如何使用JPA和Hibernate映射PostgreSQL枚举

Question

我正在尝试将名为transmission_result的postgres自定义类型映射到Hibernate/JPA POJO.postgres自定义类型或多或少是枚举类型的字符串值.

我创建了一个名为PGEnumUserType的自定义EnumUserType以及一个表示postgres枚举值的枚举类.当我对一个真正的数据库运行时,我收到以下错误:'错误:列"状态"的类型为transmission_result,但表达式的类型为字符变化提示:您需要重写或转换表达式.位置:135'

看到这个,我想我需要将我的SqlTypes更改为Types.OTHER.但这样做会破坏我的集成测试(在内存数据库中使用HyperSQL)和消息:'引起:java.sql.SQLException:在语句中找不到表[select enrollment0 _."id"as id1_47_0_,enrollment0 _."tpa_approval_id"as tpa2_47_0_ ,enrollment0 _."tpa_status_code"as tpa3_47_0_,enrollment0 _."status_message"as status4_47_0_,enrollment0 _."approval_id"as approval5_47_0_,enrollment0 _."transmission_date"as transmis6_47_0_,enrollment0 _."status"as status7_47_0_,enrollment0 _."transmit"as transmit8_47_0_ from"传输"enrollment0_ where enrollment0 _."id"=？]'

我不确定为什么更改sqlType会导致此错误.任何帮助表示赞赏.

JPA/Hibernate实体:

@Entity
@Access(javax.persistence.AccessType.PROPERTY)
@Table(name="transmissions")
public class EnrollmentCycleTransmission {

// elements of enum status column
private static final String ACCEPTED_TRANSMISSION = "accepted";
private static final String REJECTED_TRANSMISSION = "rejected";
private static final String DUPLICATE_TRANSMISSION = "duplicate";
private static final String EXCEPTION_TRANSMISSION = "exception";
private static final String RETRY_TRANSMISSION = "retry";

private Long transmissionID;
private Long approvalID;
private Long transmitterID;
private TransmissionStatusType transmissionStatus;
private Date transmissionDate;
private String TPAApprovalID;
private String TPAStatusCode;
private String TPAStatusMessage;


@Column(name = "id")
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
public Long getTransmissionID() {
    return transmissionID;
}

public void setTransmissionID(Long transmissionID) {
    this.transmissionID = transmissionID;
}

@Column(name = "approval_id")
public Long getApprovalID() {
    return approvalID;
}

public void setApprovalID(Long approvalID) {
    this.approvalID = approvalID;
}

@Column(name = "transmitter")
public Long getTransmitterID() {
    return transmitterID;
}

public void setTransmitterID(Long transmitterID) {
    this.transmitterID = transmitterID;
}

@Column(name = "status")
@Type(type = "org.fuwt.model.PGEnumUserType" , parameters ={@org.hibernate.annotations.Parameter(name = "enumClassName",value = "org.fuwt.model.enrollment.TransmissionStatusType")} )
public TransmissionStatusType getTransmissionStatus() {
    return this.transmissionStatus ;
}

public void setTransmissionStatus(TransmissionStatusType transmissionStatus) {
    this.transmissionStatus = transmissionStatus;
}

@Column(name = "transmission_date")
public Date getTransmissionDate() {
    return transmissionDate;
}

public void setTransmissionDate(Date transmissionDate) {
    this.transmissionDate = transmissionDate;
}

@Column(name = "tpa_approval_id")
public String getTPAApprovalID() {
    return TPAApprovalID;
}

public void setTPAApprovalID(String TPAApprovalID) {
    this.TPAApprovalID = TPAApprovalID;
}

@Column(name = "tpa_status_code")
public String getTPAStatusCode() {
    return TPAStatusCode;
}

public void setTPAStatusCode(String TPAStatusCode) {
    this.TPAStatusCode = TPAStatusCode;
}

