How to remove a folder and some links from start-menu
Hi!
An application (Kaspersky Antivirus) is installed on quite
some Win2K-SP-Workstations by the Kaspersky Administration
Kit (with its own installer).
With this installation an entry to the start-menu is made
and this entry shall be removed by Zfd 3.2-SP3. I ran
through some tests with an application-object (run as secure
system user), which only intent was to remove this entry.
First I tried by "Distribution Options - Application Files"
to remove the folder "C:\Dokumente und Einstellungen\All
Users\Startmen\Programme\Kaspersky Anti-Virus 6.0 for
Windows Workstations". Then I tried a "Distribution Script"
with "RD /S /Q "C:\Dokumente und Einstellungen\All
Users\Startmen\Programme\Kaspersky Anti-Virus 6.0 for
Windows Workstations"", which also failed.
Do I miss something? Where could I check the results of
those tries to get a better feeling on what is going wrong
here?
Thanks in advance.
Karl
When doing the job with an *.bat-file it works, when going
with the distribution script it does not. Are all the
windows/dos/batch commands available in the distribution
script option?
Karl
>>> Rolf Lidvall<[email protected]> 25.03.2009 11:37 >>>
>> Does that mean that, in case there are some files in
>that
>> folder, I have to delete those files first?
>
>Yes.
>
>> Can this whole
>> procedere be done by one app-object with keeping the
>order
>> like delete files first and then the folder as last
>thing?
>
>I think it's safest to do that with a script. I'd use the
>distribution
>script option.
>
>
>Regards
>Rolf Lidvall
>Swedish Radio (Ltd)
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TRY.
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CATCH cx_send_req_bcs INTO lc_send_exception.
ENDTRY.
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TRY.
lc_bcs->send( ).
CATCH cx_send_req_bcs INTO lc_send_exception.
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COMMIT WORK.
WRITE:/ 'Mail sent'.
ENDFORM. "call_smartforms
End Code --->
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