Scala Play upload file within a form

问题

How would I upload a file within a form defined with Scala Play's play.api.data.Forms framework. I want the file to be stored under Treatment Image.

  val cForm: Form[NewComplication] = Form(    
mapping(    
  "Name of Vital Sign:" -> of(Formats.longFormat),    
  "Complication Name:" -> text,    
  "Definition:" -> text,    
  "Reason:" -> text,    
  "Treatment:" -> text,    
  "Treatment Image:" -> /*THIS IS WHERE I WANT THE FILE*/,                
  "Notes:" -> text,    
  "Weblinks:" -> text,    
  "Upper or Lower Bound:" -> text)    
  (NewComplication.apply _ )(NewComplication.unapply _ ))  

is there a simple way to do this? By using built in Formats?

回答1:

I think you have to handle the file component of a multipart upload separately and combine it with your form data afterwards. You could do this several ways, depending on what you want the treatment image field to actually be (the file-path as a String, or, to take you literally, as a java.io.File object.)

For that last option, you could make the treatment image field of your NewComplication case class an Option[java.io.File] and handle it in your form mapping with ignored(Option.empty[java.io.File]) (so it won't be bound with the other data.) Then in your action do something like this:

def createPost = Action(parse.multipartFormData) { implicit request =>
  request.body.file("treatment_image").map { picture =>
    // retrieve the image and put it where you want...
    val imageFile = new java.io.File("myFileName")
    picture.ref.moveTo(imageFile)

    // handle the other form data
    cForm.bindFromRequest.fold(
      errForm => BadRequest("Ooops"),

      complication => {
        // Combine the file and form data...
        val withPicture = complication.copy(image = Some(imageFile))

        // Do something with result...

        Redirect("/whereever").flashing("success" -> "hooray")
      }
    )
  }.getOrElse(BadRequest("Missing picture."))
}

A similar thing would apply if you wanted just to store the file path.

There are several ways to handle file upload which will usually depend on what you're doing with the files server-side, so I think this approach makes sense.

来源：https://stackoverflow.com/questions/23982817/scala-play-upload-file-within-a-form