我们的问题是,由于源自Hibernate的StringIndexOutOfBoundsExceptiion
,我们无法从旧数据库中获取数据(包括长度为0的空字符串) CharacterTypeDescriptor
。我们想改变Hibernate的行为以正确解析空字符串。
示例数据:
1, 'Berlin', 17277, '', 'aUser'
2, 'London', 17277, '', 'anotherUser'
我们将hibernate与javax.persistence.Query
一起使用。
String sql = "SELECT * FROM table";
Query query = entityManager.createNativeQuery(sql);
List resultList = query.getResultList();
这导致StringIndexOutOfBoundsException
,其根是来自Hibernate的以下代码:
if ( String.class.isInstance( value ) ) {
final String str = (String) value;
return Character.valueOf( str.charAt(0) ); // this fails, as there is no char at position 0
}
a post on the hibernate forums证实了这一点。
我们没有选择从这个错误的版本升级hibernate,并寻找一种方法来改变Hibernate的映射。
我们不能使用PreparedStatements或普通的JDBC-Connections,也不能使用JPA-Entities。
也无法更改旧数据库。 SQL语句使用DBVisualizer完美运行。
是否可以改变Hibernate映射字符串的方式?
答案 0 :(得分:6)
East peasy!
事实上,因为这是一个非常好的问题,我决定写an article to demonstrate how easily you can achieve this goal using a custom Hibernate Type。
首先,您需要定义ImmutableType
:
public abstract class ImmutableType<T> implements UserType {
private final Class<T> clazz;
protected ImmutableType(Class<T> clazz) {
this.clazz = clazz;
}
@Override
public Object nullSafeGet(
ResultSet rs,
String[] names,
SharedSessionContractImplementor session,
Object owner)
throws SQLException {
return get(rs, names, session, owner);
}
@Override
public void nullSafeSet(
PreparedStatement st,
Object value,
int index,
SharedSessionContractImplementor session)
throws SQLException {
set(st, clazz.cast(value), index, session);
}
protected abstract T get(
ResultSet rs,
String[] names,
SharedSessionContractImplementor session,
Object owner) throws SQLException;
protected abstract void set(
PreparedStatement st,
T value,
int index,
SharedSessionContractImplementor session)
throws SQLException;
@Override
public Class<T> returnedClass() {
return clazz;
}
@Override
public boolean equals(Object x, Object y) {
return Objects.equals(x, y);
}
@Override
public int hashCode(Object x) {
return x.hashCode();
}
@Override
public Object deepCopy(Object value) {
return value;
}
@Override
public boolean isMutable() {
return false;
}
@Override
public Serializable disassemble(Object o) {
return (Serializable) o;
}
@Override
public Object assemble(
Serializable cached,
Object owner) {
return cached;
}
@Override
public Object replace(
Object o,
Object target,
Object owner) {
return o;
}
}
现在,我们可以开始定义实际的CharacterType
:
public class CharacterType
extends ImmutableType<Character> {
public CharacterType() {
super(Character.class);
}
@Override
public int[] sqlTypes() {
return new int[]{Types.CHAR};
}
@Override
public Character get(
ResultSet rs,
String[] names,
SharedSessionContractImplementor session,
Object owner)
throws SQLException {
String value = rs.getString(names[0]);
return (value != null && value.length() > 0) ?
value.charAt(0) : null;
}
@Override
public void set(
PreparedStatement st,
Character value,
int index,
SharedSessionContractImplementor session)
throws SQLException {
if (value == null) {
st.setNull(index, Types.CHAR);
} else {
st.setString(index, String.valueOf(value));
}
}
}
实体映射如下所示:
@Entity(name = "Event")
@Table(name = "event")
public class Event {
@Id
@GeneratedValue
private Long id;
@Type(type = "com.vladmihalcea.book.hpjp.hibernate.type.CharacterType")
@Column(name = "event_type")
private Character type;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public Character getType() {
return type;
}
public void setType(Character type) {
this.type = type;
}
}
让我们说我们有这些表格行:
INSERT INTO event (id, event_type) VALUES (1, 'abc');
INSERT INTO event (id, event_type) VALUES (2, '');
INSERT INTO event (id, event_type) VALUES (3, 'b');
阅读所有实体时:
doInJPA(entityManager -> {
List<Event> events = entityManager.createQuery(
"select e from Event e", Event.class)
.getResultList();
for(Event event : events) {
LOGGER.info("Event type: {}", event.getType());
}
});
您将获得预期的输出:
Event type: a
Event type:
Event type: b
查看GitHub上的源代码。
答案 1 :(得分:1)
考虑到你所拥有的所有限制,我认为最简单(但最骇客)的解决方案是重新编写SELECT查询以检查列值是否为空,在这种情况下返回一些占位符文本..或{ {1}}如果Hibernate可以处理它。