ScalaFileUpload

Handling file upload

Uploading files in a form using multipart/form-data

The standard way of uploading files in a Web application is to use a form with a special multipart/form-data encoding. It allows to mix standard form data with file attachment.

Start by writing an Html form:

@form(action = @routes.Application.upload, 'enctype -> "multipart/form-data") {
    
    <input type="file" name="picture">
    
    <p>
        <input type="submit">
    </p>
    
}

Now let's define the upload action using a multipartFormData body parser:

def upload = Action(parse.mutipartFormData) { request =>
  request.body.file("picture").map { picture =>
    val filename = picture.filename 
    val contentType = picture.contentType
    picture.ref.moveTo("/tmp/picture")
    Ok("File uploaded")
  }.getOrElse {
    Redirect(routes.Application.index).flashing(
      "error" -> "Missing file"
    )
  }
}

The ref attribute give you a reference to a TemporaryFile. This is the default way the mutipartFormData parser handle file upload.

Note: As always you can also use the anyContent body parser and retrieve it as request.asMultipartFormData.

Direct file upload

Another way to send files to the server is to use Ajax to upload file asynchronously in a form. In this case the request body with not been encoded as multipart/form-data, but will just contain the plain file content.

In this case we can just use a body parser that store the request body content to a file. In this example let's use the temporaryFile body parser:

def upload = Action(parse.temporaryFile) { request =>
  request.body.moveTo("/tmp/picture")
  Ok("File uploaded")
}

Writing your own body parser

If you want to handle the file upload directly without buffering it in a temporary file, you can just write your own BodyParser. In this case you will receive chunks of data you are free to push anywhere you want.

If you want to use the multipart/form-data encoding, you can still use the default mutipartFormData parser by providing your own PartHandler[FilePart[A]]. You receive the part headers, and you have to give an Iteratee[Array[Byte], FilePart[A]] that will produce the right FilePart.

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