为什么我在python中循环索引无法给我正确的输出

Question

I am new learner for python.I am trying to make a basic login accout check, and somehow, my code is not showing what I want. I have defined three user name, and when I run my code, if the first time I put an incorrect user name, the code show account not exit,but it is not shoing. I do not know why

I believe it is my for loop problem, because when I input a wrong account, my i index start at 0 and keep loop until the end index and compare the input username and the exist username in the list. Then after compare all the index if not found user name, then print account not exist, I try to fix this issue, but not find a correct way.

user1=[  
   {'id':'0001','name':'123','password':'a123', 'balance':0.00},
   {'id':'0002','name':'456','password':'a456', 'balance':0.00},  
   {'id':'0003','name':'789','password':'a789', 'balance':0.00}
]

for x in range(0,4):

    name = input('User Name：')
    for i in range(len(user1)):
        if name == user1[i]['name']:  
            password = input('Password：')
            if password == user1[i]['password']:  
                print("Success login")
        continue
        if name != user1[i]['name']:
            print("Account not exist, input new one")

If I input wrong user name; it should show account not exist input new one, then I put user name 456 then will ask the correct password.

Answer 1

Look at the logic of your loop body:

for i in range(len(user1)):
    if name == user1[i]['name']:  
        password = input('Password：')
        if password == user1[i]['password']:  
            print("Success login")
    continue
    if name != user1[i]['name']:
        print("Account not exist, input new one")

Regardless of the input, you will never get to the second if statement: any time your program gets to that point, you tell it to continue with the next loop iteration. Try this instead:

for i in range(len(user1)):
    if name == user1[i]['name']:  
        password = input('Password：')
        if password == user1[i]['password']:  
            print("Success login")
        else:
            print("Account not exist, input new one")

Note that this will work better if you put the all of the accounts into a single dict, so you can access them directly:

user1 = {
   '123': {'id':'0001', 'password':'a123', 'balance':0.00},
   '456': {'id':'0002', 'password':'a456', 'balance':0.00},  
   '789': {'id':'0003', 'password':'a789', 'balance':0.00}
}

This allows you to directly access each account by name, rather than searching the entire list for the user trying to log in.