Issue

If the "LIKE ... ESCAPE" clause is defined in a Hibernate filter, for example:

 @Entity
 @FilterDef(name="ignoreSome", parameters={@ParamDef(name="youraddress", type="string")})
 @Filter(name="ignoreSome", condition="address like :youraddress escape '\'")
 public class Member implements Serializable {

when the filter is enabled, the following errors are thrown:

Caused by: org.h2.jdbc.JdbcSQLException: Syntax error in SQL statement "SELECT THIS_.ID AS ID1_0_0_, THIS_.ADDRESS AS ADDRESS2_0_0_, THIS_.EMAIL AS EMAIL3_0_0_, THIS_.NAME AS NAME4_0_0_, THIS_.PHONE_NUMBER AS PHONE5_0_0_, THIS_.STARTDATE AS STARTDAT6_0_0_ FROM MEMBERHIBERNATE4DEMO THIS_ WHERE THIS_.ADDRESS LIKE ? THIS_[*].ESCAPE '' ORDER BY THIS_.NAME ASC "; SQL statement:
select this_.id as id1_0_0_, this_.address as address2_0_0_, this_.email as email3_0_0_, this_.name as name4_0_0_, this_.phone_number as phone5_0_0_, this_.startDate as startDat6_0_0_ from MemberHibernate4Demo this_ where this_.address like ? this_.escape '' order by this_.name asc [42000-168]

Please note that the key word "escape" is interpreted as a column name:
where this_.address like ? this_.escape ''

A similar query works fine with JPA Query object.
For example:

  Query query = em.createQuery("from Member where address like :youraddress escape '\'");
  query.setParameter("youraddress", "Brno%");