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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