Python 如何打印fetchone()函数,但获取非类型错误?Sqlite3
我正试图打印fetchone()值,但它只给了我一个错误,Python 如何打印fetchone()函数,但获取非类型错误?Sqlite3,python,sqlite,Python,Sqlite,我正试图打印fetchone()值,但它只给了我一个错误,“非类型”对象不可编辑,有人知道为什么吗 这是我的密码: def viewemaillook(): global realusername global realpassword viewemail = input("Would you like to see/change your email") if viewemail == "y" or viewemail == "y
“非类型”对象不可编辑
,有人知道为什么吗
这是我的密码:
def viewemaillook():
global realusername
global realpassword
viewemail = input("Would you like to see/change your email")
if viewemail == "y" or viewemail == "yes":
c.execute("SELECT email, * FROM stuffToPlot WHERE username = ? and password = ?", (realusername,realpassword,))
grab = c.fetchone()
for i in grab:
print(i)
elif viewemail == "n" or "no":
print("Okay")
else:
print("Invalid option")
viewemaillook()
viewemaillook()
else:
print("Not a valid option")
change()
change()
如果查询未找到任何结果,则
fetchone()
将返回None
。当您尝试在grab
上循环时,它等于None
,您将得到错误。首先尝试在抓取中检查是否有结果。它正在打印(“无”),知道原因吗?我在数据库中保存了一封电子邮件不确定,您的光标是否指向正确的数据库?表名正确吗?可能有多种原因。首先尝试通过硬编码值来运行查询。