how to put #notrayicon inside of an IF?

[supadodger 0]

if $cmdline[1] <> "icon" then

#notrayicon

EndIf

[supadodger 0]

If $cmdline[1] <> "icon" then
Opt ("TrayIconHide",1)
EndIf

[ZacUSNYR 4]

You could always hide the icon with #NoTrayIcon then in your code where you'd want it to show use

AutoItSetOption("TrayIconHide", 0)

You could also change the 0 to 1 and put that in your if statement, but the icon would flash and then go away there.

[Steveiwonder 0]

Hello,

Look up TraySetState()

if $cmdline[1] == "icon" Then
    TraySetState(1) ; Show
Else
    TraySetState(2) ; Hide
EndIf


sleep(10000)

Stevo

edit: typo

Edited December 20, 2010 by Steveiwonder

[ZacUSNYR 4]

#NoTrayIcon

if $cmdline[1] <> "icon" then
;code
Else
AutoItSetOption("TrayIconHide", 0)
;else code and show the tray icon.
EndIf

Edited December 20, 2010 by ZacUSNYR

[kaotkbliss 138]

No, I don't think that will work because

Windows needs to know what it is doing with a program (and it's icon) the instant the program is started. (does it show an icon, which icon? does it not show an icon?)

trying to put #NoTrayIcon later in the script means that windows would have to run your program partially before finding out what it needs to do with the icon.

(my guess as to how it would work)

You may have to use

Opt("TrayIconHide", 1) to hide

Opt("TrayIconHide", 0) to unhide

But the icon will show briefly until this line is executed.

*edit*

Wow... Way too slow.

Edited December 20, 2010 by kaotkbliss