Question

def main():
    choice = pickone() #picking the shape or to quit
    if choice not in quitlist:

        low, high = getLoHiInt() #picking the range of points

        shapes = [ball,bowlingPin,ellipse,tableLeg]

        combolist = zip(picklist,shapes) #zipped list of the shapes with the corresponding choice

        analyzeSolid(combolist[int(choice)-1][1], low, high)

    return showTermination()

pickone（）函数工作得很好，问题是当我输入终止数时，我的函数显示终止但是继续通过if循环，即使选择在quitlist中。

quitlist = ['5']

不幸的是我需要这种方式，因为我的代码的其他部分依赖于此。我还需要我的if语句在它通过if语句后在pickone（）函数重新启动，但它只是显示我的终止而是结束程序。

因为有人说我的pickone功能不能正常工作

picklist = ["1","2","3","4"]
quitlist = ["5"] #couldn't get it to work with just one list, but this works fine
def pickone():
    while True:
        print "\nPick a solid to analyze: \n1: ball\n2: bowlingPin\n3: ellipse\n4: tableleg\n5: quit"
        theinput = raw_input("What is the number of your choice?: ")
        #if theinput not in zip(picklist, quitlist):
        #    print"\nChoice %s is not a valid choice.\n" %theinput
        try:
            theinput
        except ValueError:
            # So the program will continue if the input is wrong
            print "choice must be from 1 to 5" #message doesn't show up but the program still works properly
            continue
        if theinput in picklist:
            return theinput
        if theinput in quitlist:
            return theinput

编辑，我的pickone函数出现问题，它应该返回输入而不是showTermination（）

Answer 1

为了获得一个可以持续到终止的循环，你可以尝试类似的东西：

def main():
    choice = pickone() #picking the shape or to quit
    while choice not in quitlist:

        low, high = getLoHiInt() #picking the range of points

        shapes = [ball,bowlingPin,ellipse,tableLeg]

        combolist = zip(picklist,shapes) #zipped list of the shapes with the corresponding choice

        analyzeSolid(combolist[int(choice)-1][1], low, high)

        choice = pickone()
    return showTermination()

为了弄清楚为什么你的循环在继续选择退出值时，请尝试打印出choice中的内容，也许它会获得integer值，而不是string }？

如果是这种情况，可以尝试：

choice = str(pickone())

或

while str(choice) not in quitlist:
    ...

Answer 2

我喜欢制作这样的菜单

def do_menu(menu,error="Invalid Choice Try Again!"):
    while True:
        for k,(msg,action) in menu.items():
            print msg
        resp = raw_input("Make a Choice:")
        if resp in menu:
            return menu[resp][1]()
        print error

import random,sys

#####JUST SOME GENERIC MENU ACTIONS
something = []

def add_something():
    something.append(random.randint(1,10))
    print "ADDED %d"%something[-1]

def print_something():
    print something

#DEFINE THE MENU
menu = {
'A':("[A]dd Something",add_something),
'P':("[P]rint Something",print_something),
'Q':("[Q]uit",sys.exit)
}
while True:
    #print menu and get user response and act upon it
    print "\n#####[ MENU ]####"
    result = do_menu(menu)

定义将执行各种任务的函数，但以某个输入终止

2 个答案: