Vista - Findout shortcut from which script was launched ?

Hello,

In a VBScript I wish to know if the script was called from a shortcut and if
so what is the path of this shortcut. Is this possible ?

In case you wish to know, here is why : i have a script which renames files
in a folder.
Currently i do the renames in the folder where the script itself is
situated.
I would like this to work also with a shortcut to the script file : the
rename should apply to the folder where the shortcut is situated, not the
folder where the script file is situated.

Any idea ?

Thanks a lot

Jean-Marie

Jean-Marie wrote:
I don't believe there is any way for a script to know that it was launched
by a shortcut, or the location of the shortcut. Wscript.ScriptPath returns
the name and path of the script file itself, not any shortcut that launched
it.

However, objFSO.GetAbsolutePathName(".") returns the current directory. You
can modify the "Start in" property of the shortcut so this returns the value
you want:

Set objFSO = CreateObject("Scripting.FileSystemObject")
Wscript.Echo objFSO.GetAbsolutePathName(".")

You would need to modify this property of the shortcut for each folder. It
actually would not matter where the shortcut is saved.

--
Richard Mueller
MVP Directory Services
Hilltop Lab - http://www.rlmueller.net
--

"jm.almeras" <nospam@xxxxxx> wrote in message
news:ecaJQ8czJHA.1716@xxxxxxNo.
Change your script to accept the name of a folder as its only parameter.
Then run it by dragging the desired folder onto either the script itself, or
a shortcut to the script.

/Al

On May 6, 12:15 am, "jm.almeras" <nos...@xxxxxx> wrote:This is a slippery problem, but it does have a solution...
if you ask the question in the right way. I have a more
complex situation because I want to apply the .vbs
script only to the selected files of a given directory.

There are two situations. The first, which I never use
anymore, is that I have a shortcut. The shortcut needs
to know what directory it was started in. But that's
not good enough. It actually needs to know what
Explorer window is active so for that I had to make
a mini .OCX file which uses the winAPI
GetForegroundWindowHandle() and then I run through
the .windows collection of the Shell.Application
object to see if there's a match on the .HWND.
Richard Meuller's suggestion probably suffices
for your situation.

The other thing that I do (independent of the prior
paragraph) is to make a change in the registry
so that for all .jpg files, I get a context
menu item (ie. when I right click) that will
execute my script when it's selected. The
script gets passed the name of the file
that it was started on, from which it knows
the directory.

Csaba Gabor from Vienna

Thank you Richard, Al, urkec and Csaba for your advice.

Richard's solution works but i find it inconvenient, because editing the
shortcut property to write the folder path as startin directory is
cumbersome (i have tested, it does not work if i put "." as startin
property)

I took Al's suggestion and introduced the possibility of dropping a folder
on the script. It works fine. Also, urkec's idea to put the script in the
SendTo Folder is good.

Your answer is interesting to me because i also wish to make it possible to
apply the script to selected files. It works with a small number of files
which i drop on the script (or its shortcut), but there is a limit : the
maximum size of the command line (pity !)

I would be interested in knowing what change you make in the registry. But
is this solution not equivalent to putting the script in the SendTo Folder
and when selecting a file or several, selecting Send To the script in the
contextual menu ?

Thank you

Jean-Marie

"Csaba Gabor" <danswer@xxxxxx> a écrit dans le message de news:
32a8e31d-e8f8-4be2-809f-69e170270593@xxxxxx...

On May 8, 12:02 pm, "jm.almeras" <nos...@xxxxxx> wrote:Here's an example:
In HKCR\*\shell\ place:
Cmd (REG_SZ): &Cmd window in this directory

Then make a subkey: In HKCR\*\shell\Cmd\ place command (REG_SZ):
cmd /K cd "%0\.." & title Context Menu Cmd Window & mode con lines=60
cols=100
or
cmd "c:\phplib\Context Cmd window.lnk"

If you do the second one (which is preferable for me) my
Context Cmd window.lnk shortcut has the following properties set:
under General: Context Cmd window
under Shortcut:
Target: %windir%\system32\cmd.exe
Shortcut key: None
Run: Normal window
under Options:
Cursor size: Small
Display options: Window
Edit options: QuickEdit mode, Insert mode
under Layout:
Screen buffer size: Width: 90, Height: 300
Window size: Width: 90, Height: 70
Window position: Let system position window

If you have a custom editor you like, you could do something like:
\HKCR\*\shell\Edit2: &Edit (with emacs)
\HKCR\*\shell\Edit2\command: C:\Program Files\emacs\bin\runemacs.exe -
geometry 155x59 "%1"

For a specific file type it's a bit more involved. Let's say .jpg
Go to
\HKCR\.jpg\(Default) and see the value listed there. It's jpegfile
Now make
\HKCR\jpegfile\shell\Resize: Res&ize
\HKCR\jpegfile\shellResize\command: c:\phpdir\php-win.exe "C:\Photo
\Util\Resize.php" "%1" %*

That %0 and %1 are the file that you are right clicking on.
The quotes (") are because of possible spaces in file names
%* means any remaining arguments
I've never worked with "SendTo" and I don't know the answer to that.