#include <$var>

[Naufuge 0]

Its it possible to include a file thats location is stored in a var? Like

$File_OpenDialog = FileOpenDialog("Open File...",@ScriptDir,"Application (*.exe)",1+2+8+16)
#include <$File_OpenDialog>

Thanks

Brian

[erebus 0]

Quoting from the helpfile:

The filename of the current script to include. Path is optional. This must be a string--it cannot be a variable.

So the answer is no, the same limitation exist with FileInstall() too.

[Naufuge 0]

Ok Thanks

(Darn...)

[SmOke_N 192]

Are you trying to include an executable? If so, use FileIntall()

Edit: Guess not on the variable... good catch erebus

Edit2: Note to self! Only answer questions when not multi-tasking!

Edited December 16, 2005 by SmOke_N

[erebus 0]

(Darn...)

Use your head; there is _always_ a different approach to do what you want to do.

[MSLx Fanboy 0]

Because #Include is a preprocessor directive (notice the hash #), it is executed before any actual script code is processed.

Also, if the program is compiled and distributed, the #Include is replaced with the text of the file that is included (File.au3, Misc.au3, etc) and nothing else can be included 'on-the-fly'.

If you are looking for the functionality I suspect you want, maybe FileMove or FileCopy would do the trick.

[LxP 2]

Use your head; there is _always_ a different approach to do what you want to do.

Like initially #Include-ing all of the possible script pieces and then using Call() or Execute() to invoke the appropriate functionality.

Edit: Added context.

Edited December 16, 2005 by LxP

[erebus 0]

I think that he wants to use #include's filename dynamically.

A good idea is to #include everything of course as LxP said.