Play框架加入并建立模型 [英] Play Framework joins and the model

问题描述

这不是问题，更多的是这是正确的方法"这类交易.

This isn't a problem, its more of a "is this the right approach" kind of deal.

所以可以说我有一个这样的模型

So lets say I have a model like this

class A extends Model{
    @OneToMany(cascade = CascadeType.ALL)
    public B;
}

class B extends Model{

   String c;
}

现在，我要访问其B对象中所有具有特定c值的A对象.

Now I want to access all objects of A that have a particular value of c in their B objects.

所以我应该

1. 获取具有特定值c的B的所有对象，然后找到与这些对象相对应的A(如果这样，则感到困惑)
2. 使用find.all()获取A的所有对象，然后浏览列表(这似乎是一个坏主意，因为会有很多A，而没有那么多B).

1. Get all objects of B with a certain value c and then find corresponding A's with those objects(if so how, getting confused)
2. Get all objects of A with find.all() then look through the list (seems a bad idea as there will be a large amount of A's and not so many B's).

任何帮助将不胜感激(哦，假设我已经写了@Entity和@Required，并且其余所有都适当地写了)

Any help would be appreciated (oh and assume I've written @Entity and @Required and all the rest appropriately)

推荐答案

选项1是正确的选择.您可以使用诸如

Option 1 is the right way to go. You can use a query such as

A.find().where().eq("b.id", yourBId).findList();

如果未自动获取，则可能添加fetch("b").

Possibly adding a fetch("b"), if it isn't fetched automatically.

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