是否有一种很好的，安全的，快速的方法将InputStream写入Scala中的文件？

Question

具体来说，我将文件上传保存到Lift Web应用程序中的本地文件。

Answer 1

With Java 7 or later you can use Files from the new File I/O :

Files.copy(from, to)

where from and to can be Path s or InputStream s. This way, you can even use it to conveniently extract resources from applications packed in a jar.

Answer 2

If it's a text file, and you want to limit yourself to Scala and Java, then using scala.io.Source to do the reading is probably the fastest--it's not built in, but easy to write:

def inputToFile(is: java.io.InputStream, f: java.io.File) {
  val in = scala.io.Source.fromInputStream(is)
  val out = new java.io.PrintWriter(f)
  try { in.getLines().foreach(out.println(_)) }
  finally { out.close }
}

But if you need other libraries anyway, you can make your life even easier by using them (as Michel illustrates).

(PS--in Scala 2.7, getLines should not have a () after it.)

(PPS--in old versions of Scala, getLines did not remove the newline, so you need to print instead of println .)

Answer 3

I don't know about any Scala specific API, but since Scala is fully compatible to Java you can use any other library like Apache Commons IO and Apache Commons FileUpload .

Here is some example code (untested):

//using Commons IO:
val is = ... //input stream you want to write to a file
val os = new FileOutputStream("out.txt")
org.apache.commons.io.IOUtils.copy(is, os)
os.close()

//using Commons FileUpload
import javax.servlet.http.HttpServletRequest
import org.apache.commons.fileupload.{FileItemFactory, FileItem}
import apache.commons.fileupload.disk.DiskFileItemFactory
import org.apache.commons.fileupload.servlet.ServletFileUpload
val request: HttpServletRequest = ... //your HTTP request
val factory: FileItemFactory = new DiskFileItemFactory()
val upload = new ServletFileUpload(factory)
val items = upload.parseRequest(request).asInstanceOf[java.util.List[FileItem]]
for (item <- items) item.write(new File(item.getName))

Answer 4

The inputToFile method given above doesn't work well with binary files like .pdf files. It throws a runtime exception while attempting to decode the file into string. What worked for me was this:

def inputStreamToFile(inputStream: java.io.InputStream, file: java.io.File) = {
    val fos = new java.io.FileOutputStream(file)
    fos.write(
      Stream.continually(inputStream.read).takeWhile(-1 !=).map(_.toByte).toArray
    )
    fos.close()
}