Hibernate JPA Criteria Query

Question

I need to query an object that contains a list.

@Entity @Table(name = "userAccount") public class UserAccounts implements Serializable { @Id @GeneratedValue(strategy = GenerationType.AUTO) private int id; @Column(name = "email", nullable = false, unique = true) private String username; @Column(name = "hash", nullable = false, length = 60) private String hash; @ManyToMany(cascade = CascadeType.ALL) @JoinTable(name = "user_roles", joinColumns = @JoinColumn(name = "userID"), inverseJoinColumns = @JoinColumn(name = "roleID")) private List<Role> roles; @Enumerated(EnumType.ORDINAL) private Status status; //getters and setters }

I need to get objects that contains role "ROLE_USER". how can I achieve that using CriteriaBuilder?

 CriteriaBuilder cb = em.getCriteriaBuilder(); CriteriaQuery<UserAccounts> cq = cb.createQuery(UserAccounts.class); Root<UserAccounts> ua = cq.from(UserAccounts.class);

What Should I put here?

// cq.where(cb.and(cq.equals(ua.get("roles")), // cq.equals(ua.get("status"), Status.ACTIVE)));

TIA.

Neil Stockton · Accepted Answer · 2016-01-10 09:05:43Z

1

The equivalent JPQL could be like this

SELECT ua FROM UserAccount ua JOIN ua.roles r WHERE r.name = :roleName

so the Criteria would be something like

Join<UserAccounts, Role> roleRoot = ua.join(UserAccounts_.roles); roleRoot.alias("r"); ParameterExpression<String> param = qb.parameter(String.class); cq.where(roleRoot.get(Role_.name).equals(param));

answered Jan 10, 2016 at 9:05

Neil Stockton

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3 Comments

pat3ck029 Over a year ago

Hi, using ua.join(UserAccounts_.roles); requires me to change roles to public. i don't think i should change that to public though.

Neil Stockton Over a year ago

No it doesn't require public. This is the static metamodel class! You can use the equivalent String based getter if you insist on using Strings

pat3ck029 Over a year ago

Finally got it to work. turns out I was mising Hibernate-metamodel-generator.jar. Thank You So Much!

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