Thursday, July 12, 2007

Configure Windows Application in SSGD

Before configuring a Windows Application make sure the application is running on the Windows Application Server (Windows Terminal Server).

  1. First connect to the Windows Terminal Server directly via RDP with an RDP Client.
  2. Start a command prompt (Start->Run->cmd.exe)
  3. cd to the directory from where the application must be started (Startup Directory)
  4. start the program (Application Command Line)
If the application starts you are ready to configure the application in SSGD. First prepare the parameters used in SSGD:
  • Split the Application Command Line in "Application Command" and "Arguments for Command"
    • The Application Command is the first part of the Application Command Line until the extention exe, bat, com, vbs (and alike)
    • The remainder of the Application Command Line are the Arguments for the Command
    • Replace in both the \ with a double \\, So c:\Program Files\Internet Explorer\iexplore.exe will be c:\\Program Files\\Internet Explorer\\iexplore.exe
  • The protocol arguments are: -dir "Startup Directory" (with the \ replaced by \\)
    For example: -dir "c:\\Program Files\\Internet Explorer\\"

Configure the application in SSGD. This can be done in several ways. In this example I will use the command line:

tarantella object new_windowsapp \
--name ".../_ens/o=organization/cn=WindowsApplication" \
--width 1000 --height 800 \
--app Application Command \
--args Arguments for Command \
--protoargs "-dir Startup Directory" \
--appserv ".../_ens/o=organization/cn=ApplicationServerName"

1 comment:

rob tunru said...

Hi Remold,

Wellicht een idee om ook iets dergelijks te veretellen voor b.v. open office op een linux server ??
Voor newbies zoals ik zijn veel dingen niet vanzelfsprekend en een werken voorbeeld zou heel fijn zijn.
Trouwens hoe kan ik dat doen met object manager ??. Ik zie geen mogelijkheid om de IP aan te geven van de linux apps server. Mail me op rob.tunru@raketnet.nl