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Katherine
01-09-2003, 09:26 PM
I need to Create a New Launch Condition under System Software requirements.
I have to check if registry entry, that contains a folder is exists on computer, then install my *.dll file into this folder. If not, then exit from installation.
Must to be obvious and intuitive to select this options from provided wizard as I did but it's not working in my situation.
If anybody know what a secret or can direct me to any doc or info, please inform or contact me.
kgogeizel@hotmail.com

Thanks a lot.

Chandima
01-10-2003, 09:49 AM
We can accomplish the first part of this problem without too much difficulty.

1. Go to the Reqruirements View
2. Launch the System Search Wizard and press "Next"
3. Select "Registry entry, that contains a folder"
4. Press "Next"
5. Suppose the folder path is in straing value named "FolderPath" in HKLM\Software\Test. Then the panel should be configured as follows:

Registry Root: HKEY_LOCAL_MACHINE
Registry Key: Software\Test
Registry Value: FolderPath
Search 64 bit portion: unchecked

6. Press "Next"
7. Select "Found" radio button.
8. Enter message as something like "A required folder was not found on your system. The setup will now exit."

Build and install. The setup will launch and search the registry key you specified. It'll then look on the local machine for that particular folder. If it finds it, the setup will continue, if not, it will gracefully exit.

As for putting a dll in this location there are two options. One would be to go to the General Information view and set the value of the INSTALLDIR property to "HKEY_LOCAL_MACHINE\Software\Test\FolderPath". Then in the files view put the dll in INSTALLDIR.

The other option would be to use a Custom Action to set this property.