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得到org.hibernate.MappingException异常:没有JDBC类型1111的方言映射,使用弗拉德米哈尔恰:休眠类型-5

hibernate-types spring-data-jpa jpa hibernate java

Drew Wills · 技术社区 · 7 年前

我正试图使用此库在JPA&Hibernate中持久化JSON:

https://github.com/vladmihalcea/hibernate-types

我下面(每一个有意义的细节)是这篇文章的例子:

https://vladmihalcea.com/how-to-store-schema-less-eav-entity-attribute-value-data-using-json-and-hibernate/

这是我(Gradle)的依赖:

compile 'com.vladmihalcea:hibernate-types-5:2.2.2'

下面是我的实体类的源代码:

package org.apereo.portal.fbms.data;

import com.fasterxml.jackson.databind.JsonNode;
import com.vladmihalcea.hibernate.type.json.JsonStringType;
import org.hibernate.annotations.Type;
import org.hibernate.annotations.TypeDef;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;

@Entity
@Table(name = "FBMS_FORM")
@TypeDef(
        name = "json",
        typeClass = JsonStringType.class
)
public class FormEntity {

    @Id
    @GeneratedValue
    private Long id;

    @Column(name = "FNAME", nullable = false, unique = true)
    private String fname; // TODO:  Regex-based validator

    @Column(name = "VERSION", nullable = false)
    private int version;

    @Column(name = "SCHEMA", nullable = false)
    @Type(type = "json")
    private JsonNode schema;

    @Column(name = "METADATA")
    @Type(type = "json")
    private JsonNode metadata;

    public Long getId() {
        return id;
    }

    // Additional getters and setters omitted for brevity...

}

我有个例外:

Caused by: javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory; nested exception is org.hibernate.MappingException: No Dialect mapping for JDBC type: 1111
        at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:396) ~[spring-orm-4.3.16.RELEASE.jar:4.3.16.RELEASE]
        at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:371) ~[spring-orm-4.3.16.RELEASE.jar:4.3.16.RELEASE]
        at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.afterPropertiesSet(LocalContainerEntityManagerFactoryBean.java:336) ~[spring-orm-4.3.16.RELEASE.jar:4.3.16.RELEASE]
        at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1688) ~[spring-beans-4.3.16.RELEASE.jar:4.3.16.RELEASE]
        at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1626) ~[spring-beans-4.3.16.RELEASE.jar:4.3.16.RELEASE]
        ... 16 common frames omitted
Caused by: org.hibernate.MappingException: No Dialect mapping for JDBC type: 1111
        at org.hibernate.dialect.TypeNames.get(TypeNames.java:70) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.dialect.TypeNames.get(TypeNames.java:101) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.dialect.Dialect.getTypeName(Dialect.java:341) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.mapping.Column.getSqlType(Column.java:231) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.mapping.Table.sqlAlterStrings(Table.java:464) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.tool.schema.internal.SchemaMigratorImpl.migrateTable(SchemaMigratorImpl.java:254) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.tool.schema.internal.SchemaMigratorImpl.doMigrationToTargets(SchemaMigratorImpl.java:170) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.tool.schema.internal.SchemaMigratorImpl.doMigration(SchemaMigratorImpl.java:60) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.tool.hbm2ddl.SchemaUpdate.execute(SchemaUpdate.java:134) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.tool.hbm2ddl.SchemaUpdate.execute(SchemaUpdate.java:101) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:472) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:444) ~[hibernate-core-5.0.12.Final.jar:5.0.12.Final]
        at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:879) ~[hibernate-entitymanager-5.0.12.Final.jar:5.0.12.Final]
        at org.springframework.orm.jpa.vendor.SpringHibernateJpaPersistenceProvider.createContainerEntityManagerFactory(SpringHibernateJpaPersistenceProvider.java:60) ~[spring-orm-4.3.16.RELEASE.jar:4.3.16.RELEASE]
        at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:360) ~[spring-orm-4.3.16.RELEASE.jar:4.3.16.RELEASE]
        at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:384) ~[spring-orm-4.3.16.RELEASE.jar:4.3.16.RELEASE]
        ... 20 common frames omitted

如果有帮助的话。。。

我没有包含关于RDBMS平台或JDBC驱动程序的信息;这是有意的——解决方案需要独立于数据库平台。
我将对这个实体使用spring data jpa。

1 回复 | 直到 6 年前

GROX13 6 年前

解决这个问题的方法,在我的例子中,就是添加自定义的hibernate方言。 This 资源可能会有帮助。创建自定义方言类:

// CustomPostgreSQLDialect.java
public class CustomPostgreSQLDialect extends PostgreSQL10Dialect {

    public CustomPostgreSQLDialect() {
        super();
        registerHibernateType(Types.OTHER, StringArrayType.class.getName());
        registerHibernateType(Types.OTHER, IntArrayType.class.getName());
        registerHibernateType(Types.OTHER, JsonStringType.class.getName());
        registerHibernateType(Types.OTHER, JsonBinaryType.class.getName());
        registerHibernateType(Types.OTHER, JsonNodeBinaryType.class.getName());
        registerHibernateType(Types.OTHER, JsonNodeStringType.class.getName());
    }
}

需要在属性中注册您的自定义方言

# application.yml
spring:
  jpa:
    properties:
      hibernate:
        dialect: "com.test.CustomPostgreSQLDialect"

如果你需要提供 CustomPostgreSQLDialect 通过 hibernate.dialect 配置属性:

<property
    name="hibernate.dialect"
    value="com.test.CustomPostgreSQLDialect"
/>

Vlad Mihalcea 6 年前

根据触发此异常的操作,您有不同的修复方法。

在您的例子中,您在引导期间遇到了问题,这意味着Hibernate方言没有 1111 JDBC列类型,代表 Types.OTHER . 因此,您需要使用当前方言进行注册,如下所示:

public class PostgreSQL95JsonDialect
        extends PostgreSQL95Dialect {

    public PostgreSQL95JsonDialect() {
        super();
        this.registerHibernateType(
            Types.OTHER, JsonNodeBinaryType.class.getName()
        );
    }
}

Ans使用定制的Hibernate方言:

<property
    name="hibernate.dialect"
    value="com.vladmihalcea.book.hpjp.hibernate.type.json.PostgreSQL95JsonDialect"
/>

如果在执行SQL本机查询时遇到此异常,则需要通过 addScalar :

JsonNode properties = (JsonNode) entityManager
.createNativeQuery(
    "SELECT properties " +
    "FROM book " +
    "WHERE isbn = :isbn")
.setParameter("isbn", "978-9730228236")
.unwrap(org.hibernate.query.NativeQuery.class)
.addScalar("properties", JsonNodeBinaryType.INSTANCE)
.getSingleResult();

assertEquals(
    "High-Performance Java Persistence",
    properties.get("title").asText()
);

有关此主题的详细信息,请查看 this article 也。

-1

Cepr0 7 年前

我认为Vlad的库中有一个bug-当我从 json 到 text ( @Column(columnDefinition = "text") )然后一切顺利。

但我认为你应该 use jsonb 而不是 json格式 . 对于“jsonb”,所有操作都按预期进行:

@Column(name = "METADATA", columnDefinition = "jsonb")
@Type(type = "jsonb")
private JsonNode metadata;

查看我的演示: sb-hibernate-types-demo

更新

一种解决方法-使用自定义的Hibernate方言,在这里我注册了列类型 clob 对于代码 Types.OTHER (1111),一切正常。测试时间 H2 和 HSQLD 数据库。