所以我正在尝试创建一个grails应用程序，该应用程序挂钩到主键列为UUID的现有数据库中：
Column | Type | Modifiers
-------------------+-----------------------------+-----------
uuid | uuid | not null
但是，当我设置这样的数据源时：
class DataStore {
UUID uuid
...
static mapping = {
...
id generator: 'assigned', name: 'uuid', type: 'pg-uuid'
}
它坚持认为uuid列是varchar（255）。我不知道我要做些什么才能让它认识到uuid列是一个uuid列，并且我已经尝试将一个UserType类放在src / groovy /中，但是这并没有解决任何问题。
我也试图用inet列做同样的事情，但我在这里一步一步。
这里有什么帮助吗？我现在已经结束了。
编辑：我发现了这个grails using uuid as id and mapping to to binary column，但是当我尝试现在运行它时它只会抛出此错误：
Caused by: org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'transactionManager': Cannot resolve reference to bean 'sessionFactory' while setting bean property 'sessionFactory'; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'sessionFactory': Invocation of init method failed; nested exception is org.codehaus.groovy.grails.exceptions.GrailsDomainException: Error evaluating ORM mappings block for domain [cstools.domain.DataStore]: No such property: UUIDUserType for class: org.codehaus.groovy.grails.orm.hibernate.cfg.HibernateMappingBuilder
... 23 more
UUIDUserType.groovy
import java.io.Serializable;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Types;
import java.util.UUID;
import org.hibernate.HibernateException;
public class UUIDUserType implements org.hibernate.usertype.UserType {
private static final String CAST_EXCEPTION_TEXT = " cannot be cast to a java.util.UUID.";
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#assemble(java.io.Serializable,
* java.lang.Object)
*/
public Object assemble( Serializable cached, Object owner ) throws HibernateException {
if ( !String.class.isAssignableFrom( cached.getClass() ) ) {
return null;
}
return UUID.fromString( (String) cached );
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#deepCopy(java.lang.Object)
*/
public Object deepCopy( Object value ) throws HibernateException {
if ( !UUID.class.isAssignableFrom( value.getClass() ) ) {
throw new HibernateException( value.getClass().toString() + CAST_EXCEPTION_TEXT );
}
UUID other = (UUID) value;
return UUID.fromString( other.toString() );
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#disassemble(java.lang.Object)
*/
public Serializable disassemble( Object value ) throws HibernateException {
return value.toString();
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#equals(java.lang.Object,
* java.lang.Object)
*/
public boolean equals( Object x, Object y ) throws HibernateException {
if ( !UUID.class.isAssignableFrom( x.getClass() ) ) {
throw new HibernateException( x.getClass().toString() + CAST_EXCEPTION_TEXT );
}
else if ( !UUID.class.isAssignableFrom( y.getClass() ) ) {
throw new HibernateException( y.getClass().toString() + CAST_EXCEPTION_TEXT );
}
UUID a = (UUID) x;
UUID b = (UUID) y;
return a.equals( b );
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#hashCode(java.lang.Object)
*/
public int hashCode( Object x ) throws HibernateException {
if ( !UUID.class.isAssignableFrom( x.getClass() ) ) {
throw new HibernateException( x.getClass().toString() + CAST_EXCEPTION_TEXT );
}
UUID a = (UUID) x;
return a.hashCode();
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#isMutable()
*/
public boolean isMutable() {
return false;
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#nullSafeGet(java.sql.ResultSet,
* java.lang.String[], java.lang.Object)
*/
public Object nullSafeGet( ResultSet rs, String[] names, Object owner ) throws HibernateException, SQLException {
String value = rs.getString( names[0] );
if ( value == null ) {
return null;
}
else {
return UUID.fromString( value );
}
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#nullSafeSet(java.sql.PreparedStatement,
* java.lang.Object, int)
*/
public void nullSafeSet( PreparedStatement st, Object value, int index ) throws HibernateException, SQLException {
if ( value == null ) {
st.setNull( index, Types.VARCHAR );
return;
}
if ( !UUID.class.isAssignableFrom( value.getClass() ) ) {
throw new HibernateException( value.getClass().toString() + CAST_EXCEPTION_TEXT );
}
st.setString( index, value.toString() );
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#replace(java.lang.Object,
* java.lang.Object, java.lang.Object)
*/
public Object replace( Object original, Object target, Object owner ) throws HibernateException {
if ( !UUID.class.isAssignableFrom( original.getClass() ) ) {
throw new HibernateException( original.getClass().toString() + CAST_EXCEPTION_TEXT );
}
return UUID.fromString( original.toString() );
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#returnedClass()
*/
@SuppressWarnings( "unchecked" )
public Class returnedClass() {
return UUID.class;
}
/*
* (non-Javadoc)
*
* @see org.hibernate.usertype.UserType#sqlTypes()
*/
public int[] sqlTypes() {
return int[] { Types.CHAR };
}
}
答案 0 :(得分：5)
编辑：此解决方案需要PostgreSQL 8.4或更高版本，相应的JDBC驱动程序8.4-703或更高版本以及Hibernate 3.6（对于内置的 pg-uuid 类型，自定义UserType应该使用先前的Hibernate版本）
刚刚发现您不需要自定义UserType将java.util.UUID映射到PostgreSQL uuid。从Hibernate 3.6开始，您可以使用内置类型pg-uuid，这是org.hibernate.type.PostgresUUIDType的快捷方式（请参阅Hibernate 3.6 docs about basic types）。这种内置类型应该完全按照自定义UserType执行。
但我认为你必须使用 uuid2 生成器而不是 uuid 生成器。请参阅Hibernate 3.6 docs about generators。 uuid生成器创建UUID的String表示，而uuid2生成器能够生成值java.util.UUID，java.lang.String或长度为16的字节数组（byte [16]）。不过，我不知道如何配置uuid2生成器。我不使用这个生成器，只是在我的Entity基类的构造函数中分配一个随机的UUID。
因此，我认为您必须将代码更改为以下内容：
class DataStore {
UUID uuid
...
static mapping = {
...
id generator: 'assigned', name: 'uuid2', type: 'pg-uuid'
}