我可以将这个if语句转换为Python中的函数吗？

我将重复这段代码，所以我尝试将其转换为函数。 但我总是出错。 编辑 否我收到错误：缺少1个位置参数：“连接” 在方法中放置“连接”的最佳方式是什么

def sqlfirst(self, connection):
    firstname = str(self.first_entry.get())
    lastname = str(self.last_entry.get())     
    license = str(self.lic_entry.get())
    if (firstname and not lastname and not license):  # "You entered first name."
        try:
            connection = pypyodbc.connect('Driver={SQL Server};Server=;Database=;Trusted_Connection=yes;')
        except pypyodbc.Error as ex:
            sqlstate = ex.args[0]
            if sqlstate == '28000':
                self.answer_label['text'] = "You do not have access." 
        cursor = connection.cursor() 
        SQLCommand = ("SELECT LASTNAME, FIRSTNAME, L_LICNUMBER "      
            "FROM dbo. "   # table name
            "with (nolock)"
            "WHERE FIRSTNAME = ?")
        Values = [firstname]
        cursor.execute(SQLCommand,Values)
        return cursor.fetchmany(10)




def calculate2(self):
    results = self.sqlfirst()
    if results:

        self.output1.delete(0, END)
        self.output1.insert(0,results[2])
        connection.close()
        traceback.print_exc()

下面是我的旧代码：

我是否可以将上述内容转换为调用的函数，以便随时调用sqlfirst（self）？我试过：

def sqlfirst(self):
    firstname = str(self.first_entry.get())
    lastname = str(self.last_entry.get())     
    if (firstname and not lastname):  # "You entered first name."
        try:
            connection = pypyodbc.connect('Driver={SQL Server};Server=;Database=;Trusted_Connection=yes;')
        except pypyodbc.Error as ex:
            sqlstate = ex.args[0]
            if sqlstate == '28000':
                self.answer_label['text'] = "You do not have access." 
        cursor = connection.cursor() 
        SQLCommand = ("SELECT LASTNAME "      
            "FROM dbo.NAME "   # table name
            "with (nolock)"
            "WHERE FIRSTNAME = ?")
        Values = [firstname]
        cursor.execute(SQLCommand,Values)
        results = cursor.fetchmany(10)



def calculate(self):
    self.sqlfirst()
        if results:

            self.output2.delete(0, END)
            self.output2.insert(0,results[2])
            connection.close()

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在函数末尾返回cursor.fetchmany（10）

，而不是将其赋给局部变量，该局部变量在函数退出时丢失。然后像这样调用函数：

results=sqlfirst（）

。好的，我试试看。顺便说一句，我认为我的大部分问题都来自我的IDE以及它如何使用空格和制表符。我现在正试图搞乱设置。首先，这些不是函数，而是方法。第二，如果您需要有关错误的帮助，请向我们显示错误的回溯。您的类定义在哪里？