How to map PostgreSQL enum with JPA and Hibernate

Question

I am trying to map a postgres custom type,named transmission_result, to a Hibernate/JPA POJO. The postgres custom type is more or less an enum type of string values.

I have created a custom EnumUserType called PGEnumUserType as well as a enum class representing the postgres enumerated values. When I run this against a real database, I receive the following error: '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 '

Upon seeing this, I figured I needed to change my SqlTypes to Types.OTHER. But doing so breaks my integration tests (using HyperSQL in memory DB) with the message : '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"=?]'

I'm not sure why changing the sqlType results in this error. Any help is appreciated.

JPA/Hibernate Entity:

@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;
}
}

Custom 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();
}
}

Enum class:

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

Answer 1

I figured it out. I needed to use setObject instead of setString in the nullSafeSet function and pass in the Types.OTHER as the java.sql.type to let jdbc know that it was a postgres type.

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 2

If you have following post_status_info enum type in PostgreSQL:

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

You can easily map Java Enum to a PostgreSQL Enum column type using the following custom Hibernate Type:

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 
            );
        }
    }
}

To use it, you need to annotate the field with the Hibernate @Type annotation as illustrated in the following example:

@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
}

That's it, it works like a charm. Here's a test on GitHub that proves it .

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)而不需要@Type 。

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

Answer 4

A quick solution will be

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

?stringtype=unspecified is the answer

Answer 5

build.gradle.kts

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

In Kotlin it is important to make a generic extension with EnumType<Enum<*>>()

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
}

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A simpler option that I don't recommend is the first option in Arthur's answer which adds a parameter in the connection URL to the db so that the enum data type is not lost. I believe that the responsibility of mapping the data type between the backend server and the database is precisely the backend.

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

Source

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