if else声明 - 如何跳过其他？

Question

下面我的代码很有用,但是,如果我选择'1',代码将运行if语句并打印"success",但它也将运行else语句并退出.如果用户选择1,2或3,如何停止运行else语句？

谢谢!

print "\nWelcome to the tool suite.\nWhich tool would you like to use?\n"
print "SPIES - press 1"
print "SPADE - press 2"
print "SPKF - press 3"
print "For information on tools - press i\n"

choice = raw_input("")
if choice == "1":
    execfile("lot.py")
    print ("success")

if choice == "2":
    #execfile("spade.py")
    print ("success")

if choice == "3":
    #execfile("spkf.py")
    print ("success")

if choice == "i":
    print "Google Maps is awesome.\n"
    print "SPADE\n"
    print "SPKF\n\n"

else:
    print "Error, exiting to terminal"
    exit(1)

Answer 1

你想要这个elif结构.

if choice == "1":
    ...
elif choice == "2":
    ...
else: # choice != "1" and choice != "2"
    ...

否则,不同的if语句彼此断开.我添加了一个空白行强调:

if choice == "1":
    ...

if choice == "2":
    ...
else: # choice != 2
    ...