在我的程序中,我有一个如下所示的菜单:
MenuChoice = ''
while MenuChoice != 'x':
print("Type 1 to enter the first option")
print("Type 2 to enter the second option")
print("Type 3 to enter the third option")
print("Type 4 to enter the fourth option")
print("Press x to quit")
try:
MenuChoice = str(input("Please enter your choice here ----------------->>>>>"))
except ValueError:
print("Please enter one of the menu choices above, TRY AGAIN ")
我只是想知道一种方法,我可以确保只接受数字1到5,如果输入了其他内容,那么程序会再次询问问题。
请不要烤我。
由于
答案 0 :(得分:1)
您使用while
循环是正确的,但请考虑您想要的条件。你只想要数字1-5吗?所以这样做是有道理的:
MenuChoice = 0
print("Type 1 to enter the first option")
print("Type 2 to enter the second option")
print("Type 3 to enter the third option")
print("Type 4 to enter the fourth option")
print("Press x to quit")
while not (1 <= MenuChoice <= 4):
MenuChoice = input("Please enter your choice here ----------------->>>>>")
if MenuChoice == 'x' : break
try:
MenuChoice = int(MenuChoice)
except ValueError:
print("Please enter one of the menu choices above, TRY AGAIN ")
MenuChoice = 0 # We need this in case MenuChoice is a string, so we need to default it back to 0 for the conditional to work
我们输入一个整数,以便我们可以看到它是否在1-5之间。此外,您应该将您的开始打印语句放在循环之外,这样它就不会不断地向读者发送垃圾邮件(除非这是您想要的)。
答案 1 :(得分:0)
我认为他需要一个真正的循环。使用python 2.X
import time
print("Type 1 to enter the first option")
print("Type 2 to enter the second option")
print("Type 3 to enter the third option")
print("Type 4 to enter the fourth option")
print("Press x to quit")
while True:
try:
print ("Only Use number 1 to 4 or x to Quit... Thanks please try again")
MenuChoice = raw_input("Please enter your choice here ----------------->>>>> ")
try:
MenuChoice = int(MenuChoice)
except:
MenuChoice1 = str(MenuChoice)
if MenuChoice1 == 'x' or 1 <= MenuChoice <= 4:
print "You selected %r Option.. Thank you & good bye"%(MenuChoice)
time.sleep(2)
break
except:
pass