According to the FAQ we are suppose to handle file uploads using the
given example. I added some code in the isFormField block to attempt
to capture input parameters. However this caused
org.apache.commons.fileupload.FileItemStream.ItemSkippedException.
Anyone is able to this successfully?
import org.apache.commons.fileupload.FileItemStream;
import org.apache.commons.fileupload.FileItemIterator;
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import java.io.InputStream;
import java.io.IOException;
import java.util.logging.Logger;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class FileUpload extends HttpServlet {
private static final Logger log =
Logger.getLogger(FileUpload.class.getName());
public void doPost(HttpServletRequest req, HttpServletResponse res)
throws ServletException, IOException {
try {
ServletFileUpload upload = new ServletFileUpload();
res.setContentType("text/plain");
FileItemIterator iterator = upload.getItemIterator(req);
while (iterator.hasNext()) {
FileItemStream item = iterator.next();
InputStream stream = item.openStream();
if (item.isFormField()) {
log.warning("Got a form field: " + item.getFieldName());
//i added this
if(item.getName().equals("myInputParam")) {
String myInputParam = Streams.asString(stream);
}
} else {
log.warning("Got an uploaded file: " + item.getFieldName() +
", name = " + item.getName());
// You now have the filename (item.getName() and the
// contents (which you can read from stream). Here we just
// print them back out to the servlet output stream, but you
// will probably want to do something more interesting (for
// example, wrap them in a Blob and commit them to the
// datastore).
int len;
byte[] buffer = new byte[8192];
while ((len = stream.read(buffer, 0, buffer.length)) != -1)
{
res.getOutputStream().write(buffer, 0, len);
}
}
}
} catch (Exception ex) {
throw new ServletException(ex);
}
}
}
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