具有CONTAINS功能的JPA Criteria API

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();

但是出现异常：org.hibernate.hql.internal.ast.QuerySyntaxException：意外的AST节点：

有什么帮助吗？

Answer 1

如果您要坚持使用CONTAINS，则应该是这样的：

//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();

您的问题似乎缺少一些代码，因此在此代码段中我做了一些假设。