Question

我使用带有spring-boot-starter-data-jpa的1.2.0版本创建了一个Spring Boot应用程序，我正在使用MySQL。我正确地在application.properties文件中配置了MySQL属性。

我有一个简单的JPA实体，Spring Data JPA Repository和Service如下：

@Entity
class Person
{
    @Id @GeneratedValue(strategy=GenerationType.AUTO)
    private Integer id;
    private String name;
    //setters & getters
}

@Repository
public interface PersonRepository extends JpaRepository<Person, Integer>{

}


import java.util.List;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Service;
import org.springframework.transaction.annotation.Transactional;

@Service
@Transactional
class PersonService
{
    @Autowired PersonRepository personRepository;

    @Transactional
    void save(List<Person> persons){
        for (Person person : persons) {         
            if("xxx".equals(person.getName())){
                throw new RuntimeException("boooom!!!");
            }
            personRepository.save(person);          
        }
    }
}

我有以下JUnit测试：

import org.junit.Test;
import org.junit.runner.RunWith;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.boot.test.SpringApplicationConfiguration;
import org.springframework.test.context.junit4.SpringJUnit4ClassRunner;

@RunWith(SpringJUnit4ClassRunner.class)
@SpringApplicationConfiguration(classes = Application.class)
public class ApplicationTests {

    @Autowired
    PersonService personService;

    @Test
    public void test_logging() {
        List<Person> persons = new ArrayList<Person>();
        persons.add(new Person(null,"abcd"));
        persons.add(new Person(null,"xyz"));
        persons.add(new Person(null,"xxx"));
        persons.add(new Person(null,"pqr"));

        personService.save(persons);
    }

}

这里的期望是它不应该在PERSON表中插入任何记录，因为它会在插入第三个人对象时抛出异常。但它没有被回滚，前两个记录被插入并提交。

然后我想到快速尝试使用JPA EntityManager。

@PersistenceContext
private EntityManager em;

em.save(person);

然后我得到 javax.persistence.TransactionRequiredException：没有可用的事务性EntityManager 异常。

谷歌搜索了一段时间后，我在同一主题上遇到了这个JIRA帖子https://jira.spring.io/browse/SPR-11923。

然后我将Spring Boot版本更新为 1.1.2 ，以使Spring版本早于 4.0.6 。

然后 em.save（person）按预期工作并且Transaction工作正常（意味着它在发生RuntimeException时回滚所有db插入）。

但是，即使使用Spring 4.0.5 + Spring Data JPA 1.6.0版本，当使用 personRepository.save（person）而不是 em.persist（person）时，事务也无法正常工作/强>

似乎Spring Data JPA存储库正在提交事务。

我错过了什么？如何使服务水平 @Transactional 注释有效？

PS：

Maven pom.xml

<?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>com.sivalabs</groupId> <artifactId>springboot-data-jpa</artifactId> <version>1.0-SNAPSHOT</version> <packaging>jar</packaging> <name>springboot-data-jpa</name> <description>Spring Boot Hello World</description> <parent> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-parent</artifactId> <version>1.2.0.RELEASE</version> <relativePath /> </parent> <properties> <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding> <start-class>com.sivalabs.springboot.Application</start-class> <java.version>1.7</java.version> </properties> <build> <plugins> <plugin> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-maven-plugin</artifactId> </plugin> </plugins> </build> <dependencies> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-actuator</artifactId> </dependency> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-data-jpa</artifactId> </dependency> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-web</artifactId> </dependency> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-test</artifactId> <scope>test</scope> </dependency> <dependency> <groupId>mysql</groupId> <artifactId>mysql-connector-java</artifactId> </dependency> </dependencies> </project>

Application.java

@EnableAutoConfiguration @Configuration @ComponentScan public class Application { public static void main(String[] args) { SpringApplication.run(Application.class, args); } }

Answer 1

将交易注释更改为 @Transactional（的rollbackFor = Exception.class）

Answer 2

来自comment @ m-deinum：

制作您的PersonService public以及您要调用的方法。

这似乎为几个用户做了诀窍。 this answer也引用了同样的事情，引用了manual saying：

使用代理时，应该应用@Transactional注释仅限具有公众可见度的方法。如果你做注释保护，使用@Transactional注释的私有或包可见方法，没有引发错误，但带注释的方法没有出现错误配置的事务设置。考虑使用AspectJ（参见如果你需要注释非公开方法。

