Question

我正在尝试将grails-project转换为playframework。在Grails中，您可以定义一个id，以便将id存储在数据库中（请参阅release notes中的增强的枚举支持）。我看到similar question，但没有可接受的解决方案。如果我更改CRUD模块的类型会出现问题，因为应该显示Enum的信息会丢失。

所以我想知道是否存在一个很好的解决方案，它基于Hibernate。也许是在破解JPAPlugin？

[更新1]我开始尝试使用@type - 注释的第二个解决方案。不幸的是，这已经被hibernate 3.6（播放1.2.2使用）破坏了。 TypeFactory.basic()不再是available。但是根据文档我无法找到解决方法。

[更新2] hibernate 3.6.1有solution，但在每次使用枚举时定义类型真的很笨拙。

@Type(type="hibernatehelper.GenericEnumUserType", 
            parameters= {
            @Parameter(
                    name  = "enumClass",                      
                    value = "models.Geschlecht"),
        })
public Geschlecht geschlecht = Geschlecht.WEIBLICH;

Answer 1

不确定它是否真的有效，但可能的解决办法如下：

写了一个泛型类型映射器：

package hibernatehelper;

import java.io.Serializable;
import java.lang.reflect.Method;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.util.Properties;
import org.hibernate.HibernateException;
import org.hibernate.type.AbstractSingleColumnStandardBasicType;
import org.hibernate.type.TypeResolver;
import org.hibernate.usertype.ParameterizedType;
import org.hibernate.usertype.UserType;


public class GenericEnumUserType implements UserType, ParameterizedType {

    private Class <? extends Enum> enumClass;

    private Class <?> identifierType;

    private Method identifierMethod;

    private Method valueOfMethod;

    private static final String defaultIdentifierMethodName = "getId";

    private static final String defaultValueOfMethodName = "parseId";

    private AbstractSingleColumnStandardBasicType type;

    private int[] sqlTypes;

    @Override
    public void setParameterValues(Properties parameters) {
        String enumClassName = parameters.getProperty("enumClass");
        try {
            enumClass = Class.forName(enumClassName).asSubclass(Enum.class);
        } catch (ClassNotFoundException exception) {
            throw new HibernateException("Enum class not found", exception);
        }

        String identifierMethodName =
                parameters.getProperty("identifierMethod",
                        defaultIdentifierMethodName);

        try {
            identifierMethod =
                    enumClass.getMethod(identifierMethodName, new Class[0]);
            identifierType = identifierMethod.getReturnType();
        } catch (Exception exception) {
            throw new HibernateException("Failed to optain identifier method",
                    exception);
        }

        TypeResolver tr = new TypeResolver();
        type =
                (AbstractSingleColumnStandardBasicType) tr.basic(identifierType
                        .getName());
        if (type == null) {
            throw new HibernateException("Unsupported identifier type "
                    + identifierType.getName());
        }
        sqlTypes = new int[] {type.sqlType()};

        String valueOfMethodName = parameters.getProperty("valueOfMethod",
                defaultValueOfMethodName);
        try {
            valueOfMethod = enumClass.getMethod(valueOfMethodName,
                            new Class[] {identifierType});
        } catch (Exception exception) {
            throw new HibernateException("Failed to optain valueOf method",
                    exception);
        }
    }

    @Override
    public Class returnedClass() {
        return enumClass;
    }

    @Override
    public Object nullSafeGet(ResultSet rs, String[] names, Object owner)
            throws HibernateException, SQLException {
        Object identifier = type.get(rs, names[0]);
        if (identifier == null) {
            return null;
        }

        if (valueOfMethod == null) {

        }

        try {
            return valueOfMethod.invoke(enumClass, new Object[] {identifier});
        } catch (Exception exception) {
            throw new HibernateException(
                    "Exception while invoking valueOfMethod of enumeration class: ",
                    exception);
        }
    }

    public void nullSafeSet(PreparedStatement st, Object value, int index)
            throws HibernateException, SQLException {
        try {
            Object identifier =
                    value != null ? identifierMethod.invoke(value,
                            new Object[0]) : null;
            st.setObject(index, identifier);
        } catch (Exception exception) {
            throw new HibernateException(
                    "Exception while invoking identifierMethod of enumeration class: ",
                    exception);

        }
    }

    @Override
    public int[] sqlTypes() {
        return sqlTypes;
    }

    @Override
    public Object assemble(Serializable cached, Object owner)
            throws HibernateException {
        return cached;
    }

    @Override
    public Object deepCopy(Object value) throws HibernateException {
        return value;
    }

    @Override
    public Serializable disassemble(Object value) throws HibernateException {
        return (Serializable) value;
    }

    @Override
    public boolean equals(Object x, Object y) throws HibernateException {
        return x == y;
    }

    @Override
    public int hashCode(Object x) throws HibernateException {
        return x.hashCode();
    }

    public boolean isMutable() {
        return false;
    }

    public Object replace(Object original, Object target, Object owner)
            throws HibernateException {
        return original;
    }
}

编写了一个Enhancer，它查找来自类型Enum的每个属性，并查看此类型是否具有静态方法parseId。比add the following annotation with javaassist：

@Type(type="hibernatehelper.GenericEnumUserType", 
    parameters= {
    @Parameter(
            name  = "enumClass",                      
            value = "<fullqualified classname of the enum class>"),
})

但我不确定这对于这样的问题是否有多大的魔力。也许有人可以给我一个建议。

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1 个答案: