将 PostgreSQL JSON 列映射到 Hibernate 实体属性

2022-08-31 11:31:04

我的PostgreSQL DB(9.2)中有一个带有JSON类型列的表。我很难将此列映射到 JPA2 实体字段类型。

我尝试使用String,但是当我保存实体时,我得到一个异常,它无法将字符转换为JSON。

处理 JSON 列时要使用的正确值类型是什么?

@Entity
public class MyEntity {

    private String jsonPayload; // this maps to a json column

    public MyEntity() {
    }
}

一个简单的解决方法是定义一个文本列。


答案 1

如果您有兴趣,这里有一些代码片段可以获取Hibernate自定义用户类型。首先扩展PostgreSQL方言以告诉它json类型,感谢Craig Ringer JAVA_OBJECT指针:

import org.hibernate.dialect.PostgreSQL9Dialect;

import java.sql.Types;

/**
 * Wrap default PostgreSQL9Dialect with 'json' type.
 *
 * @author timfulmer
 */
public class JsonPostgreSQLDialect extends PostgreSQL9Dialect {

    public JsonPostgreSQLDialect() {

        super();

        this.registerColumnType(Types.JAVA_OBJECT, "json");
    }
}

接下来实现 org.hibernate.usertype.UserType.下面的实现将 String 值映射到 json 数据库类型,反之亦然。请记住,字符串在 Java 中是不可变的。可以使用更复杂的实现将自定义Java Bean映射到存储在数据库中的JSON。

package foo;

import org.hibernate.HibernateException;
import org.hibernate.engine.spi.SessionImplementor;
import org.hibernate.usertype.UserType;

import java.io.Serializable;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import java.sql.Types;

/**
 * @author timfulmer
 */
public class StringJsonUserType implements UserType {

    /**
     * Return the SQL type codes for the columns mapped by this type. The
     * codes are defined on <tt>java.sql.Types</tt>.
     *
     * @return int[] the typecodes
     * @see java.sql.Types
     */
    @Override
    public int[] sqlTypes() {
        return new int[] { Types.JAVA_OBJECT};
    }

    /**
     * The class returned by <tt>nullSafeGet()</tt>.
     *
     * @return Class
     */
    @Override
    public Class returnedClass() {
        return String.class;
    }

    /**
     * Compare two instances of the class mapped by this type for persistence "equality".
     * Equality of the persistent state.
     *
     * @param x
     * @param y
     * @return boolean
     */
    @Override
    public boolean equals(Object x, Object y) throws HibernateException {

        if( x== null){

            return y== null;
        }

        return x.equals( y);
    }

    /**
     * Get a hashcode for the instance, consistent with persistence "equality"
     */
    @Override
    public int hashCode(Object x) throws HibernateException {

        return x.hashCode();
    }

    /**
     * Retrieve an instance of the mapped class from a JDBC resultset. Implementors
     * should handle possibility of null values.
     *
     * @param rs      a JDBC result set
     * @param names   the column names
     * @param session
     * @param owner   the containing entity  @return Object
     * @throws org.hibernate.HibernateException
     *
     * @throws java.sql.SQLException
     */
    @Override
    public Object nullSafeGet(ResultSet rs, String[] names, SessionImplementor session, Object owner) throws HibernateException, SQLException {
        if(rs.getString(names[0]) == null){
            return null;
        }
        return rs.getString(names[0]);
    }

