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Cmd Line Switch/argument Help! Win Xp


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#1 DaveMcKeen

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Posted 08 November 2007 - 04:53 PM

I hope this is on topic for this forum! I want to create a Start Menu shortcut that runs a program (application) and opens a specified file when that program runs.

The Program:
E:\Ap\Adobe\Photoshop CS\Photoshop.exe

The File:
D:\Data\Photography\Projects\Promotional Documents - Biz Comp Flyer\CD Label\Photoshop\9-step scale.jpg

I think I'm looking for the same thing as a command line syntax, something like:

C:\Documents and Settings\dmckeen> program filespec /file filespec

I want to enter the syntax I'm looking for in the Target line of the Properties dialog of the Start Menu Shortcut I'm creating. I'm sure this is simple, but I've been going nuts trying to tell Google what I want.

Thanks in advance,

- Dave

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#2 Andrew

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Posted 08 November 2007 - 05:26 PM

Hi DaveMcKeen, welcome to BC! :thumbsup:

The Syntax of the command should be like this:

"E:\Ap\Adobe\Photoshop CS\Photoshop.exe" "D:\Data\Photography\Projects\Promotional Documents - Biz Comp Flyer\CD Label\Photoshop\9-step scale.jpg"

That's the full path of the application, followed by the full path of the file to be opened. Since the paths have spaces in the, enclose each one in quotes with a space between them.

#3 DaveMcKeen

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Posted 08 November 2007 - 05:57 PM

That's it! Thanks, AA. That gets me going.

- Dave




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