persist.Load('someshortcut.lnk')
print sh.GetPath(shell.SLGP_RAWPATH)[0]
hth
Roger
"Steve M" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED]
> Below is some code adapted from something I think was written by Mark
> Hammond. Originally I needed to create a Windows shortcut (link), and
> this code does the trick, requiring only the target filename and the
> desired shortcut name.
> Now, I find I need to open a shortcut and extract the target filename,
> and I don't have a clue how that is achieved. To be clear, I mostly
> don't understand the gory (Windows API) details of this code. So, can
> anyone show how to open an existing shortcut file (given its name) and
> discover the name of the file to which it is a shortcut?
>
>
> import os
> from win32com.shell import shell
> import pythoncom
>
> # Get the shell interface.
> sh = pythoncom.CoCreateInstance(shell.CLSID_ShellLink, None, \
> pythoncom.CLSCTX_INPROC_SERVER, shell.IID_IShellLink)
>
> # Get an IPersist interface
> persist = sh.QueryInterface(pythoncom.IID_IPersistFile)
>
> target_of_link = os.path.abspath('target.doc')
> link_name = 'shortcut_to_target.doc.lnk'
>
> sh.SetPath(target_of_link)
> persist.Save(link_name, 1)
>
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