언제쯤 검색없이 코딩을 짤까요.

Q. 이름, 전화번호, 주소를 입력받아 주소록을 만들자.

members = []
titles = ['name:', 'tel:', 'address:']

#등록함수
def addMember():
    m = [0,0,0]
    print('새 멤버 등록')
    for i in range(0, len(m)):
        if i == 0:
            while True:
                m[i] = input(titles[i])
                res = getByName(m[i])
                if res==None:
                    break
        else:
            m[i] = input(titles[i])

    members.append(m)

def getByName(name):
    for idx, i in enumerate(members):
        if name == i[0]:
            return idx, i  #튜플로 반환(idx, i) => res[0]:idx, res[1]:i

def printMember():
    print('멤버 검색')
    name = input('검색할 이름:')
    res = getByName(name)
    if res == None:
        print('not found name')
    else:
        i = res[1]
        cnt = 0
        for j in i:
            print(titles[cnt],j)
            cnt += 1
        print('----------')

def printAll():
    print('모든 멤버')
    for i in members:
        cnt = 0
        for j in i:
            print(titles[cnt], j)
            cnt += 1
        print('----------')

def editMember():
    print('멤버 정보 수정')
    name = input('수정할 이름:')
    res = getByName(name)
    if res == None:
        print('not found name')
    else:
        i = res[1]
        i[1] = input('new tel:')
        i[2] = input('new address:')

def delMember():
    print('멤버 정보 삭제')
    name = input('삭제할 이름:')
    res = getByName(name)
    if res == None:
        print('not found name')
    else:
        del members[res[0]]


def main():
    while True:
        menu = input('1.등록 2.검색 3.수정 4.삭제 5.전체출력 6.종료')
        if menu == '1':
            addMember()
        elif menu =='2':
            printMember()
        elif menu == '3':
            editMember()
        elif menu == '4':
            delMember()
        elif menu == '5':
            printAll()
        elif menu == '6':
            break

main()

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