如何让这个程序在启动前等待屏幕点击？

Question

我正在努力完成一个课程，我做得相当好。这是一个简单的python海龟图形游戏，在这个游戏中，您可以尝试避开毒点并导航到一个正方形。但是，我的程序会在用户点击屏幕之前立即启动。我该怎么解决这个问题？谢谢！在

我的代码：

# This game involves avoiding red poison blobs while attempting to navigate to
# a square. If you hit the blob, you begin to speed up, making it more difficult
# not to hit more. Additionally, you lose a point. If you reach the square you
# get a point.

import turtle
import math
import random

# screen
wn = turtle.Screen()
wn.bgcolor("black")
wn.tracer(3)

# Draw border
pen1 = turtle.Turtle()
pen1.color("white")
pen1.penup()
pen1.setposition(-275,-275)
pen1.pendown()
pen1.pensize(5)
for side in range(4):
    pen1.forward(550)
    pen1.left(90)
pen1.hideturtle()

# player
player = turtle.Turtle()
player.color("dark green")
player.shape("turtle")
player.penup()

# poisonBlob
maxpoisonBlob = 15
poisonBlob = []

for a in range(maxpoisonBlob):
    poisonBlob.append(turtle.Turtle())
    poisonBlob[a].color("dark red")
    poisonBlob[a].shape("circle")
    poisonBlob[a].shapesize(4, 4, 4)
    poisonBlob[a].penup()
    poisonBlob[a].speed(0)
    poisonBlob[a].setposition(random.randint(-255, 255), random.randint(-255, 255))

maxfood = 1
food = []

for a in range(maxfood):
        food.append(turtle.Turtle())
        food[a].color("light blue")
        food[a].shape("square")
        food[a].penup()
        food[a].speed(0)
        food[a].setposition(random.randint(-240, 240), random.randint(-240, 240))

# speed variable
speed = 6.5

def turnleft():
    player.left(30)

def turnright():
    player.right(30)

def increasespeed():
    global speed
    speed += 1

def touchPoison(t1, t2):
    d = math.sqrt(math.pow(t1.xcor()-t2.xcor(),2) + math.pow(t1.ycor()-t2.ycor(),2))
    if d < 50:
        return True
    else:
        return False

def touchfood(z1, z2):
        d = math.sqrt(math.pow(z1.xcor()-z2.xcor(),2) + math.pow(z1.ycor()-z2.ycor(),2))
        if d < 20:
                return True
        else:
                return False


turtle.listen()
turtle.onkey(turnleft, "Left")
turtle.onkey(turnright, "Right")


def main():
    print("Help your turtle navigate red poison blobs while attempting to navigate to the\n"
          "food! If you hit the poison, you begin to speed up, making it more difficult\n"
          "not to hit more. Additionally, you lose a point. If you reach the square you\n"
          "get a point. To navigate, click the screen, and then use the right and left\n"
          "arrow keys. Quickly, your turtle is running away!")

    score = 0

    while True:
        player.forward(speed)

        # turtle boundary
        if player.xcor() > 260 or player.xcor() < -260:
            player.right(180)

        # turtle boundary
        if player.ycor() > 260 or player.ycor() < -260:
            player.right(180)

        # move poison
        for a in range(maxpoisonBlob):
            poisonBlob[a].forward(3)

           # Poison boundaries
            if poisonBlob[a].xcor() > 220 or poisonBlob[a].xcor() < -220:
                poisonBlob[a].right(180)

            # Poision boundaries
            if poisonBlob[a].ycor() > 220 or poisonBlob[a].ycor() < -220:
                poisonBlob[a].right(180)     

            # Poison touching
            if touchPoison(player, poisonBlob[a]):
                increasespeed()
                poisonBlob[a].setposition(random.randint(-230, 230), random.randint(-230, 230))
                poisonBlob[a].right(random.randint(0,360))
                score -= 1
                #Draw score
                pen1.undo()
                pen1.penup()
                pen1.hideturtle()
                pen1.setposition(-260, 280)
                scorestring = "Score: %s" %score
                pen1.write(scorestring, font=("Calibri", 12))

        for a in range(maxfood):

            #Positive Point Checking
            if touchfood(player, food[a]):
                food[a].setposition(random.randint(-230, 230), random.randint(-230, 230))
                score += 1
                #Draw Score
                pen1.undo()
                pen1.penup()
                pen1.hideturtle()
                pen1.setposition(-260, 280)
                scorestring = "Score: %s" %score
                pen1.write(scorestring, font=("Calibri", 12))

if __name__ == "__main__":
    main()

Answer 1

一般来说，我们尽量不直接回答作业问题，但我会给你一些指导。在

我不熟悉您正在使用的库，但基本思想是您需要等到用户单击屏幕后再继续。我的理解是，您希望在main中的print（）函数之后执行此操作，以便用户阅读消息，然后单击继续。在

快速搜索找到了这个链接：Turtle in python- Trying to get the turtle to move to the mouse click position and print its coordinates

这将详细介绍onscreenclick（）事件。从onkey（）方法来看，您已经熟悉绑定到事件，因此您应该能够使用这些知识来绑定onscreenclick（）事件，然后您只需要将代码分解一点，这样在onscreenclick（）事件发生之前它不会执行您的游戏。在

如果这还不够清楚，请帮助我了解你的困惑在哪里，我将进一步协助。谢谢！在

Answer 2

您可以先定义一个函数，在该函数中调用“if”语句，然后通过在其中插入新定义的函数来使用turtle.onscreenclick()，如下所示：

def Game(x, y):
    if __name__ == "__main__":
        main()

turtle.onscreenclick(Game)

确保在末尾插入所有这些内容！

现在，在函数定义中的函数名后面加上括号中的（x，y）将为用户在图形窗口中单击的点指定相应的（x，y）坐标，而该坐标又会由于单击屏幕而激活该函数。我成功地做到了，所以我可以向你保证它是有效的！我希望这有帮助！：）