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java - How to write in a SqlDescriptor with a jdbc prepared statement into a tsrange column?

问题描述

I followed the excellent guide How to map Java and SQL arrays with JPA and Hibernate to map the special sql type tsrange to hibernate. I decided to use Java and SQL descriptors and not a user type, because the jdbc sql handling should be better.

When I try to persist an entity with a column named time range and the type tsrange, I got always: ERROR [org.hibernate.engine.jdbc.spi.SqlExceptionHelper] (default task-2) ERROR: column "time_range" is of type tsrange but expression is of type character varying Hinweis: You will need to rewrite or cast the expression.

My understanding is, that I need to write a special/non standard sql type with the setObject method and the type Type.OTHER or Type.JAVA_OBJECT. What is the prefered way, to put a range sql type into a PreparedStatement?

The BasicBinder source, where I fill the jdbc prepared statement, sqlString contains "[2019-01-14 13:06:26, 2019-01-14 13:12:39]":

@Override
public <X> ValueBinder<X> getBinder(JavaTypeDescriptor<X> javaTypeDescriptor) {
    return new BasicBinder<X>(javaTypeDescriptor, this) {

        @Override
        protected void doBind(PreparedStatement st, X value, int index, WrapperOptions options) throws SQLException {
            String sqlString = javaTypeDescriptor.toString(value);
            // I tried the following statements:
            //st.setObject(index, sqlString, getSqlType());
            //st.setObject(index, sqlString);
            //st.setString(index, sqlString+"::tsrange");
            st.setString(index, sqlString);
        }

        @Override
        protected void doBind(CallableStatement st, X value, String name, WrapperOptions options) throws SQLException {
            st.setObject(name, javaTypeDescriptor.toString(value));
        }
    };
}

Here are the column definition of the entity:

@Id
@Column(name = "id")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;

@Basic
@Column(nullable = false, name = "time_range", columnDefinition = "tsrange")
@Type(type="com.example.model.types.TsRange")
private PgTsRange timeRange;

标签： javapostgresqlhibernatejdbc

解决方案

Hibernate Types 项目提供了一个PostgreSQLRangeType支持 PostgreSQLTSRANGE列类型。

您需要做的第一件事是使用以下 Maven 依赖项：

<dependency>
    <groupId>com.vladmihalcea</groupId>
    <artifactId>hibernate-types-52</artifactId>
    <version>${hibernate-types.version}</version>
</dependency>

之后，您可以像这样映射您的 PostgreSQL 范围：

@Entity(name = "Book")
@Table(name = "book")
@TypeDef(
    typeClass = PostgreSQLRangeType.class,
    defaultForType = Range.class
)
public class Book {
 
    @Id
    @GeneratedValue
    private Long id;
 
    @NaturalId
    private String isbn;
 
    private String title;
 
    @Column(
        name = "price_cent_range",
        columnDefinition = "numrange"
    )
    private Range<BigDecimal> priceRange;
 
    @Column(
        name = "discount_date_range",
        columnDefinition = "daterange"
    )
    private Range<LocalDate> discountDateRange;
 
    //Getters and setters omitted for brevity
}

和Range类PostgreSQLRangeType来自 Hibernate Types 项目。

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