Doctrine ORM：如何使用 nn 连接表定义 1-n 关系
所以，这有点复杂：我有两个表，例如 cats 和 dogs。 它们处于多对多关系（可以称为友谊或其他），因此 Doctrine 会自动为我创建一个带有适当字段的表cats_dogs。 （默认情况下，即 rowid、cat_id、dog_id。）
现在，假设我有第三个表 award，我想在其中奖励这些友谊之一。因此，这里我需要一个引用 cats_dogs 中的一行的字段。然而，由于这个表在我的模型之间并不真正存在，（Doctrine 为我处理它）最优雅的解决方案是什么？
最后，我希望在我的奖励模型中包含两个字段，一个猫和一个狗，它们需要处于友谊之中。
我正在使用注释驱动程序。
So, this is a bit complicated: I have two tables, say cats and dogs.
They are in a many-to-many relationship (could be called friendships or whatever), so that Doctrine automatically creates a table cats_dogs for me with the appropriate fields. (that is rowid, cat_id, dog_id per default.)
Now, imagine I have a third table, award, where I want to award one of these friendships. Here I therefore need a field that references one row in cats_dogs. However, since this table does not really exist between my models, (Doctrine handles it for me) what would be the most elegant solution for this?
In the end, I want in my award model two fields, a cat and a dog, who need to be in a friendship.
I am using the annotation driver.
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是什么阻止您手动创建 m:n 表而不是让教条为您创建？
What stops you from manually creating the m:n table instead of having doctrine do it for you?
该 Doctrine 的目标是从 E/R 模式映射对象并更轻松地访问对象连接。因此我相信Doctrine自动提供的表
cats_dogs是必要的。它简洁并达到了其目的，即它提供了猫的所有狗的列表，反之亦然，狗的所有猫的列表。因此，我可以得出结论，最好创建一个名为
Award的第三个实体（除了Cat和Dog），它提供一对一的一种与Cat的关系，另一种与Dog的一对一关系。使其与cats_dogs表一致只能由您决定，默认情况下不是 Doctrine 任务。例如，您可以使用一些级联持久选项。我相信这是 Doctrine 最有效的解决方案。
最后一点，请考虑每个表都应该映射一个或多个实体之间的特定关系，事实上，表
cats_dogs表示友谊关系，而表cats_dogs表示友谊关系。 >Award 将代表两个朋友之间的获奖关系关系。The Doctrine aims is to map objects from an E/R schema and to make easier the access to object connections. Therefore I believe that the table
cats_dogsautomatically provided by Doctrine is necessary as it is. It is concise and hits its purposes, i.e. it provides a list of all dogs of a cat or, vice versa, all the cats of a dog.Thus, I can conclude that it is preferable to create a third entity (besides
CatandDog) namedAwardwhich provides a one-to-one relationship withCatand another one-to-one relationship withDog. Making it consistent with thecats_dogstable is only up to you, and is not a Doctrine task by default. E.g., you can use some cascade persist option.I believe that this is the most effective solution with Doctrine.
As a final remark, consider that each table should map a specific relationship between one or more entities, and in fact the table
cats_dogsrepresents the friendship relationships, while the tableAwardwill represent the awarded relationship relationship between two friends.