내 코드가 작동하지 않는 것 같습니다. 최대 1 개의 사전을 생성합니다. 다른 이름과 번호를 만들면 사전을 대체하여 현재 이름을 대체합니다. 이 바보 같은 질문 인 경우 간단한 파이썬 연락처 목록
list = {}
def start():
print "Welcome to Contact+ \n \nPlease enter your name: ",
name = raw_input()
print "Hi " + name + " would you like to check your current contacts or make new ones? \nTo make new contacts type in 'New' \nTo check current contacts type in 'Contacts'"
print "Go to: ",
choose = ""
choose = raw_input()
valid = False
while(not valid):
if choose == "'New'" or choose == "'new'" or choose == "New" or choose == "new":
new_function()
elif choose == "'Contacts'" or choose == "'contacts'" or choose == "Contacts" or choose == "contacts":
contacts_function()
def new_function():
global list
list = {}
print "\nPlease input the name: ",
contact_name = raw_input()
print "Please input the number: ",
contact_number = raw_input()
list.update({contact_name:contact_number})
print "Contact created \n\nWould you like to make more contacts or check current contacts? \nTo make new contacts type in 'New' \nTo check current contacts type in 'Contacts'"
print "Go to: ",
choose = ""
choose = raw_input()
valid = False
while(not valid):
if choose == "'New'" or choose == "'new'" or choose == "New" or choose == "new":
new_function()
elif choose == "'Contacts'" or choose == "'contacts'" or choose == "Contacts" or choose == "contacts":
contacts_function()
def contacts_function():
global list
for keys,values in list.items():
print "\n---------------------------------------------------------"
print str("Name: ") + str(keys)
print str("Number: ") + str(values)
print "---------------------------------------------------------\n"
print "Would you like to make more contacts or check current contacts? \nTo make new contacts type in 'New' \nTo check current contacts type in 'Contacts'"
print "Go to: ",
choose = ""
choose = raw_input()
valid = False
while(not valid):
if choose == "'New'" or choose == "'new'" or choose == "New" or choose == "new":
new_function()
elif choose == "'Contacts'" or choose == "'contacts'" or choose == "Contacts" or choose == "contacts":
contacts_function()
start()
미안 해요, 난 프로그래밍에 새로운 여전히 이니
왜 list라는 사전이 있습니까? – jonrsharpe
'raw_input' 문은 while 루프 안에 있어야합니다. 명령의 철자를 잘못 입력하면 프로그램은 'while'상태로 무한 루프됩니다. 또한,'raw_input(). lower()'는'if '비교를 제한하는데 도움을 줄 것입니다. –