JpaRepository将不会保存数据 [英] JpaRepository won't save data

问题描述

我使用JpaRepository保存数据，但是hibernate.show_sql显示选择"并且不会保存数据.以下是我的服务:

I use JpaRepository to save data, but the hibernate.show_sql shows "select" and won't save data. Following is my service:

@Autowired
private UserRepository userRepository;

@PostConstruct
public void init() {
    User admin = new User();
    admin.setDisplayName("admin");
    admin.setEmailAddress("admin@admin");
    admin.setPassword("admin___");
    admin.setRegisteredAt(new Date());
    admin.setLastAccessAt(new Date());
    admin.setUuid(UUID.randomUUID().toString());

    try {
        System.out.println("before save");
        userRepository.save(admin);
        System.out.println("after save");
    } catch (Exception e) {
        System.out.println(e.getMessage());
    }
}

输出看起来像这样:

=========保存之前======

========before save======

休眠:选择user0_.uuid作为uuid1_0_0_，将user0_.display_name作为display_2_0_0_，将user0_.email_address作为email_ad3_0_0_，将user0_.last_access_at作为last_acc4_0_0_，将user0_.password作为密码5_0_0_0，将user0_.u__uid_作为用户0从寄存器06.

Hibernate: select user0_.uuid as uuid1_0_0_, user0_.display_name as display_2_0_0_, user0_.email_address as email_ad3_0_0_, user0_.last_access_at as last_acc4_0_0_, user0_.password as password5_0_0_, user0_.registered_at as register6_0_0_ from User user0_ where user0_.uuid=?

========保存后=======

========after save=======

以下是我的applicationContext.xml:

Following is my applicationContext.xml:

<context:component-scan base-package="test">
    <context:exclude-filter type="annotation"
        expression="org.springframework.stereotype.Controller" />
</context:component-scan>

<bean id="myDataSource" class="org.apache.commons.dbcp.BasicDataSource">
    <property name="driverClassName" value="com.mysql.jdbc.Driver" />
    <property name="url" value="jdbc:mysql://localhost:3306/helloworld" />
    <property name="username" value="root" />
    <property name="password" value="password" />
</bean>

<bean id="myEmf"
    class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="packagesToScan" value="test.entity"></property>
    <property name="dataSource" ref="myDataSource" />
    <property name="jpaProperties">
        <props>
            <prop key="hibernate.show_sql">true</prop>
            <prop key="hibernate.hbm2ddl.auto">update</prop>
        </props>
    </property>
    <property name="persistenceProvider">
        <bean class="org.hibernate.jpa.HibernatePersistenceProvider"></bean>
    </property>
</bean>

<tx:annotation-driven transaction-manager="transactionManager" />

<bean id="transactionManager"
    class="org.springframework.jdbc.datasource.DataSourceTransactionManager">
    <property name="dataSource" ref="myDataSource" />
</bean>

<jpa:repositories base-package="test.repository"
    entity-manager-factory-ref="myEmf" transaction-manager-ref="transactionManager"></jpa:repositories>

我的课程是由JPA工具生成的:

Attached is my class generated by JPA Tools:

@Entity
@NamedQuery(name="User.findAll", query="SELECT u FROM User u")
public class User implements Serializable {
  private static final long serialVersionUID = 1L;

  @Id
  private String uuid;

  @Column(name="display_name")
  private String displayName;

  @Column(name="email_address")
  private String emailAddress;

  @Temporal(TemporalType.TIMESTAMP)
  @Column(name="last_access_at")
  private Date lastAccessAt;

  private String password;

  @Temporal(TemporalType.TIMESTAMP)
  @Column(name="registered_at")
  private Date registeredAt;

  public   User() {
  }

  public String getUuid() {
      return this.uuid;
  }

  public void setUuid(String uuid) {
      this.uuid = uuid;
  }

  public String getDisplayName() {
      return this.displayName;
  }

  public void setDisplayName(String displayName) {
      this.displayName = displayName;
  }

  public String getEmailAddress() {
      return this.emailAddress;
  }

  public void setEmailAddress(String emailAddress) {
      this.emailAddress = emailAddress;
  }

  public Date getLastAccessAt() {
      return this.lastAccessAt;
  }

  public void setLastAccessAt(Date lastAccessAt) {
      this.lastAccessAt = lastAccessAt;
  }

  public String getPassword() {
      return this.password;
  }

  public void setPassword(String password) {
      this.password = password;
  }

  public Date getRegisteredAt() {
      return this.registeredAt;
  }

  public void setRegisteredAt(Date registeredAt) {
      this.registeredAt = registeredAt;
  }

}

推荐答案

由于您使用的是JPA，因此事务管理器应为JpaTransactionManager，而不是DataSourceTransactionManager.

Since you're using JPA, the transaction manager should be a JpaTransactionManager, not a DataSourceTransactionManager.

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