JPA / Criteria API - Like&同样的问题 [英] JPA/Criteria API - Like & equal problem
问题描述
我正在尝试在我的新项目中使用Criteria API:
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();
}
现在当我改变这一行时:
Now when I change this line:
criteria.add(cb.equal(emp.get("name"), p));
to:
criteria.add(cb.like(emp.get("name"), p));
我收到错误消息:
CriteriaBuilder类型中的(Expression,Expression)方法不适用于参数(Path,ParameterExpression)
The method like(Expression, Expression) in the type CriteriaBuilder is not > applicable for the arguments (Path, ParameterExpression)
有什么问题?
推荐答案
也许你需要
criteria.add(cb.like(emp.<String>get("name"), p));
因为的第一个参数如()
是 Expression< String>
,而非 Expression<?>
,如 equal()
。
另一种方法是启用静态元模型的生成(请参阅JPA实现的文档)并使用typesafe Criteria API:
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));
(请注意,您无法从 em.getMetamodel获取静态元模型( )
,你需要通过外部工具生成它。)
(Note that you can't get static metamodel from em.getMetamodel()
, you need to generate it by external tools).
这篇关于JPA / Criteria API - Like&同样的问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!