@Column(name = "status_message")
public String getTPAStatusMessage() {
    return TPAStatusMessage;
}

public void setTPAStatusMessage(String TPAStatusMessage) {
    this.TPAStatusMessage = TPAStatusMessage;
}
}

自定义EnumUserType:

public class PGEnumUserType implements UserType, ParameterizedType {

private Class<Enum> enumClass;

public PGEnumUserType(){
    super();
}

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

}

public int[] sqlTypes() {
    return new int[] {Types.VARCHAR};
}

public Class returnedClass() {
    return enumClass;
}

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

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

public Object nullSafeGet(ResultSet rs, String[] names, Object owner) throws HibernateException, SQLException {
    String name = rs.getString(names[0]);
    return rs.wasNull() ? null: Enum.valueOf(enumClass,name);
}

public void nullSafeSet(PreparedStatement st, Object value, int index) throws HibernateException, SQLException {
    if (value == null) {
        st.setNull(index, Types.VARCHAR);
    }
    else {
        st.setString(index,((Enum) value).name());
    }
}

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

public boolean isMutable() {
    return false;  //To change body of implemented methods use File | Settings | File Templates.
}

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

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

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

public Object fromXMLString(String xmlValue) {
    return Enum.valueOf(enumClass, xmlValue);
}

public String objectToSQLString(Object value) {
    return '\'' + ( (Enum) value ).name() + '\'';
}

public String toXMLString(Object value) {
    return ( (Enum) value ).name();
}
}

枚举类:

public enum TransmissionStatusType {
accepted,
rejected,
duplicate,
exception,
retry}

Answer 1

如果您post_status_info在 PostgreSQL 中有以下枚举类型：

CREATE TYPE post_status_info AS ENUM (
    'PENDING', 
    'APPROVED', 
    'SPAM'
)

您可以使用以下自定义 Hibernate Type 轻松将 Java Enum 映射到 PostgreSQL Enum 列类型：

public class PostgreSQLEnumType extends org.hibernate.type.EnumType {
     
    public void nullSafeSet(
            PreparedStatement st, 
            Object value, 
            int index, 
            SharedSessionContractImplementor session) 
        throws HibernateException, SQLException {
        if(value == null) {
            st.setNull( index, Types.OTHER );
        }
        else {
            st.setObject( 
                index, 
                value.toString(), 
                Types.OTHER 
            );
        }
    }
}

要使用它，您需要使用 Hibernate@Type注释来注释该字段，如下例所示：

@Entity(name = "Post")
@Table(name = "post")
@TypeDef(
    name = "pgsql_enum",
    typeClass = PostgreSQLEnumType.class
)
public static class Post {
 
    @Id
    private Long id;
 
    private String title;
 
    @Enumerated(EnumType.STRING)
    @Column(columnDefinition = "post_status_info")
    @Type( type = "pgsql_enum" )
    private PostStatus status;
 
    //Getters and setters omitted for brevity
}

就是这样，它就像一个魅力。这是GitHub 上的一个测试，证明了这一点。

Answer 2

我想到了.我需要在nullSafeSet函数中使用setObject而不是setString,并将Types.OTHER作为java.sql.type传入,让jdbc知道它是postgres类型.

public void nullSafeSet(PreparedStatement st, Object value, int index) throws HibernateException, SQLException {
    if (value == null) {
        st.setNull(index, Types.VARCHAR);
    }
    else {
//            previously used setString, but this causes postgresql to bark about incompatible types.
//           now using setObject passing in the java type for the postgres enum object
//            st.setString(index,((Enum) value).name());
        st.setObject(index,((Enum) value), Types.OTHER);
    }
}

Answer 3

以下内容也可能有助于让 Postgres 以静默方式将字符串转换为您的 SQL 枚举类型，以便您可以使用@Enumerated(STRING)并且不需要@Type.

CREATE CAST (character varying as post_status_type) WITH INOUT AS IMPLICIT;