Answer 3

我使用SpringBoot v1.2.0和v1.5.2对此进行了测试，所有这些都按预期工作，您不需要使用实体管理器@Transactional（rollbackFor = Exception.class）。

我看不出你错误配置的部分，乍一看似乎都很好，所以我只是将所有配置作为参考发布，带有更新的注释和H2嵌入式内存DB：

<强> application.properties

# Embedded memory database
spring.jpa.hibernate.ddl-auto=create
spring.datasource.url=jdbc:h2:~/test;AUTO_SERVER=TRUE

<强>的pom.xml

<project xmlns="http://maven.apache.org/POM/4.0.0"
         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
    <modelVersion>4.0.0</modelVersion>

    <groupId>com.demo</groupId>
    <artifactId>jpaDemo</artifactId>
    <version>1.0-SNAPSHOT</version>

    <properties>
        <maven.compiler.source>1.8</maven.compiler.source>
        <maven.compiler.target>1.8</maven.compiler.target>
    </properties>

    <parent>
        <groupId>org.springframework.boot</groupId>
        <artifactId>spring-boot-starter-parent</artifactId>
        <version>1.5.2.RELEASE</version>
    </parent>

    <dependencies>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-data-jpa</artifactId>
        </dependency>
        <dependency>
            <groupId>com.h2database</groupId>
            <artifactId>h2</artifactId>
        </dependency>
        <dependency>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-starter-test</artifactId>
            <scope>test</scope>
        </dependency>
        <dependency>
            <groupId>junit</groupId>
            <artifactId>junit</artifactId>
        </dependency>
    </dependencies>
</project>

<强> Person.java

@Entity
public class Person
{
    @Id
    @GeneratedValue(strategy= GenerationType.AUTO)
    private Integer id;
    private String name;

    public Person() {}
    public Person(String name) {this.name = name;}
    public Integer getId() {return id;}
    public void setId(Integer id) {this.id = id;}
    public String getName() {return name;}
    public void setName(String name) {this.name = name;}
}

<强> PersonRepository.java

@Repository
public interface PersonRepository extends JpaRepository<Person, Integer> {}

<强> PersonService.java

@Service
public class PersonService
{
    @Autowired
    PersonRepository personRepository;

    @Transactional
    public void saveTransactional(List<Person> persons){
        savePersons(persons);
    }

    public void saveNonTransactional(List<Person> persons){
        savePersons(persons);
    }

    private void savePersons(List<Person> persons){
        for (Person person : persons) {
            if("xxx".equals(person.getName())){
                throw new RuntimeException("boooom!!!");
            }
            personRepository.save(person);
        }
    }
}

<强> Application.java

@SpringBootApplication
public class Application {
    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }
}

<强> PersonServiceTest.java

@RunWith(SpringRunner.class)
@SpringBootTest
public class PersonServiceTest {

    @Autowired
    PersonService personService;

    @Autowired
    PersonRepository personRepository;

    @Test
    public void person_table_size_1() {
        List<Person> persons = getPersons();
        try {personService.saveNonTransactional(persons);}
        catch (RuntimeException e) {}

        List<Person> personsOnDB = personRepository.findAll();
        assertEquals(1, personsOnDB.size());
    }

    @Test
    public void person_table_size_0() {
        List<Person> persons = getPersons();
        try {personService.saveTransactional(persons);}
        catch (RuntimeException e) {}

        List<Person> personsOnDB = personRepository.findAll();
        assertEquals(0, personsOnDB.size());
    }

    public List<Person> getPersons(){
        List<Person> persons = new ArrayList() {{
            add(new Person("aaa"));
            add(new Person("xxx"));
            add(new Person("sss"));
        }};

        return persons;
    }
}

*清除数据库并为每个测试重新初始化，以便我们始终针对已知状态进行验证

Spring Boot + Spring Data JPA +交易无法正常工作

3 个答案: