[python-win32] Vista and listening sockets

[Tim Roberts]

le dahut wrote:
> Hello,
>
> When launching an app that opens a socket, windows Vista complains 
> that I have to accept it before it is launched and then that the 
> firewall accepts the opening of the port.
>
> Is there a way to register my app to be launched by users without UAC 
> warning ?
>
> Is there a way to open the port on the firewall inside the program or 
> its installer ?

You can tweak the firewall settings, but you have to be UAC to do so, 
because you have to write to the HKEY_LOCAL_MACHINE registry hive.  If 
you have a setup application that has "setup" in the file name, this 
should work.  BTW, this may be the longest registry path you will ever 
encounter:

    HKEY_LOCAL_MACHINE\
      System\
        CurrentControlSet\
          Services\
            SharedAccess\
              Parameters\
                FirewallPolicy\
                  DomainProfile\
                    AuthorizedApplications\
                      List\

Within that key, create a REG_SZ value whose name is the path to your 
application, whose value is a 4-part string separated by colons:
    <path-to-app>:*:Enabled:<friendly name>

Put the same thing into
                  ...
                  StandardProfile\
                    AuthorizedApplications\
                      List\

-- 
Tim Roberts, timr at probo.com
Providenza & Boekelheide, Inc.