How to get the upload file with other inputs in play2?

Question

In html, a form with multipart data:

<form action="@routes.Files.upload" method="post" enctype="multipart/form-data">     <input type="hidden" name="groupId" value="1" />     <input type="hidden" name="tagId" value="2" />     <input type="file" name="file"/>     <input type="submit" value="upload it"/> </form> 

How to write the action Files upload?

I know how to get a uploaded file:

request.body.file("file") map {     filepart => filepart.ref.moveTo(newFile); } 

And how to get submitted inputs:

Form(tuple("groupId" -> text, "tagId" -> text)).bindFromRequest.fold(     errors => ...,     params => .... ) 

But how to combine them together?

I don't find a suitable type for file can be used in Form(tuple(...)), and neither a way to get input value in request.body.

Answer 1

This answer is for Java, but you should be able to adapt it to Scala fairly easily.

What you need to do is define a Model for all the fields in your form except the file. Then use the file-upload API as normal to retrieve the file.

For example, this is what I did:

The Form (in upload.scala.html):

@form(action = routes.UploadResourceController.doUpload(), 'enctype -> "multipart/form-data") {      @inputText(uploadForm("lang"))     @inputText(uploadForm("country"))     @inputFile(uploadForm("resourceFile"))      <p>         <input type="submit">     </p> } 

The Model (models/UploadResource.java):

public class UploadResource {     @Required     public String lang;      @Required     public String country;      /* notice a field for the file is missing */ } 

The Controller (controllers/UploadResourceController.java):

public static Result doUpload() {     Form<UploadResource> filledForm = uploadForm.bindFromRequest();      if (filledForm.hasErrors()) {         return badRequest(views.html.upload.render(filledForm));     } else {         UploadResource resource = filledForm.get();         MultipartFormData body = request().body().asMultipartFormData();         FilePart resourceFile = body.getFile("resourceFile");          /* Check resourceFile for null, then extract the File object and process it */      } } 

I hope this helps.

Answer 2

An example in Scala where the form field is required:

Model:

case class Specs (userid: String) 

Controller:

object Upload extends Controller {    val uploadForm = Form(          mapping(                "userid" -> nonEmptyText          )(Specs.apply)(Specs.unapply)    )    def upload = Action(parse.multipartFormData) { implicit request =>       val sp : Option[Specs] = uploadForm.bindFromRequest().fold (             errFrm => None,             spec => Some(spec)       )       request.body.file("file").map { f =>          sp.map { spec =>             val filePath = ... // incorporate userid             // XXX: file read to memory b4 writing to disk. bad for large files             f.ref.moveTo(new File(filePath), replace=true)             Ok("File uploaded")          }.getOrElse{             BadRequest("Form binding error.")          }       }.getOrElse {          BadRequest("File not attached.")       }    } }