Detect a double key press in AutoHotkey

2019-01-23 12:40发布

I'd like to trigger an event in AutoHotkey when the user double "presses" the esc key. But let the escape keystroke go through to the app in focus if it's not a double press (say within the space of a second).

How would I go about doing this?

I've come up with this so far, but I can't work out how to check for the second escape key press:

~Esc::

    Input, TextEntry1, L1 T1
    endKey=%ErrorLevel%

    if( endKey != "Timeout" )
    {
        ; perform my double press operation
        WinMinimize, A
    }
return

2条回答
2楼-- · 2019-01-23 12:45

Found the answer in the AutoHotkey documentation!

; Example #4: Detects when a key has been double-pressed (similar to double-click).
; KeyWait is used to stop the keyboard's auto-repeat feature from creating an unwanted
; double-press when you hold down the RControl key to modify another key.  It does this by
; keeping the hotkey's thread running, which blocks the auto-repeats by relying upon
; #MaxThreadsPerHotkey being at its default setting of 1.
; Note: There is a more elaborate script to distinguish between single, double, and
; triple-presses at the bottom of the SetTimer page.

~RControl::
if (A_PriorHotkey <> "~RControl" or A_TimeSincePriorHotkey > 400)
{
    ; Too much time between presses, so this isn't a double-press.
    KeyWait, RControl
    return
}
MsgBox You double-pressed the right control key.
return

So for my case:

~Esc::
if (A_PriorHotkey <> "~Esc" or A_TimeSincePriorHotkey > 400)
{
    ; Too much time between presses, so this isn't a double-press.
    KeyWait, Esc
    return
}
WinMinimize, A
return
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啃猪蹄的小仙女
3楼-- · 2019-01-23 12:58

With the script above, i found out that the button i wanted to detect was being forwared to the program (i.e. the "~" prefix).

This seems to do the trick for me (i wanted to detect a double "d" press)

d::
keywait,d
keywait,d,d t0.5 ; Increase the "t" value for a longer timeout.
if errorlevel
{
    ; pretend that nothing happened and forward the single "d"
    Send d
    return
}
; A double "d" has been detected, act accordingly.
Send {Del}
return

Source

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