如何使用 JpaRepository 进行批量（多行）插入？

Question

当从服务层使用长 List<Entity> 调用我的 JpaRepository 的 saveAll 方法时，Hibernate 的跟踪日志记录显示每个实体发出单个 SQL 语句.

我可以强制它执行批量插入（即多行）而不需要手动 fiddle 和 EntityManger、事务等甚至原始 SQL 语句字符串吗？

对于多行插入，我的意思是不只是从以下位置过渡：

start transaction
INSERT INTO table VALUES (1, 2)
end transaction
start transaction
INSERT INTO table VALUES (3, 4)
end transaction
start transaction
INSERT INTO table VALUES (5, 6)
end transaction

至：

start transaction
INSERT INTO table VALUES (1, 2)
INSERT INTO table VALUES (3, 4)
INSERT INTO table VALUES (5, 6)
end transaction

而是：

start transaction
INSERT INTO table VALUES (1, 2), (3, 4), (5, 6)
end transaction

在 PROD 中，我使用的是 CockroachDB，性能差异很大。

下面是重现问题的最小示例（为简单起见，H2）。

./src/main/kotlin/ThingService.kt:

package things

import org.springframework.boot.autoconfigure.SpringBootApplication
import org.springframework.boot.runApplication
import org.springframework.web.bind.annotation.RestController
import org.springframework.web.bind.annotation.GetMapping
import org.springframework.data.jpa.repository.JpaRepository
import javax.persistence.Entity
import javax.persistence.Id
import javax.persistence.GeneratedValue

interface ThingRepository : JpaRepository<Thing, Long> {
}

@RestController
class ThingController(private val repository: ThingRepository) {
    @GetMapping("/test_trigger")
    fun trigger() {
        val things: MutableList<Thing> = mutableListOf()
        for (i in 3000..3013) {
            things.add(Thing(i))
        }
        repository.saveAll(things)
    }
}

@Entity
data class Thing (
    var value: Int,
    @Id
    @GeneratedValue
    var id: Long = -1
)

@SpringBootApplication
class Application {
}

fun main(args: Array<String>) {
    runApplication<Application>(*args)
}

./src/main/resources/application.properties:

jdbc.driverClassName = org.h2.Driver
jdbc.url = jdbc:h2:mem:db
jdbc.username = sa
jdbc.password = sa

hibernate.dialect=org.hibernate.dialect.H2Dialect
hibernate.hbm2ddl.auto=create

spring.jpa.generate-ddl = true
spring.jpa.show-sql = true

spring.jpa.properties.hibernate.jdbc.batch_size = 10
spring.jpa.properties.hibernate.order_inserts = true
spring.jpa.properties.hibernate.order_updates = true
spring.jpa.properties.hibernate.jdbc.batch_versioned_data = true

./build.gradle.kts:

import org.jetbrains.kotlin.gradle.tasks.KotlinCompile

plugins {
    val kotlinVersion = "1.2.30"
    id("org.springframework.boot") version "2.0.2.RELEASE"
    id("org.jetbrains.kotlin.jvm") version kotlinVersion
    id("org.jetbrains.kotlin.plugin.spring") version kotlinVersion
    id("org.jetbrains.kotlin.plugin.jpa") version kotlinVersion
    id("io.spring.dependency-management") version "1.0.5.RELEASE"
}

version = "1.0.0-SNAPSHOT"

tasks.withType<KotlinCompile> {
    kotlinOptions {
        jvmTarget = "1.8"
        freeCompilerArgs = listOf("-Xjsr305=strict")
    }
}

repositories {
    mavenCentral()
}

dependencies {
    compile("org.springframework.boot:spring-boot-starter-web")
    compile("org.springframework.boot:spring-boot-starter-data-jpa")
    compile("org.jetbrains.kotlin:kotlin-stdlib-jdk8")
    compile("org.jetbrains.kotlin:kotlin-reflect")
    compile("org.hibernate:hibernate-core")
    compile("com.h2database:h2")
}

运行:

./gradlew bootRun

触发数据库插入：

curl http://localhost:8080/test_trigger

日志输出：

Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: select thing0_.id as id1_0_0_, thing0_.value as value2_0_0_ from thing thing0_ where thing0_.id=?
Hibernate: call next value for hibernate_sequence
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)
Hibernate: insert into thing (value, id) values (?, ?)

Answer 1

您可以配置 Hibernate 来执行批量 DML。看看 Spring Data JPA - concurrent Bulk inserts/updates。我认为答案的第 2 部分可以解决您的问题：

Enable the batching of DML statements Enabling the batching support would result in less number of round trips to the database to insert/update the same number of records.

Quoting from batch INSERT and UPDATE statements:

hibernate.jdbc.batch_size = 50

hibernate.order_inserts = true

hibernate.order_updates = true

hibernate.jdbc.batch_versioned_data = true

更新：您必须在 application.properties 文件中以不同方式设置休眠属性。它们在命名空间下：spring.jpa.properties.*。示例如下所示：

spring.jpa.properties.hibernate.jdbc.batch_size = 50
spring.jpa.properties.hibernate.order_inserts = true
....

