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下面只是提供错误的示例代码，我希望在修复时获得帮助，或者了解更好的编写方法。我有一个名为 mysql_connection 的 mysql“超级”类。在这个类中，建立到数据库的连接。我也有一些方法。一个简单地运行“选择版本（）”以显示连接/查询有效。然后我有一个“chktable”方法，在这个例子中实例化了一个名为“table”的新子类，它继承了超类。实例化类后，我在子类中调用一个方法，该方法尝试使用超类中的查询方法来运行“显示像'tbl name'这样的表”。这是我得到错误的地方。

import mysql.connector
from mysql.connector import errorcode
from mysql.connector.cursor import MySQLCursor

class mysql_connection(object):
    def __init__(self, **kwargs):
        self.connection_options = {}
        self.connection_options['user'] = 'root'
        self.connection_options['password'] = ''
        self.connection_options['host'] = '192.168.33.10'
        self.connection_options['port'] = '3306'
        self.connection_options['database'] = "test"
        self.connection_options['raise_on_warnings'] = True
        self.connect()

    def connect(self):
        try:
            self.cnx = mysql.connector.connect(**self.connection_options)
        except mysql.connector.Error as err:
            if err.errno == errorcode.ER_ACCESS_DENIED_ERROR:
                print "Something is wrong with your user name or password"
            elif err.errno == errorcode.ER_BAD_DB_ERROR:
                print "Database does not exists" 
            else:
                print err

    def query(self, statement, data=''):
        cursor = MySQLCursor(self.cnx)
        cursor.execute(statement)
        result = cursor.fetchall()
        cursor.close
        return result

    def get_version(self):
        print self.query("select version()")

    def chktable(self, tb_name):
        tab = table(name=tb_name)
        tab.check_table()

class table(mysql_connection):
    def __init__(self, **kwargs):
        self.name = kwargs['name']

    def check_table(self):
        return super(table, self).query("show tables like '{}".format(self.name))

conn = mysql_connection()
conn.get_version()
conn.chktable("test")

我得到的错误是：

$ python example.py
[(u'5.1.73',)]
Traceback (most recent call last):
  File "example.py", line 50, in <module>
    conn.chktable("test")
  File "example.py", line 39, in chktable
    tab.check_table()
  File "example.py", line 46, in check_table
    return super(table, self).query("show tables like '{}".format(self.name))
  File "example.py", line 28, in query
    cursor = MySQLCursor(self.cnx)
    AttributeError: 'table' object has no attribute 'cnx'

我不完全理解回调超类以及它如何与子类一起工作，所以这可能是我的问题。我也想知道是否有更好的方法来实现这一点。我在想我可以完全摆脱子类，但我喜欢我制作的子类，所以我觉得应该有办法绕过它。我可以尝试的第二件事是将子类放在主类中，但我认为这是不正确的。

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1 回答 1

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As Jon points out, this is not an appropriate use of inheritance. That is for "is-a" relationships: ie Dog inherits from Animal, because a dog is an animal. But a table is not a connection: a table might use a connection, but that simply means you should assign an instance of the connection to an instance variable in table.

Also, in an inheritance relationship there is usually no good reason for a superclass to know about its subclasses, as you have in the chktable method.

(The actual bug you're seeing is because you haven't called the superclass method in table's __init__, but it's better to fix your structure.)

于 2014-12-06T09:55:56.387 回答