Question

我有：

ClassAttribute:
    def __init__(self, string):
        ### example of string : 
        ###     "c1 int not null"
        splitted_string = string.split(' ', 2)
        self.col_name = splitted_string[0]
        self.col_def = splitted_string[1]
        if len(splitted_string) == 3:
            self.col_option = splitted_string[2]
        else:
            self.col_option = ""

ClassTable:
    def __init__(self, string):
        ### example of string : 
        ###     "create table TABLE_A (c1 int not null, c2 string)"
        string_splitted = string.split(' ', 3)
        self.table_name = string_splitted[2]
        ### In real life, I've a list of my attributes
        ### Example : 
        ###     list_attributes = ['c1 int not null', 'c2 string']
        self.table_attributes = []
        for attr in list_attributes:
            self.table_attributes.append(ClassAttribute(attr))

问题：

如何在不迭代list_attributes的情况下调用表中的属性对象？例如，我想添加＆＃34; not null＆＃34;第n个属性的选项，我不知道它的位置，但我知道它的名字。
从属性对象中，如何获取表名？

Answer 1

Evert的建议对我有用：

ClassAttribute:
    def __init__(self, table, string):
        splitted_string = string.split(' ', 2)
        self.table_name = table.table_name
        self.col_name = splitted_string[0]
        self.col_def = splitted_string[1]
        if len(splitted_string) == 3:
            self.col_option = splitted_string[2]
        else:
            self.col_option = ""

ClassTable:
    def __init__(self, string):
        string_splitted = string.split(' ', 3)
        self.table_name = string_splitted[2]
        self.table_attributes = []
        for attr in list_attributes:
            self.table_attributes.append(ClassAttribute(self, attr))

python中的类之间的链接

1 个答案: