Java Hibernate-无法在where子句中使用UserType执行查询
我定义了一个Hibernate用户类型,在数据进入数据库之前对其进行转换,然后在从数据库读回数据时取消转换。当我插入一行或使用该行的ID获取行或以其他方式查询该行时,这种方法非常有效。但是,当我尝试使用查询查找记录时,参数绑定似乎失败:Java Hibernate-无法在where子句中使用UserType执行查询,java,hibernate,jpa,orm,usertype,Java,Hibernate,Jpa,Orm,Usertype,我定义了一个Hibernate用户类型,在数据进入数据库之前对其进行转换,然后在从数据库读回数据时取消转换。当我插入一行或使用该行的ID获取行或以其他方式查询该行时,这种方法非常有效。但是,当我尝试使用查询查找记录时,参数绑定似乎失败: org.springframework.dao.InvalidDataAccessApiUsageException: Parameter value [thisIsTheSearchString] did not match expected type [co
org.springframework.dao.InvalidDataAccessApiUsageException: Parameter value [thisIsTheSearchString] did not match expected type [com.xxx.MyUserType (n/a)]; nested exception is java.lang.IllegalArgumentException: Parameter value [thisIsTheSearchString] did not match expected type [com.xxx.MyUserType (n/a)]
我尝试实现LiteralType和objectToSQLString方法,但似乎从未调用过此方法
作为一个简化的例子:
public class MyUserType implements UserType, LiteralType {
@Override
public int[] sqlTypes() {
return new int[] {
Types.VARCHAR
};
}
@Override
public Class returnedClass() {
return MyUserType.class;
}
@Override
public boolean equals(Object x, Object y) throws HibernateException {
return ObjectUtils.equals(x, y);
}
@Override
public int hashCode(Object x) throws HibernateException {
assert (x != null);
return x.hashCode();
}
@Override
public Object nullSafeGet(
ResultSet rs,
String[] names,
SessionImplementor session,
Object owner)
throws HibernateException, SQLException
{
assert names.length == 1;
return untransform( rs.getString( names[0] ); );
}
String transform(String untransformed) {
//...
}
String untransform(String transformed) {
//...
}
@Override
public void nullSafeSet(
PreparedStatement st,
Object value,
int index,
SessionImplementor session)
throws HibernateException, SQLException
{
if ( value == null ) {
st.setNull(index, Types.VARCHAR);
} else {
final String untransformed = (String)value;
return transform(untransformed);
}
}
@Override
public Object deepCopy(Object value) throws HibernateException {
if ( value == null ) {
return null;
}
return (String)value;
}
@Override
public boolean isMutable() {
return true;
}
@Override
public Serializable disassemble(Object value) throws HibernateException {
return (Serializable) deepCopy(value);
}
@Override
public Object assemble(Serializable cached, Object owner)
throws HibernateException {
return deepCopy(cached);
}
@Override
public Object replace(Object original, Object target, Object owner)
throws HibernateException {
return deepCopy(original);
}
// THIS NEVER GETS CALLED
@Override
public String objectToSQLString(Object value, Dialect dialect)
throws Exception
{
if ( value == null ) {
return null;
}
String transformed = transform((String)value);
StringType stringType = new StringType();
String sqlString = stringType.objectToSQLString(transformed, dialect);
return sqlString;
}
}
实体看起来像:
@Entity
@Table(name = "blah_blah")
@TypeDefs(value = { @TypeDef(name = "transformedText", typeClass = MyUserType.class)})
public class BlahBlah implements Serializable, Persistable<Long> {
//...
@Column(name = "transformed")
@Type(type = "transformedText")
String transformed;
//...
}
我的问题是:
@Query(value =
"select b " +
"from BlahBlah b " +
"where b.transformed = ?1 ")
public List<BlahBlah> findTransformed(String text);
使用Spring数据,您可以实现,并且可以将EntityManager打开到Hibernate会话,然后将您创建的自定义类型提供给查询:
@PersistenceContext
private EntityManager entityManager;
public List<BlahBlah> findTransformed(String text) {
Session session = entityManager.unwrap(Session.class);
TypeHelper typeHelper = session.getSessionFactory().getTypeHelper();
List<BlahBlah> result = (List<BlahBlah>) session.createQuery(
"select b " +
"from BlahBlah b " +
"where b.transformed = :transformed ")
.setParameter("transformed", text, typeHelper.custom(MyUserType.class))
.list();
}
我认为您需要更改返回的类:
@Override
public Class returnedClass() {
return MyUserType.class;
}
应该是:
@Override
public Class returnedClass() {
return String.class;
}
在文件中:
返回舱
返回的班级
nullSafeGet返回的类。
返回:
阶级
您的nullSafeGet似乎返回了一个字符串。在您的查询中,b.transformed=?1在问号后面,您有1,这是发布此问题时的一个打字错误,还是您的代码中确实存在Rin,密码里有1个。我相信这是正确的-这就是参数替换的工作原理。这是有效的。Hibernate正在使用nullSafeSet获取它需要抛出到查询中的字符串。它似乎根本没有使用objectToSQLString方法,因此可能无法实现LiteralType。