SpringBoot+Kotlin+Postgres 和 JSONB:“org.hibernate.MappingException:JDBC 类型没有方言映射"； [英] SpringBoot+Kotlin+Postgres and JSONB: "org.hibernate.MappingException: No Dialect mapping for JDBC type"

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本文介绍了SpringBoot+Kotlin+Postgres 和 JSONB:“org.hibernate.MappingException:JDBC 类型没有方言映射"；的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

为了处理运行 Kotlin/SpringBoot 应用程序时出现的以下错误(完整堆栈跟踪)，我一直在咨询许多方法/帖子/stackoverflow 问题:

I have been consulting a number of approaches/posts/stackoverflow questions in order to deal with the following error (full stack trace) when running a Kotlin/SpringBoot application:

2020-04-22 18:33:56.823 ERROR 46345 --- [  restartedMain] o.s.boot.SpringApplication               : Application run failed

org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [org/springframework/boot/autoconfigure/orm/jpa/HibernateJpaConfiguration.class]: Invocation of init method failed; nested exception is javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory; nested exception is org.hibernate.MappingException: No Dialect mapping for JDBC type: 2118910070
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1803)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:595)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.createBean(AbstractAutowireCapableBeanFactory.java:517)
    at org.springframework.beans.factory.support.AbstractBeanFactory.lambda$doGetBean$0(AbstractBeanFactory.java:323)
    at org.springframework.beans.factory.support.DefaultSingletonBeanRegistry.getSingleton(DefaultSingletonBeanRegistry.java:222)
    at org.springframework.beans.factory.support.AbstractBeanFactory.doGetBean(AbstractBeanFactory.java:321)
    at org.springframework.beans.factory.support.AbstractBeanFactory.getBean(AbstractBeanFactory.java:202)
    at org.springframework.context.support.AbstractApplicationContext.getBean(AbstractApplicationContext.java:1108)
    at org.springframework.context.support.AbstractApplicationContext.finishBeanFactoryInitialization(AbstractApplicationContext.java:868)
    at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:550)
    at org.springframework.boot.web.servlet.context.ServletWebServerApplicationContext.refresh(ServletWebServerApplicationContext.java:141)
    at org.springframework.boot.SpringApplication.refresh(SpringApplication.java:747)
    at org.springframework.boot.SpringApplication.refreshContext(SpringApplication.java:397)
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:315)
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:1226)
    at org.springframework.boot.SpringApplication.run(SpringApplication.java:1215)
    at app.ApplicationKt.main(Application.kt:13)
    at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at java.base/jdk.internal.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at java.base/jdk.internal.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.base/java.lang.reflect.Method.invoke(Method.java:566)
    at org.springframework.boot.devtools.restart.RestartLauncher.run(RestartLauncher.java:49)
Caused by: javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory; nested exception is org.hibernate.MappingException: No Dialect mapping for JDBC type: 2118910070
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:403)
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:378)
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.afterPropertiesSet(LocalContainerEntityManagerFactoryBean.java:341)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1862)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1799)
    ... 21 common frames omitted
Caused by: org.hibernate.MappingException: No Dialect mapping for JDBC type: 2118910070
    at org.hibernate.dialect.TypeNames.get(TypeNames.java:71)
    at org.hibernate.dialect.TypeNames.get(TypeNames.java:103)
    at org.hibernate.dialect.Dialect.getTypeName(Dialect.java:369)
    at org.hibernate.mapping.Column.getSqlType(Column.java:238)
    at org.hibernate.tool.schema.internal.AbstractSchemaValidator.validateColumnType(AbstractSchemaValidator.java:156)
    at org.hibernate.tool.schema.internal.AbstractSchemaValidator.validateTable(AbstractSchemaValidator.java:143)
    at org.hibernate.tool.schema.internal.GroupedSchemaValidatorImpl.validateTables(GroupedSchemaValidatorImpl.java:42)
    at org.hibernate.tool.schema.internal.AbstractSchemaValidator.performValidation(AbstractSchemaValidator.java:89)
    at org.hibernate.tool.schema.internal.AbstractSchemaValidator.doValidation(AbstractSchemaValidator.java:68)
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.performDatabaseAction(SchemaManagementToolCoordinator.java:192)
    at org.hibernate.tool.schema.spi.SchemaManagementToolCoordinator.process(SchemaManagementToolCoordinator.java:73)
    at org.hibernate.internal.SessionFactoryImpl.<init>(SessionFactoryImpl.java:320)
    at org.hibernate.boot.internal.SessionFactoryBuilderImpl.build(SessionFactoryBuilderImpl.java:462)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:1249)
    at org.springframework.orm.jpa.vendor.SpringHibernateJpaPersistenceProvider.createContainerEntityManagerFactory(SpringHibernateJpaPersistenceProvider.java:58)
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:365)
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:391)
    ... 25 common frames omitted

