Hibernate Criteria API - adding a criterion: string should be in collection

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北恋
北恋 2020-12-16 06:05

I have to following entity object


@Entity
public class Foobar {
    ...
    private List uuids;
    ...
}

Now I\'d like to

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  • 2020-12-16 06:09

    The solution with the sqlRestriction from jira http://opensource.atlassian.com/projects/hibernate/browse/HHH-869 seemed the best way to go for me since i heavily use criteria api. I had to edit Thierry's code so it worked in my case

    Model:

    @Entity
    public class PlatformData
    {
      @Id
      @GeneratedValue(strategy = GenerationType.AUTO)
      private long iID;
    
      private List<String> iPlatformAbilities = new ArrayList<String>();
    }
    

    Criteria call:

    tCriteria.add(Restrictions.sqlRestriction(
            "{alias}.id in (select e.id from platformData e, platformdata_platformabilities v"
              + " where e.id = v.platformdata_id and v.element = ? )", aPlatformAbility.toString(),
            Hibernate.STRING));
    
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  • 2020-12-16 06:11

    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.

    Please annotate uuids with JPA's @ElementCollection annotation. Don't use Hibernate's @CollectionOfElements as mentioned in some of the other answer comments. The latter has equivalent functionality but is being deprecated.

    Foobar.java will look approximately like this:

    @Entity
    public class Foobar implements Serializable {
    
        // You might have some other id
        @Id
        private Long id;
    
        @ElementCollection
        private List<String> uuids;
    
        // Getters/Setters, serialVersionUID, ...
    
    }
    

    Here's how you can build a CriteriaQuery to select all Foobars whose uuids contain "abc123".

    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
        }
    }
    

    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.

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  • 2020-12-16 06:14

    I've found this post from one year ago, and I've made this method, if it can help anybody with the same problem I had a few hours ago.

        Public List<EntityObject> getHasString(String string) {
            return getSession().createCriteria(EntityObject.class)
                                   .add(Restriction.like("property-name", string, MatchMode.ANYWHERE).ignoreCase();
                               .setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY)
                               .list();
    

    Made the same with a group of strings too.

        public List<EntityObject> getByStringList(String[] tab) {
            Criterion c = Restrictions.like("property-name", tab[tab.length-1], MatchMode.ANYWHERE).ignoreCase();
            if(tab.length > 1) {
                for(int i=tab.length-2; i >= 0 ; i--) {
                    c = Restrictions.or(Restrictions.like("property-name",tab[i], MatchMode.ANYWHERE).ignoreCase(), c);
                }
            }
            return getSession().createCriteria(EntityObject.class)
                                   .add(c)
                               .setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY)
                               .list();
        }
    

    It works with "or" statements, but can easily be replaced by "and" statements.

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  • 2020-12-16 06:20

    For starters, I don't think Hibernate can map a List<String>. However, it can map a list of other entities.

    So if your code was something like this:

    @Entity
    public class Foobar {
    
        private List<EntityObject> uuids;
        ...
    }
    

    And the EntityObject has a String-property called str, the criteria could look like this:

    List<Foobar> returns = (List<Foobar>) session
                    .createCriteria.(Foobar.class, "foobars")
                    .createAlias("foobars.uuids", "uuids")
                      .add(Restrictions.like("uuids.str", "%abc123%"))
                    .list();
    
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  • 2020-12-16 06:22

    You could use a Query as in the example below or you could convert this to a NamedQuery. Unfortunately there doesn't seem to be a way to do this with Criteria.

    List<Foobar> result = session
         .createQuery("from Foobar f join f.uuids u where u =: mytest")
         .setString("mytest", "acb123")
         .list();
    
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  • 2020-12-16 06:27

    I know this is old question, but I have just encountered this issue and found solution.

    If you want to use Hibernate Criteria you can join your uuids collection and use its property elements to match elements. Just like that:

    session.createCriteria(Foobar.class)
        .createAlias("uuids", "uuids")
        .add(Restrictions.eq("uuids.elements", "MyUUID"))
        .list() 
    
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