JPA/Criteria API - Like & equal problem

只愿长相守 提交于 2019-11-28 16:58:27

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).

user3077341

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);
}

Use :

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