java - 如何使用 JPA 和 Hibernate 映射 PostgreSQL 枚举

Question

我正在尝试将名为 transmission_result 的 PostgreSQL 自定义类型映射到 Hibernate/JPA POJO。 PostgreSQL 自定义类型或多或少是一种 enum 类型的字符串值。

我创建了一个名为 PGEnumUserType 的自定义 EnumUserType 以及一个代表 PostgreSQL 枚举值的 enum 类。当我针对真实数据库运行它时，我收到以下错误:

'ERROR: column "status" is of type transmission_result but expression is of type
character varying 
  Hint: You will need to rewrite or cast the expression.
  Position: 135 '

看到这个后，我想我需要将我的 SqlTypes 更改为 Types.OTHER。但是这样做会破坏我的集成测试(在内存数据库中使用 HyperSQL)并显示消息:

'Caused by: java.sql.SQLException: Table not found in statement
[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_."transmitter" as transmit8_47_0_
 from "transmissions" enrollment0_ where enrollment0_."id"=?]'

我不确定为什么更改 sqlType 会导致此错误。感谢您的帮助。

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;
}
}

自定义枚举用户类型:

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

如果您在 PostgreSQL 中有以下 post_status_info 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 类型轻松地将 Java 枚举映射到 PostgreSQL 枚举列类型:

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 注释对该字段进行注释，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。这是一个 test on GitHub that proves it .