JPA標準API與CONTAINS功能

Question

我想克里特標準API查詢與CONTAINS函數（MS SQL）：

SELECT * FROM com.t_person其中包含（姓氏， '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，得到的異常： org.hibernate.hql.internal.ast.QuerySyntaxException：意想不到的AST節點：

任何幫助嗎？

Answer 1

如果你想堅持使用CONTAINS，它應該是這樣的：

//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(); 

看起來你的問題中缺少一些你的代碼，所以我在這段代碼中做了一些假設。

Answer 2

你可以嘗試使用，而不是CONTAINS功能CriteriaBuilder like功能：

//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(); 

這將導致類似的查詢：

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