[英]python zelle graphics checkMouse()?
yes_box = Rectangle(Point(200, 150),Point(350,50))
yes_box.setOutline('blue')
yes_box.setWidth(1)
yes_box.draw(graphics_win)
def mouse_check(arg1):
??????
嘿,有一个简短的问题可能很明显,但是我真的很困惑。 因此,我正在编写一个程序(游戏),要求您在yes_box的框内单击(如上所示)。 我需要编写一个函数来检查鼠标单击是否在矩形的边界内,如果是,则返回“ y”,如果不是,则返回“ n”。
我知道您需要使用win.getMouse()和win.checkMouse()函数,但是我不确定如何获取python以确定该单击是否在矩形对象的边界内? 任何帮助将不胜感激!
您可以使用以下功能:
p1 = rectangle.getP1()
rx1 = p1.getX()
ry1 = p1.getY()
p2 = rectangle.getP2()
rx2 = p2.getX()
ry2 = p2.getY()
x1 = point.getX()
y1 = point.getY()
if x1>=rx1 and y1<=ry1 and x1<=rx2 and y1>= ry2:
return y
else:
return n
点应该是用户单击的点。 矩形是您要知道用户是否单击的矩形。
您只需要获取win.getMouse()
返回的点,并确保x和y值在范围内即可。 我在inside
函数中执行以下操作,然后使用此布尔值在窗口上显示“ y”或“ n”
from graphics import *
def inside(test_Point, P1, P2):
'''
determines if the test_Point is inside
the rectangle with P1 and P2 at opposite corners
assumes P1 is upper left and P2 is lower right
'''
tX = test_Point.getX()
tY = test_Point.getY()
# first the x value must be in bounds
t1 = (P1.getX() <= tX) and (tX <= P2.getX())
if not t1:
return False
else:
return (P2.getY() <= tY) and (tY <= P1.getY())
win = GraphWin("Box", 600, 600)
yes_box = Rectangle(Point(200, 150), Point(350, 50))
yes_box.setOutline('blue')
yes_box.setWidth(1)
yes_box.draw(win)
# where was the mouse clicked?
t = win.getMouse()
# is that inside the box?
if inside(t, yes_box.getP1(), yes_box.getP2()):
text = 'y'
else:
text = 'n'
# draw the 'y' or 'n' on the screen
testText = Text(Point(200,300), text)
testText.draw(win)
exitText = Text(Point(200,350), 'Click anywhere to quit')
exitText.draw(win)
win.getMouse()
win.close()
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