    /**
     * Write an instance of the mapped class to a prepared statement. Implementors
     * should handle possibility of null values. A multi-column type should be written
     * to parameters starting from <tt>index</tt>.
     *
     * @param st      a JDBC prepared statement
     * @param value   the object to write
     * @param index   statement parameter index
     * @param session
     * @throws org.hibernate.HibernateException
     *
     * @throws java.sql.SQLException
     */
    @Override
    public void nullSafeSet(PreparedStatement st, Object value, int index, SessionImplementor session) throws HibernateException, SQLException {
        if (value == null) {
            st.setNull(index, Types.OTHER);
            return;
        }

        st.setObject(index, value, Types.OTHER);
    }

    /**
     * Return a deep copy of the persistent state, stopping at entities and at
     * collections. It is not necessary to copy immutable objects, or null
     * values, in which case it is safe to simply return the argument.
     *
     * @param value the object to be cloned, which may be null
     * @return Object a copy
     */
    @Override
    public Object deepCopy(Object value) throws HibernateException {

        return value;
    }

    /**
     * Are objects of this type mutable?
     *
     * @return boolean
     */
    @Override
    public boolean isMutable() {
        return true;
    }

    /**
     * Transform the object into its cacheable representation. At the very least this
     * method should perform a deep copy if the type is mutable. That may not be enough
     * for some implementations, however; for example, associations must be cached as
     * identifier values. (optional operation)
     *
     * @param value the object to be cached
     * @return a cachable representation of the object
     * @throws org.hibernate.HibernateException
     *
     */
    @Override
    public Serializable disassemble(Object value) throws HibernateException {
        return (String)this.deepCopy( value);
    }

    /**
     * Reconstruct an object from the cacheable representation. At the very least this
     * method should perform a deep copy if the type is mutable. (optional operation)
     *
     * @param cached the object to be cached
     * @param owner  the owner of the cached object
     * @return a reconstructed object from the cachable representation
     * @throws org.hibernate.HibernateException
     *
     */
    @Override
    public Object assemble(Serializable cached, Object owner) throws HibernateException {
        return this.deepCopy( cached);
    }

    /**
     * During merge, replace the existing (target) value in the entity we are merging to
     * with a new (original) value from the detached entity we are merging. For immutable
     * objects, or null values, it is safe to simply return the first parameter. For
     * mutable objects, it is safe to return a copy of the first parameter. For objects
     * with component values, it might make sense to recursively replace component values.
     *
     * @param original the value from the detached entity being merged
     * @param target   the value in the managed entity
     * @return the value to be merged
     */
    @Override
    public Object replace(Object original, Object target, Object owner) throws HibernateException {
        return original;
    }
}

现在剩下的就是注释实体。在实体的类声明中放置类似下面的内容:

@TypeDefs( {@TypeDef( name= "StringJsonObject", typeClass = StringJsonUserType.class)})

然后对属性进行批注:

@Type(type = "StringJsonObject")
public String getBar() {
    return bar;
}

Hibernate 将负责为您创建具有 json 类型的列,并处理来回映射。将其他库注入到用户类型实现中,以实现更高级的映射。

这里有一个快速示例GitHub项目,如果有人想玩它:

https://github.com/timfulmer/hibernate-postgres-jsontype


答案 2

请参阅 PgJDBC 错误 #265

PostgreSQL对数据类型转换过于严格。令人讨厌的。它甚至不会隐式转换为类似文本的值,例如 和 。textxmljson

解决此问题的严格正确方法是编写使用 JDBC 方法的自定义休眠映射类型。这可能会很麻烦,所以你可能只想通过创建一个较弱的转换来降低PostgreSQL的严格性。setObject

正如@markdsievers在评论和这篇博客文章中指出的那样,此答案中的原始解决方案绕过了 JSON 验证。所以这不是你真正想要的。写起来更安全:

CREATE OR REPLACE FUNCTION json_intext(text) RETURNS json AS $$
SELECT json_in($1::cstring); 
$$ LANGUAGE SQL IMMUTABLE;

CREATE CAST (text AS json) WITH FUNCTION json_intext(text) AS IMPLICIT;

AS IMPLICIT告诉PostgreSQL它可以在没有明确告知的情况下进行转换,允许这样的事情工作:

regress=# CREATE TABLE jsontext(x json);
CREATE TABLE
regress=# PREPARE test(text) AS INSERT INTO jsontext(x) VALUES ($1);
PREPARE
regress=# EXECUTE test('{}')
INSERT 0 1

感谢@markdsievers指出问题。