我正在尝试检查数据库中的用户名,当我这样做是为第一个用户名工作,但在它不起作用后,我想我理解为什么但我不能找到替代方案。 这是我的代码:
import sqlite3
conn = sqlite3.connect('tutorial.db')
c = conn.cursor()
def username():
global limit
global usernameinput
usernameinput = input("What would you like your username to be?")
limit = 0
select_all_tasks(conn)
def create_table():
c.execute('CREATE TABLE IF NOT EXISTS stuffToPlot(username TEXT,
password TEXT)')
conn.close
def select_all_tasks(conn):
global limit
global passwordinput
global rows
c.execute("SELECT * FROM stuffToPlot")
rows = c.fetchall()
for row in rows:
print(row)
if usernameinput in row:
print("Username Taken")
username()
else:
if limit < 1:
passwordinput = input("What would you like your password to
be?")
limit = limit + 1
def data_entry():
global passwordinput
global row
c.execute("INSERT INTO stuffToPlot VALUES(?, ?);",(usernameinput,
passwordinput))
conn.commit()
c.close()
conn.close()
create_table()
username()
select_all_tasks(conn)
data_entry()
没有错误,它只是没有注册用户名已经在数据库中。
答案 0 :(得分:1)
您需要缩进if语句和所有以下行以使其工作,否则只测试最后一行。
# select only the field "username" from the table
c.execute("SELECT username FROM stuffToPlot")
# build a set with all names, from the results given as [('name1',), ('name2',), ('name3',)]
names = {name[0] for name in c.fetchall()}
if usernameinput in names:
print("Username Taken")
username()
else:
if limit < 1:
你也可以简单一点:
{{1}}