JPA/Criteria API - Like & equal problem

Question

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

public List findEmps(String name) {
    CriteriaBuilder cb = em.getCriteriaBuilder();
             


        

Answer 1

Better: predicate (not ParameterExpression), like this :

List<Predicate> predicates = new ArrayList<Predicate>();
if(reference!=null){
    Predicate condition = builder.like(root.<String>get("reference"),"%"+reference+"%");
    predicates.add(condition);
}

Answer 2

Use :

personCriteriaQuery.where(criteriaBuilder.like(
criteriaBuilder.upper(personRoot.get(Person_.description)), 
"%"+filter.getDescription().toUpperCase()+"%")); 

Answer 3

I tried all the proposals and although it did not give an error, it did not recover anything if I did not put all the expected text, in the end what was proposed by user3077341 worked for me but with the variant that I use '%' instead of "%", the like worked perfect.

List<Predicate> predicates = new ArrayList<Predicate>();
if(reference!=null){
    Predicate condition = builder.like(root.<String>get("reference"),'%'+reference+'%');
    predicates.add(condition);
}

Answer 4

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).