简体   繁体   English

使用Hibernate从Postgres枚举到Java

[英]Enum from Postgres to Java using Hibernate

I'm trying to have a enum type on the database and convert it to java, I wrote a EnumUserType class to do the conversion, but it doesn't recognize the PGobject class. 我试图在数据库上使用一个枚举类型并将其转换为java,我编写了一个EnumUserType类来进行转换,但是它无法识别PGobject类。

public Object nullSafeGet(ResultSet rs, String[] names, SessionImplementor si, Object owner)
        throws HibernateException, SQLException {
    Object object = rs.getObject(names[0]);
    if (rs.wasNull()) {
        return null;
    }

    if (object instanceof PGobject) {
        //code doesn't reach this line
    }

    log.info(object.getClass()); // prints class org.postgresql.util.PGobject
    return null;
}

I checked and I have the exact same postgresql driver version. 我检查了一下,然后找到了完全相同的postgresql驱动程序版本。 I saw this post: Java enum with Eclipselink . 我看到了这篇文章: Eclipselink的Java枚举 It is a solution that I will also try, but my main question is: apparently it is the same class, why it is not being recognized as such? 我也将尝试这种解决方案,但是我的主要问题是:显然它是同一个类,为什么不能这样识别它? Can I have two different classes with the same name and package? 我可以使用名称和软件包相同的两个不同的类吗? If I still have to use enums in Postgres, how can I fix it to properly map to my Java enum? 如果仍然需要在Postgres中使用枚举,如何解决它以正确映射到Java枚举?

EDIT: 编辑:

I tried to do a: 我试图做一个:

PGobject pg = (PGobject) object;

and it throws a class cast exception: 并引发类强制转换异常:

org.postgresql.util.PGobject cannot be cast to org.postgresql.util.PGobject

Thanks 谢谢

I use a generic class to use as a type to map a enum. 我使用泛型类作为映射枚举的类型。 Then, you can map all your enums using it. 然后,您可以使用它映射所有枚举。

The class is this: 该类是这样的:

public class GenericEnumUserType implements UserType, ParameterizedType {

private static final String DEFAULT_IDENTIFIER_METHOD_NAME = "name";
private static final String DEFAULT_VALUE_OF_METHOD_NAME = "valueOf";

private Class enumClass;
private Class identifierType;
private Method identifierMethod;
private Method valueOfMethod;
private NullableType type;
private int[] sqlTypes;

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

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

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

    type = (NullableType) TypeFactory.basic(identifierType.getName());

    if (type == null) {
        throw new HibernateException("Unsupported identifier type " + identifierType.getName());
    }

    sqlTypes = new int[] { type.sqlType() };

    String valueOfMethodName = parameters.getProperty("valueOfMethod", DEFAULT_VALUE_OF_METHOD_NAME);

    try {
        valueOfMethod = enumClass.getMethod(valueOfMethodName, new Class[] { identifierType });
    } catch (Exception e) {
        throw new HibernateException("Failed to obtain valueOf method", e);
    }
}

@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 (rs.wasNull()) {
        return null;
    }

    try {
        return valueOfMethod.invoke(enumClass, new Object[] { identifier });
    } catch (Exception e) {
        throw new HibernateException("Exception while invoking " + "valueOf method " + valueOfMethod.getName()
                + " of enumeration class " + enumClass, e);
    }
}

@Override
public void nullSafeSet(PreparedStatement st, Object value, int index) throws HibernateException, SQLException {
    try {
        if (value == null) {
            st.setNull(index, type.sqlType());
        } else {
            Object identifier = identifierMethod.invoke(value, new Object[0]);
            type.set(st, identifier, index);
        }
    } catch (Exception e) {
        throw new HibernateException("Exception while invoking identifierMethod " + identifierMethod.getName()
                + " of enumeration class " + enumClass, e);
    }
}

@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();
}

@Override
public boolean isMutable() {
    return false;
}

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

Enum class should be like this: 枚举类应如下所示:

public enum AccountStatus {
ACTIVE(1), BLOCKED(2), DELETED(3);

private AccountStatus(int id) {
    this.id = id;
}

private int id;

public int getId() {
    return id;
}

public void setId(int id) {
    this.id = id;
}

public static AccountStatus valueOf(int id) {
    switch (id) {
    case 1:
        return ACTIVE;
    case 2:
        return BLOCKED;
    case 3:
        return DELETED;
    default:
        throw new IllegalArgumentException();
    }
}}

The static method "valueOf" is necessary to convert from id stored in database to java object. 静态方法“ valueOf”是将数据库中存储的id转换为java对象所必需的。

Then, hibernate mapping is like this: 然后,休眠映射如下所示:

<hibernate-mapping>
<typedef class="path.to.GenericEnumUserType" name="accountStatusType">
    <param name="enumClassName">com.systemonenoc.hermes.ratingengine.persistence.constants.AccountStatus</param>
    <param name="identifierMethod">getId</param>
</typedef>
<class name="package.to.class.with.enum.Account" table="account" schema="public">
    <property name="accountStatus" type="accountStatusType" column="account_status" not-null="true" />
[...]
</hibernate-mapping>

So you have to declare as a type the class GenericEnumUserType with typedef, and a method to get the id of the enum (in this case, getId()). 因此,您必须将具有typedef的GenericEnumUserType类声明为类型,并声明一个用于获取枚举ID(在本例中为getId())的方法。 In your database wil be store the id as value in a integer column, and in java you will have the enum object. 在您的数据库中,将id作为值存储在整数列中,而在java中,您将拥有enum对象。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM