题解 | #禁止重复注册#

此问题的重点在于：不区分名字的大小写，故新老用户名单均需upper或lower后一一对比

以下是对评论区方法以及涉及知识点的整理：

方法一：两列表分别循环——先处理好老用户列表，再拿新用户名称一一对应【逻辑简单！】

current_users=['Niuniu','Niumei','GURR','LOLO']
new_users=["GurR",'Niu Ke Le','LoLo','Tuo Rui Chi']
current=[username.lower() for username in current_users]
for name in new_users:
    if name.lower() in current:
        print('The user name '+name+' has already been registered! Please change it and try again!')
    else:
        print('Congratulations, the user name '+name+' is available!')

方法二：嵌套循环

current_users=['Niuniu','Niumei','GURR','LOLO']
new_users=["GurR",'Niu Ke Le','LoLo','Tuo Rui Chi']
for name in new_users:
    for i in range(0,4):
        if name.lower()==current_users[i].lower():
            print('The user name '+name+' has already been registered! Please change it and try again!')
            break ##及时跳出循环，防止后续碰到其他不等值继续else
    else:
        print('Congratulations, the user name '+name+' is available!')

此处需要注意if-else的位置以及break的存在：

- 满足if:break可跳出当前循环，

- 当循环结束if都不曾满足才会执行else

附break知识点：

1. for循环+break

• 只有if :对循环内每个对象执行if，遇break就终止循环

• If-else且同级：对循环内每个对象执行if-else，遇break终止循环

• If-else但else在外层：对循环内每个对象执行if,遇break终止；

若整个循环都未终止，则执行else（见本题情况）

2. 利用break终止内外循环：

• 内循环

• 外循环（内循环break前加一个flag即可）

参考：https://blog.csdn.net/yinlu521wang/article/details/99461129?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522167772841916800222843342%2522%252C%2522scm%2522%253A%252220140713.130102334..%2522%257D&request_id=167772841916800222843342&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2~all~top_positive~default-2-99461129-null-null.142^v73^insert_down2,201^v4^add_ask,239^v2^insert_chatgpt&utm_term=python%20break&spm=1018.2226.3001.4187