Question

我在python中使用SQLITE3，我有几个错误。

第一个：

sqlite3.ProgrammingError：提供的绑定数量不正确。当前语句使用1，并且提供了2个。

第二个：这个问题并不是一个错误，但是它不起作用。

username= input("Username: ").lower()
password= input("Password: ")


#Get username and see if it exists
c.execute('SELECT id FROM accounts WHERE username=?', (username))


print("test")
#Insert values into the account table in the database.
c.execute("INSERT INTO accounts VALUES(NULL,?,?)", (username,password))

conn.commit()

#Grab ID
ids = c.execute("SELECT id FROM accounts WHERE username=?", (username,))

#Insert defaults into playlists
c.execute("INSERT INTO playlists VALUES(NULL,?,'NULL')", (ids))

#Save the changes.
conn.commit()

它只是说＆＃34;用户名已经存在！＆＃34;唯一的帐户是

用户名：s 密码：s

我已经在StackOverflow上四处寻找，但找不到任何有助于我的事业的事情，因此我发布了这个原因。编辑：

   def loggedIn(userid):
        def viewPlaylist(userid):
            print(userid + 1)
            c.execute("SELECT 1 FROM playlists where userid=?", (userid))

            if(c.fetchone): 
                print(c.fetchone)
            else:
                printError("ERROR: No Data found, with that UserID\nCreating Data...")
                c.execute("INSERT INTO playlists VALUES(NULL,?,NULL)", (userid))
                printError("Finished!\nYou may now edit your playlist.")
                loggedIn(userid)

Answer 1

这在我身边：

def logged_in(user_name, password):
    # Here you should add conn and cursor, or paste in parameter
    c.execute("SELECT id FROM accounts where username=? and password=?", (user_name, password)))
    user = c.fetchone()
    if user:
        print('logged in')
    else:
        print('user not found')

或者以你的方式，你应该这样做：

loggedIn(c.execute("SELECT id FROM accounts where username=?",(username)).fetchone())

sqlite3.InterfaceError和查询错误

1 个答案: