Calling remote apps on a windows server from a 'buntu client

PClOStinspace · 04-23-2010, 05:46 PM

Hi, this is perhaps a strange one!!

My desktop is running 'buntu 9.10 and I have recently aquired a couple of half decent servers running Windows server 2003. I have a few windows app's that I use regularly for work that I have had limited success in running with wine or in vm's.

I stumbled upon a how to to call apps in a vm to run seamlessly on the host desktop and did some playing and have succeeded in calling app's on my servers from my 'buntu desktop, the command I have set in my launcher for one of them (DIALux) is:-

Code:

rdesktop -A -s "c:\seamlessrdp\seamlessrdpshell.exe C:\Program Files\DIALux\DIALux.exe" <servers IP address>:3389 -u username -p password

This works fine to launch the app.

My question is, can I change the default application/file association in 'buntu with a custom command so I can double click on a file and have it open in one of these remote app's??

If I can, what is the correct custom command to be using as using the above doesn't work at all, it just tries to open the file with rdesktop not the remote app.

10110111 · 04-23-2010, 06:01 PM

Is the file located on the server or on local Ubuntu machine? If it's on the local machine, you should share it via FTP/SSHFS/SMB/NFS/etc., mount on the server as a drive, then make a wrapper script to start the program you need with the argument of file name, converted to server path, something like this:

Code:

#!/bin/bash
rdesktop -A -s "c:\seamlessrdp\seamlessrdpshell.exe C:\Program Files\DIALux\DIALux.exe `translatePath $1`" <servers IP address>:3389 -u username -p password

where translatePath is the program/script which would translate ubuntu path to server path to the share you mounted on the server.
Then, associate the file type with this wrapper script, and you're done.