问题在于将 PostgreSQL 的 JSONB 数据类型映射到 Hibernate.

The issue lies in mapping the PostgreSQL’s JSONB data type with Hibernate.

我广泛尝试和调试的 2 种方法如下:

The 2 approaches I have extensively tried and debugged are the following:

  1. 实现自定义 Hibernate 映射并为 JSONB 创建自定义 UserType.参考:这里，这里, 此处 和 这里
  2. 使用 Hibernate 类型.参考是这里，这里 和 这里
  1. Implementing a Custom Hibernate Mapping and creating a custom UserType for JSONB. References: here, here, here and here
  2. Use Hibernate types. References are here, here and here

我在这两个方面都进行了大量尝试，但没有任何运气，我很想知道我哪里出错了，以及我遗漏了什么.

I have attempted profusely at both, but with no luck whatsoever and I am keen to understand where I am going wrong and what I have been missing.

方法一

我的实体:

@Entity
@TypeDef(name = "JsonUserType", typeClass = JsonUserType::class)
@Table(name = "entity")
data class MyEntity(
  @Column(nullable = false)
  val id: UUID,
  @Column(nullable = false)
  @Enumerated(value = EnumType.STRING)
  @Column(nullable = false)
  val type: Type,
  @Type(type = "JsonUserType")
  @Column(columnDefinition = "jsonb")
  @Basic(fetch = FetchType.LAZY)
  var event_data: Event
) : SomeEntity<UUID>(), SomeOtherStuff {
  override fun getName(): String {
    return id
  }
}
 
 
enum class Type(val value: String) {
  TYPE1("Type1"),
  TYPE2("Type2")
}

我的 POJO:

data class Event(
  val someContent: String,
  val someBoolean: Boolean
) : Serializable { //equals, hashcode etc are omitted }

我的自定义 Hibernate 方言:

My Custom Hibernate dialect:

class CustomPostgreSQLDialect : PostgreSQL95Dialect {
  constructor() : super() {
    this.registerColumnType(Types.JAVA_OBJECT, "jsonb")
  }
}

我的自定义类型(抽象类)

My custom type (abstract class)

abstract class JsonDataUserType : UserType {

  override fun sqlTypes(): IntArray? {
    return intArrayOf(Types.JAVA_OBJECT)
  }

  override fun equals(value1: Any?, value2: Any?): Boolean {
    return value1 == value2
  }

  override fun hashCode(value1: Any?): Int {
    return value1!!.hashCode()
  }

  override fun assemble(value1: Serializable?, value2: Any?): Any {
    return deepCopy(value1)
  }

  override fun disassemble(value1: Any?): Serializable {
    return deepCopy(value1) as Serializable
  }

