need only Content-Type: text/plain from the form submit using mutilpart/form data

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Guan - 24 Oct 2006 03:41 GMT

Hi All,

I am trying to use NetComponents-1.3.8a to create a web based FTP
upload and download.

i am using enctype:multipart/form-data for form submission.

To obtained the contain the following code is used,

ServletInputStream inputLocal = request.getInputStream();
inputLocalBuf = new DataInputStream(inputLocal);
if (!ftp.storeFile("test.txt",inputLocalBuf))
- - - - - - - - - -- - - -- - - -- - - - -- - - -- - - - - -- - - -
-- - - - - -

For additional inform,
storefile method - storeFile(String, InputStream) where string will be
the filename, inputstream will be data obtain from inputstream.

This following is the output i obtained when i open the test.txt i
uploaded.

-----------------------------7d6762690786
Content-Disposition: form-data; name="radio1"

upload
-----------------------------7d6762690786
Content-Disposition: form-data; name="datafile"; filename="C:\Documents
and Settings\Administrator\Desktop\test.txt"
Content-Type: text/plain

This is a test
-----------------------------7d6762690786
Content-Disposition: form-data; name="Submit"

Submit
-----------------------------7d6762690786--

- -- - - - - -- - - -- - - - -- - - -- - - -- - - - - - - - - - -- - -
- - -- - - - - -- - - - - - - -- - - - - - - - -
Now my question are:

1) I only need the content - "this is a test" in my test.txt
2) i like to find out how obtain the filename i want from the
multipart/form-data rather than i hardcode.

Thanks in advance :)

Reply to this Message

Manish Pandit - 24 Oct 2006 08:17 GMT

Hi,

Use Apache FileUpload API. That will get FileItem object(s) from the
Multipart request, which should help you extract the file name and
content. You can strip out the path information and get just the file
name to store.

http://jakarta.apache.org/commons/fileupload/

-cheers,
Manish

Reply to this Message

Guan - 24 Oct 2006 09:44 GMT

Hi Manish,

Thanks for the reply. i shall read thru the API recommended.

> Hi,
>
[quoted text clipped - 7 lines]
> -cheers,
> Manish

Reply to this Message

Guan - 26 Oct 2006 05:58 GMT

Hi,

i included the following code to my program just to test it out.
However i encountered an error when running this line List items =
upload.parseRequest(request); during compliation

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
import java.util.*;
import java.util.List;
import com.oroinc.net.ftp.*;
import org.apache.commons.fileupload.FileUpload;
import org.apache.commons.fileupload.DiskFileUpload;
..............
.............
// first check if the upload request coming in is a multipart request
boolean isMultipart = FileUpload.isMultipartContent(request);

// parse this request by the handler
DiskFileUpload upload = new DiskFileUpload();

// this gives us a list of items from the request
List items = upload.parseRequest(request);

------------------------------------------------------------------------------------------------------
The error displayed in the Jcreator description :

unreported exception
org.apache.commons.fileupload.FileUploadException;must be caught or
declared to be thrown

> Hi,
>
[quoted text clipped - 7 lines]
> -cheers,
> Manish

Reply to this Message

Manish Pandit - 26 Oct 2006 06:10 GMT

You need to catch the FileUploadException. If you are not familiar with
Exception Handling in Java, I'd recommend reading some tutorials and
understanding it.

-cheers,
Manish

Reply to this Message

Guan - 04 Nov 2006 07:16 GMT

Hi Manish,

Thanks for your help. I have completed the upload part.

Is there any recommendation Class for the downloading of the file from
the server?

Thanks

> You need to catch the FileUploadException. If you are not familiar with
> Exception Handling in Java, I'd recommend reading some tutorials and
> understanding it.
>
> -cheers,
> Manish

Reply to this Message

need only Content-Type: text/plain from the form submit using mutilpart/form data

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