java OpenJPA criteriaBuilder 嵌套对象属性获取
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OpenJPA criteriaBuilder nested object property fetch
提问by quantum
Is there any way in OpenJPA to get hold of a nested object property via CriteriaBuilder?
OpenJPA 中是否有任何方法可以通过 CriteriaBuilder 获取嵌套对象属性？
Here's a small case.
这是一个小案例。
@Entity
public class X {
private Object Y;
// getters, setters...
}
@Entity
public class Y {
private String Z;
// getters, setters...
}
So, when using CriteriaBuilder, we use X as Root, i.e.:
因此，在使用 CriteriaBuilder 时，我们使用 X 作为 Root，即：
@PersistenceContext
private EntityManager entityManager;
//.....
Root<X> rootObj = criteriaBuilder.from(X.class);
CriteriaQuery<X> select;
String param1 = X.getY().getZ();
// initializing predicate, default value is TRUE
Predicate predicate1 = criteriaBuilder.isNull(null);
// construct search predicate which fails miserably due to IllegalArgumentExecption
if (X != null) {
predicate1 = criteriaBuilder.and(predicate1, criteriaBuilder.equal(rootObj.<String> get("Y.Z"), param1));}
Now, my grief is this -> get("Y.Z")
现在，我的悲伤是这样 -> get("Y.Z")
CriteriaBuilder doesn't know to fetch Z reflectively (however it can and will resolve Y). Is there any way to get hold of Z directly from get()?
CriteriaBuilder 不知道反射性地获取 Z（但是它可以并且将会解析 Y）。有什么办法可以直接从 get() 获取 Z 吗？
Apart from using JPQL, I can think of one other method - which I dislike immensely: I suppose I could have exposed Z as an @Transientproperty in X (as to prevent OpenJPA from persisting it as a column), but that sounds like a Really Bad Idea: I am essentially flattening out an object graph manually and introduce unneeded garbage inside the entity bean, not counting the time needed to flatten out a complex graph or the error-proness of this (it can go awry in so many ways).
除了使用 JPQL，我还能想到另一种方法——我非常不喜欢它：我想我可以将 Z 作为@TransientX 中的一个属性公开（以防止 OpenJPA 将它作为列持久化），但这听起来真的很糟糕想法：我基本上是手动展平对象图并在实体 bean 中引入不需要的垃圾，不计算展平复杂图所需的时间或容易出错的情况（它可能在很多方面出错）。
Is there a way to make this work? Any ideas are appreciated.
有没有办法使这项工作？任何想法表示赞赏。
回答by quantum
Heh, the solution is suprisingly simple - and it looks really ugly, but it works.
嘿，解决方案出奇的简单——它看起来真的很丑，但它确实有效。
predicate1 = criteriaBuilder.and(predicate1, criteriaBuilder.equal(rootObj.get("Y").<String> get("Z"), param1));}
I really don't know if there is a more elegant solution to this.
我真的不知道是否有更优雅的解决方案。
回答by Oleg Mikhailov
For any arbitrary nested attribute path ("relation.subRelation.attribute"):
对于任意嵌套的属性路径（“relation.subRelation.attribute”）：
private Path<T> getPath(Root<T> root, String attributeName) {
Path<T> path = root;
for (String part : attributeName.split("\.")) {
path = path.get(part);
}
return path;
}