  override fun deepCopy(p0: Any?): Any {
    return try {
      val bos = ByteArrayOutputStream()
      val oos = ObjectOutputStream(bos)
      oos.writeObject(p0)
      oos.flush()
      oos.close()
      bos.close()
      val bais = ByteArrayInputStream(bos.toByteArray())
      ObjectInputStream(bais).readObject()
    } catch (ex: ClassNotFoundException) {
      throw HibernateException(ex)
    } catch (ex: IOException) {
      throw HibernateException(ex)
    }
  }

  override fun replace(p0: Any?, p1: Any?, p2: Any?): Any {
    return deepCopy(p0)
  }

  override fun nullSafeSet(p0: PreparedStatement?, p1: Any?, p2: Int, p3: SharedSessionContractImplementor?) {
    if (p1 == null) {
      p0?.setNull(p2, Types.OTHER)
      return
    }
    try {
      val mapper = ObjectMapper()
      val w = StringWriter()
      mapper.writeValue(w, p1)
      w.flush()
      p0?.setObject(p2, w.toString(), Types.OTHER)
    } catch (ex: java.lang.Exception) {
      throw RuntimeException("Failed to convert Jsonb to String: " + ex.message, ex)
    }
  }
  override fun nullSafeGet(p0: ResultSet?, p1: Array<out String>?, p2: SharedSessionContractImplementor?, p3: Any?): Any {
    val cellContent = p0?.getString(p1?.get(0))
    return try {
      val mapper = ObjectMapper()
      mapper.readValue(cellContent?.toByteArray(charset("UTF-8")), returnedClass())
    } catch (ex: Exception) {
      throw RuntimeException("Failed to convert String to Jsonb: " + ex.message, ex)
    }
  }

  override fun isMutable(): Boolean {
    return true
  }

}

这样的类取自这个 Stackoverflow 问题

我的具体类:

class JsonType : JsonDataUserType() {
    override fun returnedClass(): Class<Event> {
      return Event::class.java
    }
}

我的 application.yml jpa 休眠属性

My application.yml jpa hibernate properties

jpa.properties.database.database-platform: org.hibernate.dialect.PostgreSQL95Dialect
jpa.properties.hibernate.dialect: org.myapp.util.CustomPostgreSQLDialect

方法二

Hibernate 属性与 PoJo 类完全相同，不包含自定义映射器.

Hibernate properties are exactly the same as well as the PoJo class, no custom mapper is included.

实体

@Entity
@TypeDef(
  name = "jsonb",
  typeClass = JsonBinaryType::class
)
@Table(name = "entity")
data class MyEntity(
  @Column(nullable = false)
  val id: UUID,
  @Column(nullable = false)
  @Enumerated(value = EnumType.STRING)
  @Column(nullable = false)
  val type: Type,
  @Type(type = "jsonb")
  @Column(columnDefinition = "jsonb")
  @Basic(fetch = FetchType.LAZY)
  var event_data: Event
) : SomeEntity<UUID>(), SomeOtherStuff {
  override fun getName(): String {
    return id
  }
}
  
  
enum class Type(val value: String) {
  TYPE1("Type1"),
  TYPE2("Type2")
}

自定义方言(使用休眠类型):

Custom Dialect (using hibernate types):

class CustomPostgreSQLDialect : PostgreSQL95Dialect {
  constructor() : super() {
    this.registerHibernateType(Types.OTHER, JsonNodeBinaryType::class.java.name)
    this.registerHibernateType(Types.OTHER, JsonStringType::class.java.name)
    this.registerHibernateType(Types.OTHER, JsonBinaryType::class.java.name)
    this.registerHibernateType(Types.OTHER, JsonNodeBinaryType::class.java.name)
    this.registerHibernateType(Types.OTHER, JsonNodeStringType::class.java.name)
  }
}

请注意，我也尝试过只使用:

Please note that I also tried using only:

this.registerHibernateType(Types.OTHER, "jsonb")

以及在我的实体或它从其扩展的基础实体中拥有所有这些(就此而言没有变化):

as well as having all this in my entity or the base entity it extended from (no change for that matter):

