Question

在Python 3.2中，我正在编写一个基本菜单程序，当输入退出选项时，该功能没有结束。当选择quit时，它会结束脚本其余部分所在的循环，并且应该终止脚本，但不管是出于何种原因？我错过了一个杀死脚本的'end'函数，还是新的Python Shell只是错误？很确定这在Python 2.7中不是必需的。

import random
choice = str(input("\nMenu:\n(I)nstructions\n(C)alculate blah\n(Q)uit\n>>>"))
while choice != "q" or choice != "Q":
    while choice != "i" and choice != "I" and choice != "c" and choice != "C" and choice != "q" and choice != "Q":
        print("Invalid menu choice.")
        choice = str(input("\nMenu:\n(I)nstructions\n(C)alculate blah\n(Q)uit\n>>>"))
    if choice == "i" or choice == "I":
        print("blahblah.")
        choice = str(input("\nMenu:\n(I)nstructions\n(C)alculate blah\n(Q)uit\n>>>"))   
    if choice == "c" or choice == "C":
        x = int(input("Please enter the number of x: "))
        while x < 0:
            x = int(input("Please enter the number of x: "))
        y = int(input("Please enter the number of y: "))
        while y < 0:
            y = int(input("Please enter the number of y: "))
        z = str(input("blah (B) or (P) z?: "))
        while z != "b" and z != "p" and z != "B" and z != "P":
            z = str(input("blah (B) or (P) z?: "))
        if z == "b" or z == "B":
            total = x*10 + y*6 + 0
            print("blah $", total, " blah ", x, " x and ", y, " y. blah!")
        #function that outputs the cost of premium z
        if z == "p" or z == "P":
            luck = random.randrange(1, 11, 1)
            if luck == 10:
                total = x*10 + y*6
                print("\nblah$", total, " blah z for ", x, " x and ", y, " y. blah!")
            #below is the normal function, for when the customer is not a lucky winner
            if luck != 10:
                total = x*12.50 + y*7.50
                print("blah $", total, " blah ", x, " x and ", y, " y. blah!")
        choice = str(input("\nMenu:\n(I)nstructions\n(C)alculate\n(Q)uit\n>>>"))

Answer 1

你的情况错了：

while choice != "q" or choice != "Q":    # this should be "and"!

始终返回True，创建无限循环。

此外，你在这里有一个令人费解的逻辑。这可以简化很多：

import random
while True:
    choice = str(input("\nMenu:\n(I)nstructions\n(C)alculate blah\n(Q)uit\n>>>")).lower()
    if choice == "i":
        print("blahblah.")
        continue
    elif choice == "q":
        break
    elif choice == "c":
        while True:
            x = int(input("Please enter the number of x: "))
            if x >= 0: break
        while True:
            y = int(input("Please enter the number of y: "))
            if y >= 0: break
        while True:
            z = str(input("blah (B) or (P) z?: ")).lower()
            if z in "bp": break
        if z == "b":
            total = x*10 + y*6 + 0
            print("blah $", total, " blah ", x, " x and ", y, " y. blah!")
        #function that outputs the cost of premium z
        else:  # z must be "p"
            luck = random.randrange(1, 11, 1)
            if luck == 10:
                total = x*10 + y*6
                print("\nblah$", total, " blah z for ", x, " x and ", y, " y. blah!")
            #below is the normal function, for when the customer is not a lucky winner
            if luck != 10:
                total = x*12.50 + y*7.50
                print("blah $", total, " blah ", x, " x and ", y, " y. blah!")
    else:
        print("Invalid menu choice.")
        continue

Answer 2

在Python中执行退出操作的一种方法是抛出自定义异常并显式捕获该异常。如果我错了，请纠正我，AFAIK这并不会对你的python程序造成太大的影响。它可以像我在下面显示的那样简单：

...
class QuitException(Exception);
...
def MyMenuProgram():
    ...
    ...
    ...

if __name__ == '__main__':
    try:
        MyMenuProgram()
    catch QuitException:
        pass
    catch Exception, e:
        raise

Answer 3

编辑：已编辑 - input()确实是在Py3k中获取用户输入的方式。 Python 2.7和Python 3.2之间的另一个光荣差异。

Answer 4

当@Tim回答时，你的循环条件是错误的。

while choice != "q" or choice != "Q":

让我们从逻辑上看一下这个：

如果选择==“q”，则选择！=“Q”，循环条件评估为false or true，即true
如果选择==“Q”，则选择！=“q”，循环条件评估为true or false，即true
如果选择！=“q”或“Q”，循环条件评估为true or true，即true

您需要将此循环条件更改为：

while choice != "q" and choice != "Q":

或

while choice.lower() != "q":

结束Python脚本

4 个答案: