如何使用复杂类型的javax.persistence.criteria.Predicate？

Question

I have a method that overrides the method "toPredicate":

@Override
public Predicate toPredicate(Root<Person> root, CriteriaQuery<?> cq, CriteriaBuilder cb) {
    List<Predicate> predicates = new ArrayList<>();

I have to build the predicates. It's easy with simple types, for example:

@Entity
@Table
public class Person {

    @Column
    private String name;

    @Column
    private String surname;

}

With this simple class I can do:

@Override
public Predicate toPredicate(Root<Person> root, CriteriaQuery<?> cq, CriteriaBuilder cb) {
    List<Predicate> predicates = new ArrayList<>();

    if (StringUtils.isNotBlank(surname)) {
        predicates.add(cb.equal(root.get("surname"), surname));
    }

    if (StringUtils.isNotBlank(name)) {
        predicates.add(cb.equal(root.get("name"), name));
    }

But if the attribute is a complex type, how can I find the simple attribute contained in the complex type?

This is a potential situation:

@Entity
@Table
public class Person {

    @Column
    private String name;

    @Column
    private String surname;

    @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    private List<Document> documents;

}

@Entity
@Table
public class Document {

    @Column
    private String type;

    @Column
    private String code;

}

What I have to do if I want to make a predicate with the complex attribute "documents"?

I tried:

@Override
public Predicate toPredicate(Root<Person> root, CriteriaQuery<?> cq, CriteriaBuilder cb) {
    List<Predicate> predicates = new ArrayList<>();

    if (StringUtils.isNotBlank(type)) {
        predicates.add(cb.equal(root.get("documents").get("type"), type));
    }

But I have this Exception:

java.lang.IllegalStateException: Illegal attempt to dereference path source [null.documents] of basic type
    at org.hibernate.jpa.criteria.path.AbstractPathImpl.illegalDereference(AbstractPathImpl.java:98)
    at org.hibernate.jpa.criteria.path.AbstractPathImpl.get(AbstractPathImpl.java:191)

My task is to find every Person with a determinate kind of document. How can I do it?

Thanks.

Answer 1

you can use join :

@Override
public Predicate toPredicate(Root<Person> root, CriteriaQuery<?> cq, CriteriaBuilder cb) {
    Predicate predicate = null;

    if (StringUtils.isNotBlank(type)) {
        predicate = cb.equal(root.join("documents").get("type"), type);

       cq.where(predicate);
       cq.distinct(true);

      }