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How to display TrayMenu programmingly?


wccppp
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Hello group,

A TrayMenu can be displayed via a user click. Is it possible to display it programmingly? I'm thinking

1. A function like TrayMenuDisplay (if there is such a function) will do;

2. If somehow the screen coordinate of the tray icon can be determined, then i can issue a mouse click on that icon to display the TrayMenu.

Atm, I don't know how to accomplish either one. Any help/hint?

Thank you!

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Hi,

Search here on the forum for SysTray_UDF.

You can click then like this:

#NoTrayIcon
#include <SysTray_UDF.au3>

$Pid = ProcessExists("App.exe")
If Not $Pid Then
    MsgBox(16, "Error", "This process not exists")
    Exit
EndIf

$Index = _SysTrayIconIndex($Pid)
If @error Then
    MsgBox(16, "Error", "This process not have tray icon")
    Exit
EndIf

_SysTrayIconclick($Index)
If @error Then
    MsgBox(16, "Error", "There was an error to click on the icon")
    Exit
EndIf

Func _SysTrayIconclick($iIndex)
    Local $TrayIcon_Pos = _SysTrayIconPos($iIndex)
    If Not IsArray($TrayIcon_Pos) Then Return SetError(1, 0, 0)
    
    Local $Old_Mouse_Pos = MouseGetPos()
    MouseClick("Right", $TrayIcon_Pos[0], $TrayIcon_Pos[1], 1, 0)
    MouseMove($Old_Mouse_Pos[0], $Old_Mouse_Pos[1], 0)
EndFunc

 

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Try GuiCtrlCreateContextMenu... you can find it in the helpfile... That requires a GUI though, I'm not sure how you could do it without a GUI.

Thanks for your suggestion. For what i'm doing now, SysTray_UDF seems to be a better choice.

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