Java 在Hibernate中使用LIKE%

如何在Hibernate中使用

LIKE%

。我想在我的

hbm.xml

文件中使用一个类似%的SQL。我有2个问题，我正在合并为1个

查询如下所示：

select * from PAY_GROUP_VW where CASE_SID=? AND CASE_TLE like %?%

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute query
    at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:630)
    at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
    at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:424)
    at org.springframework.orm.hibernate3.HibernateTemplate.executeFind(HibernateTemplate.java:343)

q.setParameter(0, csId);
q.setParameter(1, csTle);
return q.list();

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute query
    at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:630)
    at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
    at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:424)
    at org.springframework.orm.hibernate3.HibernateTemplate.executeFind(HibernateTemplate.java:343)

我还尝试了

%？%

。我使用了不带

%

的like子句，它很有效。但是

%

不起作用

包装异常：

select * from PAY_GROUP_VW where CASE_SID=? AND CASE_TLE like %?%

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute query
    at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:630)
    at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
    at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:424)
    at org.springframework.orm.hibernate3.HibernateTemplate.executeFind(HibernateTemplate.java:343)

q.setParameter(0, csId);
q.setParameter(1, csTle);
return q.list();

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute query
    at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:630)
    at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
    at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:424)
    at org.springframework.orm.hibernate3.HibernateTemplate.executeFind(HibernateTemplate.java:343)

这是我传递参数的方式：

select * from PAY_GROUP_VW where CASE_SID=? AND CASE_TLE like %?%

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute query
    at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:630)
    at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
    at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:424)
    at org.springframework.orm.hibernate3.HibernateTemplate.executeFind(HibernateTemplate.java:343)

q.setParameter(0, csId);
q.setParameter(1, csTle);
return q.list();

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute query
    at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:630)
    at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
    at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:424)
    at org.springframework.orm.hibernate3.HibernateTemplate.executeFind(HibernateTemplate.java:343)

像

%

|？一样尝试，并得到以下异常：

select * from PAY_GROUP_VW where CASE_SID=? AND CASE_TLE like %?%

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute query
    at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:630)
    at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
    at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:424)
    at org.springframework.orm.hibernate3.HibernateTemplate.executeFind(HibernateTemplate.java:343)

q.setParameter(0, csId);
q.setParameter(1, csTle);
return q.list();

org.springframework.dao.InvalidDataAccessResourceUsageException: could not execute query; nested exception is org.hibernate.exception.SQLGrammarException: could not execute query
    at org.springframework.orm.hibernate3.SessionFactoryUtils.convertHibernateAccessException(SessionFactoryUtils.java:630)
    at org.springframework.orm.hibernate3.HibernateAccessor.convertHibernateAccessException(HibernateAccessor.java:412)
    at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:424)
    at org.springframework.orm.hibernate3.HibernateTemplate.executeFind(HibernateTemplate.java:343)

---

也许用户id不是char/varchar？您必须先将带有str（）的用户_id转换为字符数据

例如：

SELECT *
FROM abc.def_vw a
WHERE user_id = ?
AND str(user_id) LIKE '%' || ?

---

使用“%”？%%之类的

和第二个参数字符串进行查询。字段也应该是字符串类型，因为
like
用于比较字符串。
需要注意的三件事：


是用户id字符串吗
就像在HQL中一样，效果很好。例如“%something%”
您的查询没有意义“where user_id=？and user_id像“%”您想做什么？仅使用其中一个条件
您能否将查询和代码或配置一起放入hbm.xml文件中？