在Play Framework 2.0中将文件上传为流

我已经能够使用以下Scala控制器代码将数据流式传输到第三方API：

def uploadFile() =     Action( parse.multipartFormData(myPartHandler) )     {      request => Ok("Done")    }def myPartHandler: BodyParsers.parse.Multipart.PartHandler[MultipartFormData.FilePart[Result]] = {        parse.Multipart.handleFilePart {          case parse.Multipart.FileInfo(partName, filename, contentType) => //Still dirty: the path of the file is in the partName... String path = partName; //Set up the PipedOutputStream here, give the input stream to a worker thread val pos:PipedOutputStream = new PipedOutputStream(); val pis:PipedInputStream  = new PipedInputStream(pos); val worker:UploadFileWorker = new UploadFileWorker(path,pis); worker.contentType = contentType.get; worker.start(); //Read content to the POS Iteratee.fold[Array[Byte], PipedOutputStream](pos) { (os, data) =>   os.write(data)   os }.mapDone { os =>   os.close()   Ok("upload done") }        }   }

UploadFileWorker是一个非常简单的Java类，其中包含对第三方API的调用。

public class UploadFileWorker extends Thread {String path;PipedInputStream pis;public String contentType = "";public UploadFileWorker(String path, PipedInputStream pis) {    super();    this.path = path;    this.pis = pis;}public void run() {    try {        myApi.store(pis, path, contentType);        pis.close();    } catch (Exception ex) {        ex.printStackTrace();        try {pis.close();} catch (Exception ex2) {}    }}

}

它不是完全完美的，因为我希望将路径恢复为Action的参数，但我无法做到这一点。因此，我添加了一段javascript，它更新了输入字段的名称（并因此更新了partName），并且可以解决问题。