How to break after response in Play (Scala) - scala

Question

I am new to Play Framework using Scala. I want to evaluate a condition and in that condition evaluates to true, I want to send a response and exit at that point. However, the code below that I am trying continues till the end.
I tried breaking with a return statement - However, I get a type mismatch. Can someone help me out with this?
def hello = Action { request =>
if (true) {
Ok("in If")
// Return at this point
}
print("This line should not be printed")
Ok("final")
}
EDIT
Assume a GET call is being made with 4 parameters - name, age, married, spouse. I want to make sure all 3 params (name, age, married) are passed in, and if married is true, check if spouse is passed in. If this validation fails, I want to respond saying Bad Request. Else, continue with logic. How do I write this?

Answer 1

Here is an alternative way to do it:
case class QueryInput(name: String, age: Int, married: Boolean, spouse: Option[String]) {
def validated = if(married && spouse.isEmpty) None else Some(this)
}
def validateInput(request: RequestHeader) = {
val input = for {
name <- request.queryString.get("name").flatMap(_.headOption)
age <- request.queryString.get("age").flatMap(_.headOption.flatMap(a=>Try(a.toInt).toOption))
married <- request.queryString.get("married").flatMap(_.headOption.map(_=="true"))
} yield {
QueryInput(name, age, married, request.queryString.get("spouse").flatMap(_.headOption))
}
input.flatMap(_.validated)
}
def hello() = Action { request =>
validateInput(request) match {
case Some(input) => Ok(input.toString())
case None => BadRequest
}
}
In fact, there are many options. You could also play with the Either class to do validation: Left value to accumulate errors and return bad request, right value to construct your validated input.

Answer 2

My recommendation would be to have a method for validating the parameters. Then do a simple if/else to check if the parameters are valid and return a success or a general error.
If you really want a specific

Answer 3

First thing:
When the block evaluates, all of its expressions and declarations are processed in order, and then the block returns the value of the last expression as its own value.
Second: don't use return.
And the third one is a Play Framework way of resolving your problem: action composition. Though I would not say that it is trivial.

Answer 4

You can do this, by putting a return Ok in but really, thats not the scala way. What you want to do is to change your mindset and imagine everything as a function. If you didnt know, if-then-else always returns a value. For example, you can actually write if this way:
def hello = Action { request =>
val result = if (true) {
Ok("foo")
} else {
Ok("bar")
}
result
}
of course, an even more scala way is to use matchers
def hello = Action { request =>
val result = true match {
case true => Ok("foo")
case _ => Ok("bar")
}
result
}
Take that one step further and you dont even need to specify the result object at all, because scala figures out the returning object based on the last object returned/created.
def hello = Action { request =>
true match {
case true => Ok("foo")
case _ => Ok("bar")
}
}
EDIT: TO answer the OP's edit, you still want to use the matcher. Assuming your vals are options, heres what you do:
def hello(nameOpt:Option[String], ageOpt:Option[String], marriedOpt:Option[String]) = Action { request =>
(nameOpt, ageOpt, marriedOpt) match {
case (Some(name), Some(age), Some(married)) => Ok("all 3 are given")
case (Some(name), Some(age), _) => Ok("all 2 are given")
// functionally same as above
// case (Some(name), Some(age), None) => Ok("all 2 are given")
// some combination of the above
case (None, None, Some(married)) => Ok("married but no name or age")
// default case
case _ => Ok("bar")
}
}