@TypeDefs({
    @TypeDef(name = "string-array", typeClass = StringArrayType.class),
    @TypeDef(name = "int-array", typeClass = IntArrayType.class),
    @TypeDef(name = "json", typeClass = JsonStringType.class),
    @TypeDef(name = "jsonb", typeClass = JsonBinaryType.class),
    @TypeDef(name = "jsonb-node", typeClass = JsonNodeBinaryType.class),
    @TypeDef(name = "json-node", typeClass = JsonNodeStringType.class),
})

我在这两种方法中都做错了什么吗?我无法让它工作，并且不确定是否有任何相关性，无 JDBC 类型的方言映射: 之后的数值总是不同的.我添加这个是因为我看到一些 ID 与某些类别的错误相关.

Is there anything evidently wrong I am doing in both approaches? I cannot get it working and, not sure if in any way relevant, the numerical value after No Dialect mapping for JDBC type: is always different. I am adding this as I have seen some ids being relevant for certain categories of errors.

你能帮忙吗?

谢谢

我想提供更多关于 jpa、postgres 和 hibernate 版本的信息.我目前正在处理以下问题:

I wanted to provide more info regarding jpa, postgres and hibernate versions. I am currently working with the following:

  1. postgres:10-alpine

  1. postgres:10-alpine

PostgreSQL JDBC 驱动程序 JDBC 4.2 » 42.2.8

PostgreSQL JDBC Driver JDBC 4.2 » 42.2.8

org.springframework.boot:spring-boot-starter-data-jpa:2.2.1.RELEASE

org.springframework.boot:spring-boot-starter-data-jpa:2.2.1.RELEASE

org.hibernate:hibernate-core:5.4.8.Final

org.hibernate:hibernate-core:5.4.8.Final

其中是否存在任何特定的版本控制问题?

Is there any particular versioning issues among them?

编辑 2我一直在尝试成功使用休眠类型(方法 2 如上所述).我根据 Postgres 版本 (10) 进行了以下更改:

EDIT 2 I have been trying to successfully use hibernate-types (Approach 2 as explained above). I have made the following change according to the Postgres version (10):

class CustomPostgreSQLDialect : PostgreSQL10Dialect {
  constructor() : super() {
    this.registerHibernateType(Types.OTHER, StringArrayType::class.java.name)
    this.registerHibernateType(Types.OTHER, IntArrayType::class.java.name)
    this.registerHibernateType(Types.OTHER, JsonStringType::class.java.name)
    this.registerHibernateType(Types.OTHER, JsonBinaryType::class.java.name)
    this.registerHibernateType(Types.OTHER, JsonNodeBinaryType::class.java.name)
    this.registerHibernateType(Types.OTHER, JsonNodeStringType::class.java.name)
  }
}

然后在我的实体中我有

@TypeDefs({
        @TypeDef(name = "string-array", typeClass = StringArrayType.class),
        @TypeDef(name = "int-array", typeClass = IntArrayType.class),
        @TypeDef(name = "json", typeClass = JsonStringType.class),
        @TypeDef(name = "jsonb", typeClass = JsonBinaryType.class)
})

和

 @Type(type = "jsonb")
 @Column(columnDefinition = "jsonb")
 @Basic(fetch = FetchType.LAZY)
 var event_data: Event

然后我调试了 TypeNames 中的 get 方法，错误来自:

I have then debugged the get method in TypeNames where the error is coming from:

public String get(final int typeCode) throws MappingException {
        final Integer integer = Integer.valueOf( typeCode );
        final String result = defaults.get( integer );
        if ( result == null ) {
            throw new MappingException( "No Dialect mapping for JDBC type: " + typeCode );
        }
        return result;
    }

