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我试图在数据库上有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，但它不识别 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;
}

我检查了，我有完全相同的 postgresql 驱动程序版本。我看到了这篇文章：Java enum with Eclipselink。这是一个我也会尝试的解决方案，但我的主要问题是：显然它是同一类，为什么它没有被识别为这样？我可以有两个具有相同名称和包的不同类吗？如果我仍然必须在 Postgres 中使用枚举，我该如何修复它以正确映射到我的 Java 枚举？

编辑：

我试图做一个：

PGobject pg = (PGobject) object;

它抛出一个类转换异常：

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

谢谢

4

1 回答 1

0

我使用泛型类作为映射枚举的类型。然后，您可以使用它映射所有枚举。

课程是这样的：

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;
}}

枚举类应该是这样的：

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();
    }
}}

静态方法“valueOf”是从存储在数据库中的 id 转换为 java 对象所必需的。

那么，hibernate映射是这样的：

<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>

因此，您必须使用 typedef 将 GenericEnumUserType 类声明为类型，以及获取枚举 id 的方法（在本例中为 getId()）。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。

于 2014-08-11T09:03:43.400 回答