Spark UDF返回类型

Question

I am trying to make a UDF like so:

import org.apache.spark.sql.functions.udf
def myFunction = udf( (input: String, modifier: Seq[String]) => {
    // Logic goes here...
    return Option(myString)
})

The function has a few exit points in the logic, but it complains about the return statement as per:

Message: <console>:69: error: method myFunction has return statement; needs result type
        return Option(myString)

I have spent the last few hours trying to work out how to declare the return type as an Option[String] but nothing seems to work.

I thought this would have been right, but it doesnt like it:

udf( (input: String, modifier: Seq[String]) => Option[String] {

Answer 1

I feel like I sort of resolved this.

As Ramesh pointed out, you dont have to use the return key word in a UDF. I had done this initially as there was a pile of IF-THEN-ELSE sort of blocks and I wanted the option to exit the function earlier with a return value.

I've now updated this:

def myFunction = udf( (input: String, modifier: Seq[String]) => Option[String] {
  var retrunStr = ""
  // Logic goes here...
    // Conditionally set a value
    retrunStr = input
  // Finally just set the result to the returnStr
  retrunStr
})

This seems to work, although it does feel messy to carry the returnStr variable. If I was really good, I should be able to get my conditional logic to return the correct value.

I removed some of the complexity in the logic, but I'm using nested matchers and foreach loops and breakables etc. but thats another topic.

Thanks again Ramesh, you were right to avoid using a return statement at all.