Question

我正在尝试使用CONTAINS函数（MS SQL）创建Criteria API查询：

从com.t_person中选择*，其中包含（last_name，'xxx'）

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Person> cq = cb.createQuery(Person.class);
Root<Person> root = cq.from(Person.class);

Expression<Boolean> function = cb.function("CONTAINS", Boolean.class, 
root.<String>get("lastName"),cb.parameter(String.class, "containsCondition"));
cq.where(function);
TypedQuery<Person> query = em.createQuery(cq);
query.setParameter("containsCondition", lastName);
return query.getResultList();

但获得例外： org.hibernate.hql.internal.ast.QuerySyntaxException：意外的AST节点：

任何帮助？

Answer 1

如果你想坚持使用CONTAINS，它应该是这样的：

//Get criteria builder
CriteriaBuilder cb = em.getCriteriaBuilder();
//Create the CriteriaQuery for Person object
CriteriaQuery<Person> query = cb.createQuery(Person.class);

//From clause
Root<Person> personRoot = query.from(Person.class);

//Where clause
query.where(
    cb.function(
        "CONTAINS", Boolean.class, 
        //assuming 'lastName' is the property on the Person Java object that is mapped to the last_name column on the Person table.
        personRoot.<String>get("lastName"), 
        //Add a named parameter called containsCondition
        cb.parameter(String.class, "containsCondition")));

TypedQuery<Person> tq = em.createQuery(query);
tq.setParameter("containsCondition", "%näh%");
List<Person> people = tq.getResultList();

您的问题似乎缺少某些代码，因此我在此代码段中做了一些假设。

Answer 2

您可以尝试使用CriteriaBuilder like函数代替CONTAINS函数：

//Get criteria builder
CriteriaBuilder cb = em.getCriteriaBuilder();
//Create the CriteriaQuery for Person object
CriteriaQuery<Person> query = cb.createQuery(Person.class);

//From clause
Root<Person> personRoot = query.from(Person.class);

//Where clause
query.where(
    //Like predicate
    cb.like(
        //assuming 'lastName' is the property on the Person Java object that is mapped to the last_name column on the Person table.
        personRoot.<String>get("lastName"),
        //Add a named parameter called likeCondition
        cb.parameter(String.class, "likeCondition")));

TypedQuery<Person> tq = em.createQuery(query);
tq.setParameter("likeCondition", "%Doe%");
List<Person> people = tq.getResultList();

这应该会产生类似于：

的查询

select p from PERSON p where p.last_name like '%Doe%';

具有CONTAINS功能的JPA Criteria api

2 个答案: