Configuring Grails/Hibernate/Postgres for UUID

Question

所以我想做一个grails应用程序挂钩到一个现有的DB，其中主键列是一个UUID:

      Column       |            Type             | Modifiers 
-------------------+-----------------------------+-----------
 uuid              | uuid                        | not null

但是，当我像这样设置数据源时:

class DataStore {
UUID     uuid
...
static mapping = {
...
     id generator: 'assigned', name: 'uuid', type: 'pg-uuid'
}

它坚持uuid列是一个varchar(255)。我不确定我要做什么才能使它识别uuid列是uuid列，我已经尝试在src/groovy/中放置UserType类，但这并没有解决任何问题。

我也试图对inet列做同样的事情，但我想一步一步来。

有什么帮助吗?我已经无计可施了。

编辑:我发现这个grails使用uuid作为id和映射到二进制列，但它只是抛出这个错误，当我现在试图运行它:

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 };
    }
}

Answer 1

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或更高版本和Hibernate 3.6(对于内置的pg-uuid类型，自定义UserType应该与以前的Hibernate版本一起工作)

刚刚发现你不需要一个自定义的UserType来映射java.util.UUID到PostgreSQL的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(参见Hibernate 3.6文档中关于基本类型的介绍)。这个内置类型应该完全执行自定义UserType可能执行的操作。

但是我认为你必须使用uuid2生成器而不是uuid生成器。请参阅Hibernate 3.6文档中有关生成器的内容。uuid生成器创建uuid的字符串表示形式，而uuid2生成器能够生成java.util.UUID、java.lang.String或长度为16 (byte[16])的字节数组。但是，我不知道如何配置uid2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。

所以，我认为你必须把你的代码改成:

class DataStore {
UUID     uuid
...
static mapping = {
...
  id generator: 'assigned', name: 'uuid2', type: 'pg-uuid'
}