CODEWITHSUNDEEP
scala
vampireLady
vampireLady

Reputation: 137

How to make a List[String] to be a list of strings

I have an Option[List[String]] where I have several fields and want to create an object with them as parameters. Is there a way of doing this other than manually?

Details:

class Foo(var1: String, var2: String, var3: String)

I can't do

foo = new Foo(datarow)

And would rather not do

foo = new Foo(datarow.get(0), datarow.get(1), datarow.get(2))

Upvotes: 3

Views: 87

Answers (1)

Travis Brown
Travis Brown

Reputation: 139058

There's a safe way to do this that's still pretty concise:

class Foo(var1: String, var2: String, var3: String)

val datarow = Option(List("a", "b", "c"))

val result: Option[Foo] = datarow.collect {
  case v1 :: v2 :: v3 :: _ => new Foo(v1, v2, v3)
}

The combination of collect and pattern matching on the first three elements of the list says "if this Option isn't empty and if it contains a list with at least three elements, use them to instantiate the Foo".

You end up with an Option[Foo], not a Foo, but that's a small price to pay to avoid runtime errors because the Option is empty or the list doesn't have enough elements.

Upvotes: 9

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