FileOpenDialog with variables

[darcaro]

I would like to use the FileOpenDialog function to pass a variable to and/or from an external program (or DOS batch file). Is there any way to assign a variable to the path, say $PATH is = "c:\windows" , then use that variable in the function and when I select the file, send back the filename in a variable? Any help would be greatly appreciated. thanks!!

[Zedna]

Example from HelpFile - see FileOpenDialog

$message = "Hold down Ctrl or Shift to choose multiple files."

$var = FileOpenDialog($message, @WindowsDir & "\", "Images (*.jpg;*.bmp)", 1 + 4 )

If @error Then
    MsgBox(4096,"","No File(s) chosen")
Else
    $var = StringReplace($var, "|", @CRLF)
    MsgBox(4096,"","You chose " & $var)
EndIf

[evilertoaster]

You do know that fileopendialog retrns the filename of the file you selected?

[darcaro]

thanks for your replies - I guess what I want to do is be able to assign the filename to a variable (which the function already does) , but use it in another program , like a DOS batch file or some other external program. Is that possible?

[evilertoaster]

There are several ways to make your data avialable to another program.

Bascialyl you have to write it to somewhere it can read, this could be -

1. A File

2. A registry Key

3. a stdin stream

4. A network socket

5. windows clipboard

6. some other location in memory (carefull...)

[darcaro]

Thank you for the great replys - I guess the best option would be to store the data in a temporary text file. Is there an easy way to just write the filename to a text file?? I could then use it from there for what I want to do.

I would also like to define the path name externally (ie, have the fileopendialog run, but use a variable for the path name, file type, etc ) Is there a way to specify a variable in the "command line", and have the function use it?