提示用户在 Python 中输入的问题

Question

import sys
import turtle
t=turtle.Pen
def what_to_draw():
    print ("What to do you want to see a sketch of that may or may not be colored?")
what_to_draw=sys.stdin.readline()
if what_to_draw=="flower/n":
    t.forward(90)
elif():
    print ("What you typed isn't in the code. Try putting a letter or letters to lowercase or uppercase. If that doesn't work, what you typed has not been set to make something happen")

我在上面输入了这段代码。它在python shell中说“花”没有定义。谁能帮我解决这个问题？

Answer 1

你的代码有几行有某种错误：

t=turtle.Pen 应该是：t = turtle.Pen()

你应该避免使用同名的函数和变量

def what_to_draw():
    ...
    what_to_draw = sys.stdin.readline()

使用rstrip()处理"\n"，使用.lower()处理大小写：

if what_to_draw == "flower/n":

elif(): 需要某种条件，否则使用else:

让我们尝试不同的方法。让我们尝试使用 Python 3 turtle 新增的 textinput() 函数在海龟内部完成所有操作，而不是将控制台窗口输入与海龟图形混合：

from turtle import Turtle, Screen

def what_to_draw():
    title = "Make a sketch."

    while True:
        to_draw = screen.textinput(title, "What do you want to see?")

        if to_draw is None:  # user hit Cancel so quit
            break

        to_draw = to_draw.strip().lower()  # clean up input

        if to_draw == "flower":
            tortoise.forward(90)  # draw a flower here
            break
        elif to_draw == "frog":
            tortoise.backward(90)  # draw a frog here
            break
        else:
            title = to_draw.capitalize() + " isn't in the code."

tortoise = Turtle()

screen = Screen()

what_to_draw()

screen.mainloop()

Answer 2

你的缩进是错误的，所以大部分语句都在what_to_draw()的函数体之外。
您实际上并没有调用该函数，因此它不会做任何事情。
另外，不要将what_to_draw 用作函数名和变量名。
elif:之后不应该有()
使用 input() 代替 print() 和 stdin。你也不会这样得到\n。

试试这个并告诉我：

import sys
import turtle
t=turtle.Pen()
def what_to_draw():
    draw_this = input("What to do you want to see a sketch of that may or may not be colored?")
    if draw_this == "flower":
        t.forward(90)
    elif:
        print ("What you typed isn't in the code. Try putting a letter or letters to lowercase or uppercase. If that doesn't work, what you typed has not been set to make something happen")

what_to_draw()