这就是我得到的:

defaults = {HashMap@12093}  size = 27
     {Integer@12124} -1 -> "text"
     {Integer@12126} 1 -> "char(1)"
     {Integer@12128} -2 -> "bytea"
     {Integer@12130} 2 -> "numeric($p, $s)"
     {Integer@12132} -3 -> "bytea"
     {Integer@12133} -4 -> "bytea"
     {Integer@12134} 4 -> "int4"
     {Integer@12136} -5 -> "int8"
     {Integer@12138} -6 -> "int2"
     {Integer@12140} 5 -> "int2"
     {Integer@12141} -7 -> "bool"
     {Integer@12143} 6 -> "float4"
     {Integer@12145} 7 -> "real"
     {Integer@12147} 8 -> "float8"
     {Integer@12149} -9 -> "nvarchar($l)"
     {Integer@12151} 12 -> "varchar($l)"
     {Integer@12153} -15 -> "nchar($l)"
     {Integer@12155} -16 -> "nvarchar($l)"
     {Integer@12156} 16 -> "boolean"
     {Integer@12158} 2000 -> "json"
     {Integer@12160} 2004 -> "oid"
     {Integer@12162} 2005 -> "text"
     {Integer@12163} 1111 -> "uuid"
     {Integer@12165} 91 -> "date"
     {Integer@12167} 2011 -> "nclob"
     {Integer@12169} 92 -> "time"
     {Integer@12171} 93 -> "timestamp"

找不到 jsonb，当我调试自定义方言时，我得到以下信息:

No jsonb can be found and when I debug my custom dialect, I am getting the following:

{Integer@10846} 1111 -> "com.vladmihalcea.hibernate.type.json.JsonStringType"
 key = {Integer@10846} 1111
 value = "com.vladmihalcea.hibernate.type.json.JsonStringType"

这是为什么?为什么我没有得到 jsonb 类型?

Why is that? Why am I not getting the jsonb type?

推荐答案

我在 拉请求

想法是将实体更改为:

import com.example.demo.pojo.SamplePojo
import com.vladmihalcea.hibernate.type.json.JsonBinaryType
import com.vladmihalcea.hibernate.type.json.JsonStringType
import org.hibernate.annotations.Type
import org.hibernate.annotations.TypeDef
import org.hibernate.annotations.TypeDefs
import javax.persistence.*

@Entity
@Table(name = "tests")
@TypeDefs(
        TypeDef(name = "json", typeClass = JsonStringType::class),
        TypeDef(name = "jsonb", typeClass = JsonBinaryType::class)
)
data class SampleEntity (
    @Id @GeneratedValue
    val id: Long?,
    val name: String?,

    @Type(type = "jsonb")
    @Column(columnDefinition = "jsonb")
    var data: Map<String, Any>?
) {

    /**
     * Dependently on use-case this can be done differently:
     * https://stackoverflow.com/questions/37873995/how-to-create-empty-constructor-for-data-class-in-kotlin-android
     */
    constructor(): this(null, null, null)
}

  1. 每个实体都应该有一个默认构造函数或为其所有参数设置默认值
  2. 不是保存POJO，而是保存为Map类型

由于我们可以完全控制业务逻辑中 POJO 中的内容，唯一缺少的部分就是将 POJO 转换为 Map 和 Map 到 POJO

Since we have a full-control what will be in POJO in business logic the only missing piece will be to convert POJO to Map and Map to POJO

SamplePojo 实现

SamplePojo implementation

data class SamplePojo(
        val payload: String,
        val flag: Boolean
)  {
    constructor(map: Map<String, Any>) : this(map["payload"] as String, map["flag"] as Boolean)

    fun toMap() : Map<String, Any> {
        return mapOf("payload" to payload, "flag" to flag)
    }
}

这是一种变通方法，但它允许我们处理任何深度级别的结构.

This is rather a workaround but it allows us to work with any depth-level structures.

附言我注意到您使用了 Serializer 并重新定义了 equals、toString、hashCode.如果使用 data class.

P.S. I noticed that you use Serializer and redefined equals, toString, hashCode. You don't need this if using data class.

