Making equivelant account to 'Local System'

G

Guest

Guest
Archived from groups: microsoft.public.windowsxp.configuration_manage,microsoft.public.windowsxp.general,microsoft.public.windowsxp.help_and_support,microsoft.public.windowsxp.security_admin (More info?)

We have some software that includes a service that, by default, runs under
the Local System account. One of our customers wants to run the service
under a different account but don't know what permissions to give that
account.

How do I find out what permissions 'Local System' has or what permissioins
our service needs? (I am new to the company and don't really know the
internals of the application).

TIA,

--- Al.
 
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Guest

Guest
Archived from groups: microsoft.public.windowsxp.configuration_manage,microsoft.public.windowsxp.general,microsoft.public.windowsxp.help_and_support,microsoft.public.windowsxp.security_admin (More info?)

You probably already tried to talk with the developer or find specs
somewhere, so I'll assume this is essentially a black box. If so, you might
try running the app with application verifier. It has settings that will
tell you what files/reg keys/etc. the app accesses.
http://www.microsoft.com/windows/appcompatibility/appverifier.mspx

Another option is using some of the tools from sysinternals like filemon,
regmon, process explorer, etc.

Or you could just try running the service as a regular user and see what
breaks.
--
Drew Cooper [MSFT]
This posting is provided "AS IS" with no warranties, and confers no rights.


"TVR Fan" <nosuch@address.com> wrote in message
news:u7x3D3MOEHA.2728@TK2MSFTNGP12.phx.gbl...
> We have some software that includes a service that, by default, runs under
> the Local System account. One of our customers wants to run the service
> under a different account but don't know what permissions to give that
> account.
>
> How do I find out what permissions 'Local System' has or what permissioins
> our service needs? (I am new to the company and don't really know the
> internals of the application).
>
> TIA,
>
> --- Al.
>
>