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java - JPA/Criteria API - Like & equal problem

问题描述:

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?

网友答案:

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).

网友答案:

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);
}
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