("abc","12222").withDepartment("HR").withId(1)
The idea here is that each emp parameter gets its own class which provides a method to get you to the next class, until the final one gives you an emp object. It's like currying but at the type level. As you can see I've added an extra parameter just as an example of how to extend the pattern past the first two parameters.
The nice thing about this approach is that, even if you're part-way through the build, the type you have so far will guide you to the next step. So if you have a WithDepartment so far, you know that the next argument you need to supply is a department name.
If you want to avoid modifying the origin classes you can use implicit class, e.g.
implicit class EmpExtensions(emp: emp) {
def withAddress(name: String, pincode: String) {
//code omitted
}
// code omitted
}
then import EmpExtensions wherever you need these methods

JSON serialization of Scala enums using Jackson

Following this article https://github.com/FasterXML/jackson-module-scala/wiki/Enumerations
The enumeration declaration is as
object UserStatus extends Enumeration {
type UserStatus = Value
val Active, Paused = Value
}
class UserStatusType extends TypeReference[UserStatus.type]
case class UserStatusHolder(#JsonScalaEnumeration(classOf[UserStatusType]) enum: UserStatus.UserStatus)
The DTO is declared as
class UserInfo(val emailAddress: String, val userStatus:UserStatusHolder) {
}
and the serialization code is
val mapper = new ObjectMapper()
mapper.registerModule(DefaultScalaModule)
def serialize(value: Any): String = {
import java.io.StringWriter
val writer = new StringWriter()
mapper.writeValue(writer, value)
writer.toString
}
The resulting JSON serialization is
{
"emailAddress":"user1#test.com",
"userStatus":{"enum":"Active"}
}
Is it possible to get it the following form ?
{
"emailAddress":"user1#test.com",
"userStatus":"Active"
}
Have you tried:
case class UserInfo(
emailAddress: String,
#JsonScalaEnumeration(classOf[UserStatusType]) userStatus: UserStatus.UserStatus
)
The jackson wiki's example is a little misleading. You don't need the holder class. Its just an example of a thing that has that element. The thing you need is the annotation

required: spray.httpx.marshalling.ToResponseMarshallable Error

Hey im pretty new to Spray and reactive mongo .
Im trying to return a list of result as json but i'm having some issue with converting the result to list of json.
this is my model
import reactivemongo.bson.BSONDocumentReader
import reactivemongo.bson.BSONObjectID
import reactivemongo.bson.Macros
case class Post(id: BSONObjectID, likes: Long, message: String, object_id: String, shares: Long)
object Post {
implicit val reader: BSONDocumentReader[Post] = Macros.reader[Post]
}
the Mongo method
def getAll(): Future[List[Post]] ={
val query = BSONDocument(
"likes" -> BSONDocument(
"$gt" -> 27))
collection.find(query).cursor[Post].collect[List]()
}
and this is the route
val route1 =
path("posts") {
val res: Future[List[Post]]= mongoService.getAll()
onComplete(res) {
case Success(value) => complete(value)
case Failure(ex) => complete(ex.getMessage)
}
}
error
type mismatch; found : List[com.example.model.Post] required: spray.httpx.marshalling.ToResponseMarshallable
thanks,
miki
You'll need to define how a Post will be serialized, which you can do via a spray-json Protocol (see the docs for more detailed information). It's quite easy to do so, but before that, you'll also need to define a format for the BSONObjectId type, since there's no built-in support for that type in spray-json (alternatively, if object_id is a string representation of the BSONObjectId, think about removing the id property from your Post class or change it to be a String):
// necessary imports
import spray.json._
import spray.httpx.SprayJsonSupport._
implicit object BSONObjectIdProtocol extends RootJsonFormat[BSONObjectID] {
override def write(obj: BSONObjectID): JsValue = JsString(obj.stringify)
override def read(json: JsValue): BSONObjectID = json match {
case JsString(id) => BSONObjectID.parse(id) match {
case Success(validId) => validId
case _ => deserializationError("Invalid BSON Object Id")
}
case _ => deserializationError("BSON Object Id expected")
}
}
Now, we're able to define the actual protocol for the Post class:
object PostJsonProtocol extends DefaultJsonProtocol {
implicit val format = jsonFormat5(Post.apply)
}
Furthermore, we'll also need to make sure that we have the defined format in scope:
import PostJsonProtocol._
Now, everything will compile as expected.
One more thing: have a look at the docs about the DSL structure of spray. Your mongoService.getAll() isn't within a complete block, which might not reflect your intentions. This ain't an issue yet, but probably will be if your route get more complex. To fix this issue simply put the future into the onComplete call or make it lazy:
val route1 =
path("posts") {
onComplete(mongoService.getAll()) {
case Success(value) => complete(value)
case Failure(ex) => complete(ex.getMessage)
}
}