更新:

如果你需要比Map更灵活的结构，你可以使用JsonNode.代码示例

If you need a more flexible structure than Map<String, Any>, you can use JsonNode. Code example

实体:

import com.fasterxml.jackson.databind.JsonNode
import com.vladmihalcea.hibernate.type.json.JsonBinaryType
import com.vladmihalcea.hibernate.type.json.JsonStringType
import org.hibernate.annotations.Type
import org.hibernate.annotations.TypeDef
import org.hibernate.annotations.TypeDefs
import javax.persistence.*

@Entity
@Table(name = "tests")
@TypeDefs(
        TypeDef(name = "json", typeClass = JsonStringType::class),
        TypeDef(name = "jsonb", typeClass = JsonBinaryType::class)
)
data class SampleJsonNodeEntity (
        @Id @GeneratedValue
        val id: Long?,
        val name: String?,

        @Type(type = "jsonb")
        @Column(columnDefinition = "jsonb")
        var data: JsonNode?
) {

    /**
     * Dependently on use-case this can be done differently:
     * https://stackoverflow.com/questions/37873995/how-to-create-empty-constructor-for-data-class-in-kotlin-android
     */
    constructor(): this(null, null, null)
}

更改存储库中的实体:

import com.example.demo.entity.SampleJsonNodeEntity
import org.springframework.data.jpa.repository.JpaRepository

interface SampleJsonNodeRepository: JpaRepository<SampleJsonNodeEntity, Long> {
}

两种方法的测试:

import com.example.demo.DbTestInitializer
import com.example.demo.entity.SampleJsonNodeEntity
import com.example.demo.entity.SampleMapEntity
import com.example.demo.pojo.SamplePojo
import com.fasterxml.jackson.module.kotlin.jacksonObjectMapper
import junit.framework.Assert.assertEquals
import junit.framework.Assert.assertNotNull
import org.junit.Before
import org.junit.Test
import org.junit.runner.RunWith
import org.springframework.beans.factory.annotation.Autowired
import org.springframework.boot.test.autoconfigure.jdbc.AutoConfigureTestDatabase
import org.springframework.boot.test.context.SpringBootTest
import org.springframework.test.context.ContextConfiguration
import org.springframework.test.context.junit4.SpringRunner


@RunWith(SpringRunner::class)
@SpringBootTest
@ContextConfiguration(initializers = [DbTestInitializer::class])
@AutoConfigureTestDatabase(replace = AutoConfigureTestDatabase.Replace.NONE)
class SampleRepositoryTest {

    @Autowired
    lateinit var sampleMapRepository: SampleMapRepository

    @Autowired
    lateinit var sampleJsonNodeRepository: SampleJsonNodeRepository

    lateinit var dto: SamplePojo
    lateinit var mapEntity: SampleMapEntity
    lateinit var jsonNodeEntity: SampleJsonNodeEntity

    @Before
    fun setUp() {
        dto = SamplePojo("Test", true)
        mapEntity = SampleMapEntity(null,
                "POJO1",
                dto.toMap()
        )

        jsonNodeEntity = SampleJsonNodeEntity(null,
            "POJO2",
                jacksonObjectMapper().valueToTree(dto)
        )
    }

    @Test
    fun createMapPojo() {
        val id = sampleMapRepository.save(mapEntity).id!!
        assertNotNull(sampleMapRepository.getOne(id))
        assertEquals(sampleMapRepository.getOne(id).data?.let { SamplePojo(it) }, dto)
    }

    @Test
    fun createJsonNodePojo() {
        val id = sampleJsonNodeRepository.save(jsonNodeEntity).id!!
        assertNotNull(sampleJsonNodeRepository.getOne(id))
        assertEquals(jacksonObjectMapper().treeToValue(sampleJsonNodeRepository.getOne(id).data, SamplePojo::class.java), dto)
    }

}

这篇关于SpringBoot+Kotlin+Postgres 和 JSONB:“org.hibernate.MappingException:JDBC 类型没有方言映射"；的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！

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