Hibernate,如何获得注册的自定义类型

Question

显然，可以创建一个TypeDef来根据方言切换实现。

package org.hibernate.type;

import org.hibernate.type.descriptor.java.JavaTypeDescriptor;
import org.hibernate.type.descriptor.java.UUIDTypeDescriptor;
import org.hibernate.type.descriptor.sql.SqlTypeDescriptor;
import org.hibernate.type.descriptor.sql.VarcharTypeDescriptor;

import java.io.IOException;
import java.util.Properties;

public class UUIDCustomType extends AbstractSingleColumnStandardBasicType {

private static final long serialVersionUID = 902830399800029445L;

private static final SqlTypeDescriptor  SQL_DESCRIPTOR;
private static final JavaTypeDescriptor TYPE_DESCRIPTOR;

static {
    Properties properties = new Properties();
    try {
        ClassLoader loader = Thread.currentThread().getContextClassLoader();
        properties.load( loader.getResourceAsStream( "database.properties" ) );
    }
    catch ( IOException e ) {
        throw new RuntimeException( "Could not load properties!", e );
    }

    String dialect = properties.getProperty( "dialect" );
    if ( dialect.equals( "org.hibernate.dialect.PostgreSQLDialect" ) ) {
        SQL_DESCRIPTOR = PostgresUUIDType.PostgresUUIDSqlTypeDescriptor.INSTANCE;
    } else if(dialect.equals("org.hibernate.dialect.H2Dialect")) {
        SQL_DESCRIPTOR = VarcharTypeDescriptor.INSTANCE;
    } else {
        throw new UnsupportedOperationException("Unsupported database!");
    }

    TYPE_DESCRIPTOR = UUIDTypeDescriptor.INSTANCE;
}

public UUIDCustomType() {
    super(SQL_DESCRIPTOR, TYPE_DESCRIPTOR);
}

@Override
public String getName() {
    return "uuid-custom";
}

}

我的问题是hibernate似乎没有识别它，值得注意的是，有一次我将“uuid-custom”作为类型中的静态字符串，并直接在@type中引用它，所以它并不是实际上不在类路径上。

Answer 1

我不确定是否有其他方法可以让它工作，但是将它添加到包中的typedef中-info.java可以解决这个问题

@TypeDef(
    name = UUIDCustomType.UUID,
    defaultForType = UUID.class,
    typeClass = UUIDCustomType.class
)
package com.xenoterracide.rpf.model;

import org.hibernate.annotations.TypeDef;
import org.hibernate.type.UUIDCustomType;

import java.util.UUID;