I found it

thanks

"Braun, Heiko" wrote:

> i think you cant access the fileitem via the
> parameter parser. but i remember something like
>
> foo.getFileItem();
>
> take a look at the API. its somewhere hidden in there...
>
> /heiko
>
> -----Urspr�ngliche Nachricht-----
> Von: Matthew Inger [mailto:[EMAIL PROTECTED]]
> Gesendet: Mittwoch, 3. April 2002 17:10
> An: Turbine Users List
> Betreff: RE: upload a file from a form
>
> You need to parse the data different.  Servlet API does
> not parse multipart/form-data.  I would reccommend o'reilly's
> org.oreilly.servlet package.  go to www.oreilly.org for more
> information.  It's a java package written by Jason Hunter
> (who also wrote the O'Reilly Java Servlet book).  You MUST
> buy the book to legally use the classes.
>
> -----Original Message-----
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED]]On Behalf Of Patrice Le Cozler
> Sent: Wednesday, April 03, 2002 9:34 AM
> To: [EMAIL PROTECTED]
> Subject: upload a file from a form
>
> Hello,
>
> I need to upload a XML file from a form and parse it to put the datas it
> contains in a database.
>
> So I wrote my form like this :
>
> |   <form method="post" ENCTYPE="multipart/form-data"
> action="$link.setPage("Index.vm").setAction("SQL")">
> |       upload :
> |         <input type=file name="upfile">
> |         <input type="submit" name="eventSubmit_doUpload"
> value="Upload">
> |   </form>
>
> ...and the corresponding SQL.java like that :
>
> |        public void doUpload(RunData data, Context context)
> |     throws Exception
> |     {
> |      Log.debug("Upload : "+data.getParameters().getString("upfile"));
> |     }
>
> ( the Log.debug method just allows me to verify that the doUpload method
> is called when I press the "Upload" button of the form )
>
> but all I get in my log file is :
>
> Upload :
>
> so where's the file name ?
>
> thanks
>
> Patrice
>
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