Can we convert option[string] to caseclass in scala?-CodePudding

When I tried to convert Some(string) into a case class I am getting an exception.

Ex:

val a:option[string]= Some("abc")
case class hello(a:string)
a.get.convertto[hello] => error it is showing but I want result

CodePudding user response：

You need to allow for the Option being empty so avoid get and use getOrElse. Then just use the String to make the case class, like this:

val a: Option[String] = Some("abc")
case class hello(a: String)

val cc: hello = hello(a.getOrElse("default"))

CodePudding user response：

The convertto you're looking for is map().

case class Hello(a: String)

val a: Option[String] = ??? //might be Some(s), might be None
val result: Option[Hello] = a.map(Hello)