Adding functionality before calling constructor in extra constructor

Is it possible to add functionality before calling constructor in extra constructor in scala ?
Lets say, I have class User, and want to get one string - and to split it into attributes - to send them to the constructor:
class User(val name: String, val age: Int){
def this(line: String) = {
val attrs = line.split(",") //This line is leading an error - what can I do instead
this(attrs(0), attrs(1).toInt)
}
}
So I know I'm not able to add a line before sending to this, because all constructors need to call another constructor as the first statement of the constructor.
Then what can I do instead?
Edit:
I have a long list of attributes, so I don't want to repeat line.split(",")
I think this is a place where companion object and apply() method come nicely into play:
object User {
def apply(line: String): User = {
val attrs = line.split(",")
new User(attrs(0), attrs(1).toInt)
}
}
class User(val name: String, val age: Int)
Then you just create your object the following way:
val u1 = User("Zorro,33")
Also since you're exposing name and age anyway, you might consider using case class instead of standard class and have consistent way of constructing User objects (without new keyword):
object User {
def apply(line: String): User = {
val attrs = line.split(",")
new User(attrs(0), attrs(1).toInt)
}
}
case class User(name: String, age: Int)
val u1 = User("Zorro,33")
val u2 = User("Zorro", "33")
Ugly, but working solution#1:
class User(val name: String, val age: Int){
def this(line: String) = {
this(line.split(",")(0), line.split(",")(1).toInt)
}
}
Ugly, but working solution#2:
class User(val name: String, val age: Int)
object User {
def fromString(line: String) = {
val attrs = line.split(",")
new User(attrs(0), attrs(1).toInt)
}
}
Which can be used as:
val johny = User.fromString("johny,35")
You could use apply in place of fromString, but this will lead to a confusion (in one case you have to use new, in the other you have to drop it) so I prefer to use different name
Another ugly solution:
class User(line: String) {
def this(name: String, age: Int) = this(s"$name,$age")
val (name, age) = {
val Array(nameStr,ageStr) = line.split(",")
(nameStr,ageStr.toInt)
}
}
But using a method of the companion object is probably better.

How to represent optional fields in spray-json?

I have an optional field on my requests:
case class SearchRequest(url: String, nextAt: Option[Date])
My protocol is:
object SearchRequestJsonProtocol extends DefaultJsonProtocol {
implicit val searchRequestFormat = jsonFormat(SearchRequest, "url", "nextAt")
}
How do I mark the nextAt field optional, such that the following JSON objects will be correctly read and accepted:
{"url":"..."}
{"url":"...", "nextAt":null}
{"url":"...", "nextAt":"2012-05-30T15:23Z"}
I actually don't really care about the null case, but if you have details, it would be nice. I'm using spray-json, and was under the impression that using an Option would skip the field if it was absent on the original JSON object.
Works for me (spray-json 1.1.1 scala 2.9.1 build)
import cc.spray.json._
import cc.spray.json.DefaultJsonProtocol._
// string instead of date for simplicity
case class SearchRequest(url: String, nextAt: Option[String])
// btw, you could use jsonFormat2 method here
implicit val searchRequestFormat = jsonFormat(SearchRequest, "url", "nextAt")
assert {
List(
"""{"url":"..."}""",
"""{"url":"...", "nextAt":null}""",
"""{"url":"...", "nextAt":"2012-05-30T15:23Z"}""")
.map(_.asJson.convertTo[SearchRequest]) == List(
SearchRequest("...", None),
SearchRequest("...", None),
SearchRequest("...", Some("2012-05-30T15:23Z")))
}
You might have to create an explicit format (warning: psuedocodish):
object SearchRequestJsonProtocol extends DefaultJsonProtocol {
implicit object SearchRequestJsonFormat extends JsonFormat[SearchRequest] {
def read(value: JsValue) = value match {
case JsObject(List(
JsField("url", JsString(url)),
JsField("nextAt", JsString(nextAt)))) =>
SearchRequest(url, Some(new Instant(nextAt)))
case JsObject(List(JsField("url", JsString(url)))) =>
SearchRequest(url, None)
case _ =>
throw new DeserializationException("SearchRequest expected")
}
def write(obj: SearchRequest) = obj.nextAt match {
case Some(nextAt) =>
JsObject(JsField("url", JsString(obj.url)),
JsField("nextAt", JsString(nextAt.toString)))
case None => JsObject(JsField("url", JsString(obj.url)))
}
}
}
Use NullOptions trait to disable skipping nulls:
https://github.com/spray/spray-json#nulloptions
Example:
https://github.com/spray/spray-json/blob/master/src/test/scala/spray/json/ProductFormatsSpec.scala
Don't know if this will help you but you can give that field a default value in the case class definition, so if the field is not in the json, it will assign the default value to it.
Easy.
import cc.spray.json._
trait MyJsonProtocol extends DefaultJsonProtocol {
implicit val searchFormat = new JsonWriter[SearchRequest] {
def write(r: SearchRequest): JsValue = {
JsObject(
"url" -> JsString(r.url),
"next_at" -> r.nextAt.toJson,
)
}
}
}
class JsonTest extends FunSuite with MyJsonProtocol {
test("JSON") {
val search = new SearchRequest("www.site.ru", None)
val marshalled = search.toJson
println(marshalled)
}
}
For anyone who is chancing upon this post and wants an update to François Beausoleil's answer for newer versions of Spray (circa 2015+?), JsField is deprecated as a public member of JsValue; you should simply supply a list of tuples instead of JsFields. Their answer is spot-on, though.