MongoDB和Python,用于存储游戏的用户名和密码
MongoDB with Python, used to store username and passwords for a game
这是用户登录游戏的部分。数据库托管在 mlab 上,有一个名为 loginInfo
的文档集合,定义为 mycol
。
我面临的问题是,当它 returns ID 时,我无法检查它是否有效,因为我无法检查任何类型的本地列表Python。我该如何解决?
import pymongo
def login():
print('Login\nRegister\nPlay as Guest')
login_input = input('What would you like to do?')
if login_input == 'Login':
username = input('Username:')
username_fromDB = mycol.find({"username": username})
if username_fromDB == username:
password = input('Password:')
password_fromDB = mycol.find({"password": password})
if password == password_fromDB:
print('Login success!\n')
mainMenu()
elif login_input == 'Register':
new_username = input('Username:')
new_password = input('Password:')
tobeaddedtodb = {"username": new_username, "password": new_password}
adding = mycol.insert_one(tobeaddedtodb)
print("Registered!\n")
elif login_input == 'Play as Guest':
mainMenu()
username_fromDB = mycol.find({"username": username})
这是错误的,因为 find()
returns 一个游标对象,而不是您期望的用户名。
假设 loginInfo
集合包含唯一的用户名和各自的密码,则无需对数据库进行 2 find()
次查询。您可以只检索用户实体并检查输入提供的密码是否与数据库中存储的密码匹配。
并且由于用户名是唯一的,您应该使用 find_one()
而不是 find()
。如果用户成功找到它 returns dict
,否则你会得到 None
.
尝试这样的事情:
import pymongo
def login():
print('Login\nRegister\nPlay as Guest')
login_input = input('What would you like to do?')
if login_input == 'Login':
username_inp = input('Username:')
pwd_inp = input('Password:')
user_found = mycol.find_one({"username": username_inp}) # query by specified username
if user_found: # user exists
if pwd_inp == user_found['password']:
print('Login success!\n')
mainMenu()
else:
print('Wrong password')
else:
print('User not found')
elif login_input == 'Register':
new_username = input('Username:')
new_password = input('Password:')
tobeaddedtodb = {"username": new_username, "password": new_password}
adding = mycol.insert_one(tobeaddedtodb)
print("Registered!\n")
elif login_input == 'Play as Guest':
mainMenu()
这是用户登录游戏的部分。数据库托管在 mlab 上,有一个名为 loginInfo
的文档集合,定义为 mycol
。
我面临的问题是,当它 returns ID 时,我无法检查它是否有效,因为我无法检查任何类型的本地列表Python。我该如何解决?
import pymongo
def login():
print('Login\nRegister\nPlay as Guest')
login_input = input('What would you like to do?')
if login_input == 'Login':
username = input('Username:')
username_fromDB = mycol.find({"username": username})
if username_fromDB == username:
password = input('Password:')
password_fromDB = mycol.find({"password": password})
if password == password_fromDB:
print('Login success!\n')
mainMenu()
elif login_input == 'Register':
new_username = input('Username:')
new_password = input('Password:')
tobeaddedtodb = {"username": new_username, "password": new_password}
adding = mycol.insert_one(tobeaddedtodb)
print("Registered!\n")
elif login_input == 'Play as Guest':
mainMenu()
username_fromDB = mycol.find({"username": username})
这是错误的,因为 find()
returns 一个游标对象,而不是您期望的用户名。
假设 loginInfo
集合包含唯一的用户名和各自的密码,则无需对数据库进行 2 find()
次查询。您可以只检索用户实体并检查输入提供的密码是否与数据库中存储的密码匹配。
并且由于用户名是唯一的,您应该使用 find_one()
而不是 find()
。如果用户成功找到它 returns dict
,否则你会得到 None
.
尝试这样的事情:
import pymongo
def login():
print('Login\nRegister\nPlay as Guest')
login_input = input('What would you like to do?')
if login_input == 'Login':
username_inp = input('Username:')
pwd_inp = input('Password:')
user_found = mycol.find_one({"username": username_inp}) # query by specified username
if user_found: # user exists
if pwd_inp == user_found['password']:
print('Login success!\n')
mainMenu()
else:
print('Wrong password')
else:
print('User not found')
elif login_input == 'Register':
new_username = input('Username:')
new_password = input('Password:')
tobeaddedtodb = {"username": new_username, "password": new_password}
adding = mycol.insert_one(tobeaddedtodb)
print("Registered!\n")
elif login_input == 'Play as Guest':
mainMenu()