Question

我有以下表格：

Campaigns
+----------------------------+-------------------------------------------------------------------+------+-----+-------------------+----------------+
| Field                      | Type                                                              | Null | Key | Default           | Extra          |
+----------------------------+-------------------------------------------------------------------+------+-----+-------------------+----------------+
| id                         | int(11) unsigned                                                  | NO   | PRI | NULL              | auto_increment |
| campaign_type_id           | int(11)                                                           
+----------------------------+-------------------------------------------------------------------+------+-----+-------------------+----------------+

CampaignsSiteList
+--------------+------------+------+-----+---------+-------+
| Field        | Type       | Null | Key | Default | Extra |
+--------------+------------+------+-----+---------+-------+
| campaign_id  | int(11)    | NO   | PRI | NULL    |       |
| site_list_id | int(11)    | NO   | PRI | NULL    |       |
+--------------+------------+------+-----+---------+-------+

我正在使用SQL Alchemy，我想创建一个关系，以便Campaign类的对象具有一个属性，该属性返回与之关联的site_list_id列表。我不希望关系返回CampaignSiteList对象列表，而是包含CampaignsSiteList列site_list_id的列表。

Answer 1

你可以在课堂上使用一个属性并自己将它们拉出来，例如：

class Campaigns():

   # column definitions here

   sites = relationship("CampaignSiteList", lazy="joined")

   @property
   def site_ids(self):
       return [d.id for d in self.sites]

SQL Alchemy。有许多关系返回特定列

1 个答案: