如何使用 JPA 和 Hibernate 映射 PostgreSQL 枚举

Posted 2023-02-26

【中文标题】如何使用 JPA 和 Hibernate 映射 PostgreSQL 枚举

【英文标题】：How to map PostgreSQL enum with JPA and Hibernate

【发布时间】：2011-11-28 01:04:18

【问题描述】：

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。当我在真实数据库上运行它时，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，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）并显示以下消息：

'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。任何帮助表示赞赏。

JPA/休眠实体：

@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

【问题讨论】：

也可能是由于没有从枚举转换为 varchar！

【参考方案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 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 注释对字段进行注释，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test on GitHub that proves it。

【讨论】：

哇！非常感谢您的贡献，这节省了我的时间！ （GitHub 示例很完美）

很高兴能帮上忙。

【参考方案2】：

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，这样您就可以使用 @Enumerated(STRING) 而不需要 @Type。

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

【参考方案3】：

build.gradle.kts

dependencies 
    api("javax.persistence", "javax.persistence-api", "2.2")
    api("org.hibernate",  "hibernate-core",  "5.4.21.Final")

在 Kotlin 中，使用 EnumType<Enum<*>>() 进行通用扩展很重要

PostgreSQLEnumType.kt

import org.hibernate.type.EnumType
import java.sql.Types

class PostgreSQLEnumType : EnumType<Enum<*>>() 

    @Throws(HibernateException::class, SQLException::class)
    override fun nullSafeSet(
            st: PreparedStatement,
            value: Any,
            index: Int,
            session: SharedSessionContractImplementor) 
        st.setObject(
                index,
                value.toString(),
                Types.OTHER
        )
    

Custom.kt

import org.hibernate.annotations.Type
import org.hibernate.annotations.TypeDef
import javax.persistence.*

@Entity
@Table(name = "custom")
@TypeDef(name = "pgsql_enum", typeClass = PostgreSQLEnumType::class)
data class Custom(
        @Id @GeneratedValue @Column(name = "id")
        val id: Int,
    
        @Enumerated(EnumType.STRING) @Column(name = "status_custom") @Type(type = "pgsql_enum")
        val statusCustom: StatusCustom
)

enum class StatusCustom 
    FIRST, SECOND

---

我不推荐的一个更简单的选项是Arthur's answer 中的第一个选项，它在 db 的连接 URL 中添加一个参数，这样枚举数据类型就不会丢失。我相信后端服务器和数据库之间的数据类型映射的责任恰恰是后端。

<property name="connection.url">jdbc:postgresql://localhost:5432/yourdatabase?stringtype=unspecified</property>

Source

---

【参考方案4】：

一个快速的解决方案将是

jdbc:postgresql://localhost:5432/postgres?stringtype=unspecified

?stringtype=unspecified就是答案

【参考方案5】：

我想通了。我需要在 nullSafeSet 函数中使用 setObject 而不是 setString，并将 Types.OTHER 作为 java.sql.type 传递，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);