Java JPA/Criteria API - Like & equal 问题
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JPA/Criteria API - Like & equal problem
提问by John Manak
I'm trying to use Criteria API in my new project:
我正在尝试在我的新项目中使用 Criteria API：
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:
现在当我改变这一行时：
criteria.add(cb.equal(emp.get("name"), p));
to:
到：
criteria.add(cb.like(emp.get("name"), p));
I get an error saying:
我收到一条错误消息：
The method like(Expression, Expression) in the type CriteriaBuilder is not > applicable for the arguments (Path, ParameterExpression)
CriteriaBuilder 类型中的 like(Expression, Expression) 方法不适用于参数 (Path, ParameterExpression)
What's the problem?
有什么问题？
采纳答案by axtavt
Perhaps you need
也许你需要
criteria.add(cb.like(emp.<String>get("name"), p));
because first argument of like()is Expression<String>, not Expression<?>as in equal().
因为第一个参数的like()就是Expression<String>，不Expression<?>作为equal()。
Another approach is to enable generation of the static metamodel (see docs of your JPA implementation) and use typesafe Criteria API:
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）并使用类型安全的 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).
（请注意，您无法从 中获取静态元模型em.getMetamodel()，您需要通过外部工具生成它）。
回答by user3077341
Better: predicate (not ParameterExpression), like this :
更好：谓词（不是ParameterExpression），像这样：
List<Predicate> predicates = new ArrayList<Predicate>();
if(reference!=null){
Predicate condition = builder.like(root.<String>get("reference"),"%"+reference+"%");
predicates.add(condition);
}
回答by Abdessamad HALLAL
Use :
用 ：
personCriteriaQuery.where(criteriaBuilder.like(
criteriaBuilder.upper(personRoot.get(Person_.description)),
"%"+filter.getDescription().toUpperCase()+"%"));
