在corona中停止监听'x‘秒

Question

我有一个基于摇动事件的淡入/淡出函数，但如果你在函数中再次摇动它，它会堆叠这些函数。我希望它等到函数结束(5秒)，然后再次侦听。它的代码是什么？

这是我目前所拥有的：

function fadeOut ( event )
    transition.to(yes1, {time=2000, alpha=0})
    transition.to(yes2, {time=2000, alpha=0})
    ...
    transition.to(funny3, {time=2000, alpha=0})
    transition.to(funny4, {time=2000, alpha=0})
    timer.performWithDelay(2000, onShakeComplete)
end

responses = {yes1, yes2, yes3, yes4, yes5, yes6, yes7, yes8, yes9, yes10,
no1, no2, no3, no4, no5, maybe1, maybe2, maybe3, maybe4, maybe5,
funny1, funny2, funny3, funny4}

local shaking = false

-- reset shaking flag after a shake is completed
local function onShakeComplete()
    shaking = false
end

local function onShake (event)
        if event.isShake and not shaking then
        shaking = true
        local object = responses[math.random(1,20)]
        transition.to(object, {time=2000, alpha=1})
        timer.performWithDelay(4000, fadeOut)
    end
end
Runtime:addEventListener("accelerometer", onShake)

Answer 1

-- true if we have a shake happening right now
local shaking = false

-- reset shaking flag after a shake is completed
local function onShakeComplete()
    shaking = false
end

local function onShake (event)
    if event.isShake and not shaking then
        local object = responses[math.random(1,20)]
        transition.to(object, {time=2000, alpha=1})
        timer.performWithDelay(4000, fadeOut)
        timer.performWithDelay(4000, onShakeComplete)
        shaking = true
    end
end