如何打印一个fetchone（）函数但却得到NoneType错误？Sqlite3公司

Question

我正试图打印fetchone（）值，但它只是给了我一个错误'NoneType' object is not iterable，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()

Answer 1

如果查询没有找到任何结果，则fetchone()将返回None。当你尝试在grab上循环，它等于None你得到了错误。请先检查grab中是否有结果。你知道吗