JPA/Criteria API - Like & equal problem

John Manak picture John Manak · Oct 25, 2010 · Viewed 79k times · Source

I'm trying to use Criteria API in my new project:

public List<Employee> findEmps(String name) {
    CriteriaBuilder cb = em.getCriteriaBuilder();
    CriteriaQuery<Employee> c = cb.createQuery(Employee.class);
    Root<Employee> emp = c.from(Employee.class);
    c.select(emp);
    c.distinct(emp);
    List<Predicate> criteria = new ArrayList<Predicate>();

    if (name != null) {
        ParameterExpression<String> p = cb.parameter(String.class, "name");
        criteria.add(cb.equal(emp.get("name"), p));
    }

    /* ... */

    if (criteria.size() == 0) {
        throw new RuntimeException("no criteria");
    } else if (criteria.size() == 1) {
        c.where(criteria.get(0));
    } else {
        c.where(cb.and(criteria.toArray(new Predicate[0])));
    }

    TypedQuery<Employee> q = em.createQuery(c);

    if (name != null) {
        q.setParameter("name", name);
    }

    /* ... */

    return q.getResultList();
}

Now when I change this line:

            criteria.add(cb.equal(emp.get("name"), p));

to:

            criteria.add(cb.like(emp.get("name"), p));

I get an error saying:

The method like(Expression, Expression) in the type CriteriaBuilder is not > applicable for the arguments (Path, ParameterExpression)

What's the problem?

Answer

axtavt picture axtavt · Oct 25, 2010

Perhaps you need

criteria.add(cb.like(emp.<String>get("name"), p));

because first argument of like() is Expression<String>, not Expression<?> as in equal().

Another approach is to enable generation of the static metamodel (see docs of your JPA implementation) and use typesafe Criteria API:

criteria.add(cb.like(emp.get(Employee_.name), p));

(Note that you can't get static metamodel from em.getMetamodel(), you need to generate it by external tools).