Digital Workshop

I am making a media kit that contains press release documents and images. Each image page is a simple photo gallery with the photos appearing in a defined area and navigation buttons to go from one picture to the next.

Is there a way to reliably open Windows Explorer to the IMAGES folder on the application CD so that the user may open the hi-res versions of the images? Or perhaps to copy the full, hi-res version of the image to the clipboard?

I don't want to automatically "launch" the files, since this could be disruptive, but being able to browse to the files' location on the CD would be nice. Thanks for any pointers.
-Tom

Hi Tom

You can launch explorer to the IMAGES folder on the CD
with this command

LaunchFile("explorer.exe",SYSTEM_PUBLICATION_DIR+"\\IMAGES")

If you have the hi res image file name stored in a variable,
you can open explorer with the file already selected using
the /select switch

assuming the file name of the hi res is stored in the variable "File_Name"

LaunchFile("explorer.exe ","/select,"+SYSTEM_PUBLICATION_DIR+"\\"+File_Name)

HTH
Chris

Chris, Thanks so much for that information. It's exactly what I need. Wow.
-T

Chris
Forgive my ignorance but I am not very good in scripting...
I have been trying to open a folder called "IMAGES" which is inside a folder called "PUBLICATION". The "PUBLICATION" folder is on "D" drive in my computer.
I have used your solution to my problem but I must be doing something wrong. Could you please tell me if this expression is ok:

LaunchFile("explorer.exe","/PUBLICATION/IMAGES")

I do not have any error messages when write in the script but when previous the message "cannot find the path" appears. Do I have to be especific on the Drive letter? Because the user may have it in another letter. I hope I am explaining properly.

Thank you for any help you can give me

Cybernaut

Hi Cybernaut

I assume from your post D:\PUBLICATION is the directory where publication "exe"
file resides.
If that is the case, you can use the script just as posted above

LaunchFile("explorer.exe",SYSTEM_PUBLICATION_DIR+"\\IMAGES")

Keep in mind, this won't work in preview because it reads SYSTEM_PUBLICATION_DIR
as the location where the project file ".imp" resides.

After you publish, Wherever the exe file is, if there is a subdirectory
called "IMAGES" this line of code will open it in explorer no matter
which drive letter the publication is on

HTH
Chris

Chris
No, want I am attempting to do is this:

1-I have created a pub.
2-This pub has one button with launch action
3-I want this button to open a subfolder called "IMAGES" located inside a folder called "PUBLICATIONS"
4-This folder contains a group of 25 images so when user click on button the folder containing the images open so the user can select the one he wants.

In my drive the folder PUBLICATION is on D drive but on the user it may have another letter.

The uses will install the pub in his computer but wants this button to launch this folder. In his computer, this folder and subfolder is on a drive called "SQS01"
I hope this is clear
Cybernaut

Still not understanding

Where will the publication .exe file be located ?

Chris
When user install the pub, this will be installed on his C drive But this particular folder he wants to access is on another drive but since the setting is on a network, for some people will be a different letter but the same folder path X:\Publication\Images\ where X is the letter assigned by the network.
I have been able to detect at least 3 different letters for people using the network but the name of the drive is SQS01.
The bottom line....This folder have to be accessed from different users on the network.
I hope this make sense because it is driving me crazy!
Cybernaut

I see now

Try this

LaunchFile("explorer.exe","\\\\SQS01\\PUBLICATION\\IMAGES")

HTH
Chris

Hi Cybernaut, try this:

LaunchFile("explorer.exe","\\PUBLICATION\\IMAGES")

I hope it works!

German

I'm using the following script as an action on a piece of hypertext to retrieve the file from disk:
LaunchFile("explorer.exe ","/select,"+SYSTEM_PUBLICATION_DIR+"\\Images"+"\\Image_01.jpg")

It works fine on the first try, but when I go to another page with another image and hypertext link, with the script changed to the new filename, the selection does not update in explorer. So, instead of seeing Image_02.jpg selected, you still see Image_01.jpg selected. Are there any errors in my syntax or something I can do to fix this? Thanks.
-Tom

Sorry guys, none of your solution works (Koala or Chris).It must be a way to path a drive and folder in a network!
Cybernaut

I'm no expert on networks, but this format works on my small network here

"\\\\Machine_Name\\Drive_Name\\Folder_Name\\Subfolder_Name"

Anybody out there with some network experience that could help
out with network path format ?

Chris

Forgot what I was going to say.
Paul

Chris
thank you very much. This post gave me an extra ensight so I contacted the network administrator requesting the details as per your path below. He gave me the details and it rocks!
Thank you
Cybernaut