[Scala] no escape method overload for `UUID`

[paul-english]

Description

The scala client code generator supports using UUID types of the form {"type": "string", "format": "uuid"}. This correctly generates models that use java.util.UUID. However when you use this API you get errors like the following:

[error] /src/main/scala/io/swagger/client/api/RoutesApi.scala:605: overloaded method value escape with alternatives:
[error]   (value: Float)String <and>
[error]   (value: Double)String <and>
[error]   (value: Long)String <and>
[error]   (value: String)String
[error]  cannot be applied to (java.util.UUID)
[error]     val path = "/api/thing/{id}".replaceAll("\\{format\\}", "json").replaceAll("\\{" + "id" + "\\}", apiInvoker.escape(id))
[error]                                                                                                                           ^

In the generated ApiInvoker.scala we have:

  def escape(value: String): String = {
    URLEncoder.encode(value, "utf-8").replaceAll("\\+", "%20")
  }

  def escape(value: Long): String = value.toString
  def escape(value: Double): String = value.toString
  def escape(value: Float): String = value.toString

All we need to add is the corresponding def escape(value: UUID): String = value.toString.

Swagger-codegen version

2.2.1

Swagger declaration file content or url

Any definition with type "string" and format "uuid" should generate this.

{"type": "string", "format": "uuid"}

Command line used for generation

              import io.swagger.codegen.DefaultGenerator
              import io.swagger.codegen.languages.ScalaClientCodegen
              import io.swagger.codegen.{ClientOptInput, ClientOpts}
              import io.swagger.parser.SwaggerParser;

              val swagger = new SwaggerParser().read("swagger.json")
              val config = new ScalaClientCodegen()
              config.setOutputDir("generated_code")
              val input = new ClientOptInput()
                .opts(new ClientOpts())
                .swagger(swagger)
                .config(config)
              new DefaultGenerator().opts(input).generate()

Steps to reproduce

Generate a Scala language client with a schema that has a definition with UUID format type.

Related issues

n/a

Suggest a Fix

Add def escape(value: UUID): String = value.toString to the right spot in the ApiInvoker.scala generation. Also need to import java.util.UUID in the same file.

[wing328]

PR merged into master. Thanks for the contribution from @log0ymxm