Answer 2

要使用 Spring Boot 和 Spring Data JPA 进行批量插入，您只需要两件事：

将选项 spring.jpa.properties.hibernate.jdbc.batch_size 设置为您需要的适当值（例如：20）。
使用您的存储库的 saveAll() 方法和准备插入的实体列表。

工作示例是 here。

关于将insert语句改成这样：

INSERT INTO table VALUES (1, 2), (3, 4), (5, 6)

PostgreSQL 中可用：您可以在 jdbc 连接字符串中将选项 reWriteBatchedInserts 设置为 true：

jdbc:postgresql://localhost:5432/db?reWriteBatchedInserts=true

然后 jdbc 驱动程序将执行 this transformation。

有关批处理的其他信息，您可以找到 here。

已更新

Kotlin 中的演示项目：sb-kotlin-batch-insert-demo

已更新

Hibernate disables insert batching at the JDBC level transparently if you use an IDENTITY identifier generator.

Answer 3

潜在的问题是 SimpleJpaRepository 中的以下代码：

@Transactional
public <S extends T> S save(S entity) {
    if (entityInformation.isNew(entity)) {
        em.persist(entity);
        return entity;
    } else {
        return em.merge(entity);
    }
}

除了批处理大小属性设置之外，您还必须确保 class SimpleJpaRepository 调用持续存在而不是合并。有几种方法可以解决此问题：使用不查询序列的 @Id 生成器，例如

@Id
@GeneratedValue(generator = "uuid2")
@GenericGenerator(name = "uuid2", strategy = "uuid2")
var id: Long

或者通过让您的实体实现 Persistable 并覆盖 isNew() 调用

来强制持久性将记录视为新记录

@Entity
class Thing implements Pesistable<Long> {
    var value: Int,
    @Id
    @GeneratedValue
    var id: Long = -1
    @Transient
    private boolean isNew = true;
    @PostPersist
    @PostLoad
    void markNotNew() {
        this.isNew = false;
    }
    @Override
    boolean isNew() {
        return isNew;
    }
}

或者覆盖 save(List) 并使用实体管理器调用 persist()

@Repository
public class ThingRepository extends SimpleJpaRepository<Thing, Long> {
    private EntityManager entityManager;
    public ThingRepository(EntityManager entityManager) {
        super(Thing.class, entityManager);
        this.entityManager=entityManager;
    }

    @Transactional
    public List<Thing> save(List<Thing> things) {
        things.forEach(thing -> entityManager.persist(thing));
        return things;
    }
}

以上代码基于以下链接：

Answer 4

所有提到的方法都有效，但会很慢，特别是如果插入数据的来源位于其他 table。首先，即使使用 batch_size>1，插入操作也会在多个 SQL 查询中执行。其次，如果源数据位于另一个 table 中，则需要使用其他查询获取数据（并且在最坏的情况下将所有数据加载到内存中），并将其转换为静态批量插入。第三，通过单独 persist() 调用每个实体（即使启用了批处理），您将使用所有这些实体实例膨胀实体管理器一级缓存。

但是 Hibernate 还有另一种选择。如果您将 Hibernate 用作 JPA 提供程序，则可以回退到 HQL，后者 supports bulk inserts 本机使用来自另一个 table 的子选择。示例：

Session session = entityManager.unwrap(Session::class.java)
session.createQuery("insert into Entity (field1, field2) select [...] from [...]")
  .executeUpdate();

这是否有效取决于您的 ID 生成策略。如果Entity.id是数据库生成的（比如MySQL自增），则执行成功。如果 Entity.id 是由您的代码生成的（对于 UUID 生成器尤其如此），它将失败并出现“不支持的 ID 生成方法”异常。

然而，在后一种情况下，这个问题可以通过自定义 SQL 函数来解决。例如在 PostgreSQL 中我使用 uuid-ossp 扩展提供 uuid_generate_v4() 功能，我最终在我的自定义对话框中注册：

import org.hibernate.dialect.PostgreSQL10Dialect;
import org.hibernate.dialect.function.StandardSQLFunction;
import org.hibernate.type.PostgresUUIDType;

public class MyPostgresDialect extends PostgreSQL10Dialect {

    public MyPostgresDialect() {
        registerFunction( "uuid_generate_v4", 
            new StandardSQLFunction("uuid_generate_v4", PostgresUUIDType.INSTANCE));
    }
}

然后我将此 class 注册为休眠对话框：

hibernate.dialect=MyPostgresDialect

我终于可以在批量插入查询中使用这个函数了：

SessionImpl session = entityManager.unwrap(Session::class.java);
session.createQuery("insert into Entity (id, field1, field2) "+
  "select uuid_generate_v4(), [...] from [...]")
  .executeUpdate();

最重要的是Hibernate生成的底层SQL来完成这个操作，它只是一个查询：

insert into entity ( id, [...] ) select uuid_generate_v4(), [...] from [...]

如何使用 JpaRepository 进行批量（多行）插入？

How to do bulk (multi row) inserts with JpaRepository?

hibernate

spring-data-jpa

kotlin

spring-boot

cockroachdb