Some code to start with FileUpload...
DiskFileItemFactory diskFileItemFactory = new DiskFileItemFactory();
File repositoryPath = new File("/temp");
diskFileItemFactory.setRepository(repositoryPath);
ServletFileUpload servletFileUpload = new
ServletFileUpload(diskFileItemFactory);
if (ServletFileUpload.isMultipartContent(request)){//injected sevlet
request goes here
// Parse the HTTP request...
List<FileItem> fileItemsList =
servletFileUpload.parseRequest(request);//injected sevlet request goes here
//do whatever you need to do with a file here (FileItems)
}
HTH.
-Arul
deniak wrote:
All right, thanks for your ideas. It's really helpful.
I tried the simplest solution just to know if it works...
My client is still the same.
I tried to add this code on the server side:
private @Resource HttpServletRequest requestObject;
but I don't know how I can get my 2 files with the HttpServletRequest...
Did I miss something??
Brad-77 wrote:
There you go, plenty of options. I'd recommend you go with Sergey's
suggestions as he wrote the JAX-RS implementation for CXF and actually
knows what he's talking about :-)
On Thu, Jul 31, 2008 at 10:52 PM, Sergey Beryozkin
<[EMAIL PROTECTED]> wrote:
This is probably even simpler, thanks, You can use an InputStream in the
signature - if you don't want to deal with HttpServletRequests
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: 31 July 2008 19:01
To: [email protected]
Subject: Re: CXF and fileupload
No Worries :-)
If you get FileUpload from apache commons then you can use its
multipart request class to parse the http request on the server side.
If I remember correctly, to get the HttpServletRequest object in your
service method include this method parameter:
@Context HttpServletRequest req
You can then parse the req object with the FileUpload multipart request
object and get access to your files.
Hope that helps.
On 7/31/08, deniak <[EMAIL PROTECTED]> wrote:
Ok, let's say I set the requestentity with a MultipartRequestEntity.
File zip = new File("/path/fileToUpload.zip");
File xml = new File("/path/fileToUpload.xml");
PostMethod filePost = new PostMethod("http://host/some_path";);
Part[] parts = {
new FilePart(xml.getName(), xml)
new FilePart(zip.getName(), zip)
};
filePost.setRequestEntity(
new MultipartRequestEntity(parts, filePost.getParams())
);
How can I deal with this MultipartRequestEntity on the server side with
CXF?
I know I may ask some stupid questions but it really drives me crazy :s
Thanks for your help
Brad-77 wrote:
Sounds like http request with an attached file to me, which I believe
requires a multipart request.
See 3.7.2 in http://www.ietf.org/rfc/rfc2616.txt
On Thu, Jul 31, 2008 at 2:26 PM, deniak <[EMAIL PROTECTED]>
wrote:
Actually, I just want to send a xml file and a zip file with my client
and then handle these 2 files on my server using CXF.
It's easy with just a XML file like described in the CXF samples but
there's
nothing about
uploading a file
Brad-77 wrote:
So if I'm reading this correctly you just want to read the data out
of
a file and send it as the body in a PUT request?
Or is there more to it than that?
On Thu, Jul 31, 2008 at 1:56 PM, deniak <[EMAIL PROTECTED]>
wrote:
I forgot to mention I need to use rest web services
Benson Margulies-4 wrote:
Well, CXF supports MTOM and SwA.
On Thu, Jul 31, 2008 at 8:34 AM, deniak <[EMAIL PROTECTED]>
wrote:
Actually, i need to use a put method and I was wondering how cxf
can
handle
it
Brad-77 wrote:
Hi,
if you mean using HTML form POSTs then yes, its pretty easy with
commons FileUpload: http://commons.apache.org/fileupload/
You need to get access the the HTTP request object which you can
pass
to the mime multipart request class in FileUploads. From there
you
can
iterate through the request parts, one of which will be your
file.
Hope that helps.
Brad.
On Thu, Jul 31, 2008 at 1:06 PM, deniak
<[EMAIL PROTECTED]>
wrote:
Hi,
Has anyone ever dealt with fileupload using cxf?
I mean send a file to the server and get it with cxf?
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