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From Steve Ruby <st...@rubysolutions.com>
Subject Re: File upload...
Date Mon, 27 Nov 2000 23:38:13 GMT

The response will be multipart mime format you will most likely want
some utils to parts the multipart format, this one works well for me:

http://www.servlets.com/resources/com.oreilly.servlet/index.html

you'll then do something like this:

// parse the content and put the file in default directory
multi = new MultipartRequest(data.getRequest(), uploaddir);

// get the other parameters like this
multi.getParameter("paramname");



"Servais, Pascal-Eric" wrote:
> 
> Is there anybody who could tell me how to process a file upload from a html
> form after the user has submitted the form ? I guess that the
> HttpServletRequest is implied but I have no idea how.
> 
> Thanks for any hint.
> 
> Pascal-Eric Servais ( http://pages.infinit.net/denethor
> <http://pages.infinit.net/denethor>  )
> Cognicase ( division Web )
> ---
> PGP Fingerprint : 8D0C FB66 CAF0 B9B3 E925  8D2E 7BDB 1D47 DC0B 4AEA
> "Engagez-vous qu'ils disaient, vous allez voir du pays...", Anonyme

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