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copied!<p>I assume you are using a version of Hibernate that implements JPA 2.0. Here's a JPA 2.0 solution that should work with any compliant implementation.</p> <p>Please annotate <code>uuids</code> with JPA's <code>@ElementCollection</code> annotation. Don't use Hibernate's <code>@CollectionOfElements</code> as mentioned in some of the other answer comments. The latter has equivalent functionality but is <a href="http://docs.jboss.org/hibernate/stable/annotations/reference/en/html_single/#d0e1821" rel="noreferrer">being deprecated</a>.</p> <p><code>Foobar.java</code> will look approximately like this:</p> <pre><code>@Entity public class Foobar implements Serializable { // You might have some other id @Id private Long id; @ElementCollection private List<String> uuids; // Getters/Setters, serialVersionUID, ... } </code></pre> <p>Here's how you can build a <code>CriteriaQuery</code> to select all <code>Foobar</code>s whose <code>uuids</code> contain "abc123". </p> <pre><code>public void getFoobars() { { EntityManager em = ... // EM by injection, EntityManagerFactory, whatever CriteriaBuilder b = em.getCriteriaBuilder(); CriteriaQuery<Foobar> cq = b.createQuery(Foobar.class); Root<Foobar> foobar = cq.from(Foobar.class); TypedQuery<Foobar> q = em.createQuery( cq.select(foobar) .where(b.isMember("abc123", foobar.<List<String>>get("uuids")))); for (Foobar f : q.getResultList()) { // Do stuff with f, which will have "abc123" in uuids } } </code></pre> <p>I made a self-contained proof-of-concept program while playing with this. I can't push it out right now. Please comment if you want the POC pushed to github.</p>

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