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我在屏幕上有两个图像重叠，称为 A 和 B。我有第三个图像称为 C，用户可以在屏幕上移动。两者中较大的一个 (A) 将 C 重置为碰撞时的生成点。我需要知道如何制作它，以便 C 可以走过图像 B 而不会重置到生成点。如图像 B 重叠图像 A 并否定碰撞功能。

这是代码：

--Hides status bar from top of page
display.setStatusBar( display.HiddenStatusBar )

--Set a test background image 
local backgroundimg = display.newImageRect("testbackground.png", 320,576)
backgroundimg.x = display.contentWidth*0.5
backgroundimg.y = display.contentHeight*0.5

--Set the position and amount of score and lives

local score = 0
local lives = 5

local showscore = display.newText("Score: "..score,0,-36,native.systemFont,25)
showscore:setTextColor(0, 0, 0)

local showlives = display.newText("Lives: "..lives,230,-36,native.systemFont,25)
showlives:setTextColor(0, 0, 0)

--Set physics for collisions, etc.
physics = require("physics")
physics.start()
physics.setGravity(0,0)

--set water
local water = display.newImageRect("water.png",320,192)
water.x = display.contentWidth*0.5
water.y = 144
physics.addBody(water,"static")
water:addEventListener("collision", function()timer.performWithDelay(50,waterCollide)end)

function waterCollide(event)
    lives = lives - 1
    display.remove(frog)
    frog = display.newImageRect("FrogTest.png",32,48)
    frog.x = display.contentWidth*0.5
    frog.y = 504
    physics.addBody(frog, "dynamic")
    frog.isFixedRotation = true
end

--Sets buttons images and positions
local forward = display.newImageRect("Forward Button.png",106,100)
forward.x = 160
forward.y = 478

local left = display.newImageRect("Left Button.png",106,100)
left.x = 53
left.y = 478

local right = display.newImageRect("Right Button.png",106,100)
right.x = 267
right.y = 478

--Set log position and movement
local log1 = display.newImageRect("log1.png", 96, 48)
log1.x = 32
log1.y = 226
physics.addBody(log1,"kinematic")
transition.to(log1, {time = 3500, x = 288})


--Set a frog sprite on the screen
frog = display.newImageRect("FrogTest.png",32,48)
frog.x = display.contentWidth*0.5
frog.y = 504
physics.addBody(frog, "dynamic",{density = 1.0, friction = 1, bounce = -1})
frog.isFixedRotation = true

--Sets motion variables
local motionX = 0
local motionY = 0
local speed = 4

--Moving forward
function forward:touch()
    motionX = 0
    motionY = -speed
end
forward:addEventListener("touch",forward)

--Moving Right
function right:touch()
    motionX = speed
    motionY = 0
end
right:addEventListener("touch",right)

--Moving left
function left:touch()
    motionX = -speed
    motionY = 0
end
left:addEventListener("touch",left)

--Moves Frog each time frame is called
function movefrog (event)
    frog.x = frog.x + motionX
    frog.y = frog.y + motionY
end
Runtime:addEventListener("enterFrame", movefrog)

--Stops frog from moving continuously 
local function stop (event)
    if event.phase == "ended" then
        motionX = 0
        motionY = 0
    end
end
Runtime:addEventListener("touch", stop)

--Making sure the frog does not go off the screen
local function stopfrog (event)
    if frog.x <= 16 then
        frog.x = 16
    end
    if frog.x >= 304 then
        frog.x = 304
    end
end
Runtime:addEventListener("enterFrame", stopfrog)
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1 回答 1

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您需要在函数下设置一个条件waterCollide以否定重生。

根据您需要的复杂程度，您可以简单地检查frog' 的位置是否在log1' 的边界内，或者您可能拥有log1并且任何未来的日志都有自己的碰撞事件，该事件在碰撞时设置一个标志以不触发respawn 然后在与任何日志的冲突结束时清除标志。

这是一个例子：

local onLog = 0

function frogDie()
    lives = lives - 1
    display.remove(frog)
    frog = display.newImageRect("FrogTest.png",32,48)
    frog.x = display.contentWidth*0.5
    frog.y = 504
    physics.addBody(frog, "dynamic")
    frog.isFixedRotation = true
end

function waterCollide(event)
    if onLog < 1 then frogDie() end
end

function logCollide(event)
    if event.phase == 'began' then
        onLog = onLog + 1
    else
        onLog = onLog - 1
    end
end

log1:addEventListener("collision", logCollide)
--log2:addEventListener("collision", logCollide)

使用数字来跟踪青蛙是否在日志上应该比布尔值更安全，因为日志可能最终会重叠并清除标志，然后才能在多次碰撞通过时正确重置。

于 2013-11-04T16:06:59.070 回答