23:29 提问
jpa+em.persist(stu)不能把stu对象持久化到数据中，程序不报错，但可以查数据
请各位大神帮忙解决一个问题：
spring+jpa+em.persist(student)不能把student对象持久化到Mysql数据中，程序运行也不报错，但可以查到数据库中原有数据，这是什么原因啊？谢谢！（我是新手，对jpa还不太熟悉）
spring配置：
&?xml version="1.0" encoding="UTF-8"?&
&beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:jdbc="http://www.springframework.org/schema/jdbc"
xsi:schemaLocation="http://www.springframework.org/schema/jdbc http://www.springframework.org/schema/jdbc/spring-jdbc-3.2.xsd
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.1.xsd"&
&bean id="jdbc"
class="org.springframework.beans.factory.config.PropertiesFactoryBean"&
&property name="locations" value="classpath:/dev/jdbc.properties"&&/property&
&bean id="dataSource"
class="org.springframework.jdbc.datasource.DriverManagerDataSource"&
&property name="driverClassName" value="#{jdbc.driverClassName}"&&/property&
&constructor-arg value="#{jdbc.url}"&&/constructor-arg&
&constructor-arg value="#{jdbc.userName}"&&/constructor-arg&
&constructor-arg value="#{jdbc.password}"&&/constructor-arg&
&bean id="emf"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean"&
&description&
..........
&/description&
&property name="dataSource" ref="dataSource" /&
&property name="persistenceUnitName" value="default" /&
&property name="packagesToScan"&
&value&com.wyiwei.entity_modle.infrastructur.shared.modle&/value&
&/property&
&property name="jpaVendorAdapter"&
&bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter"&
&property name="generateDdl" value="true" /&
&property name="showSql" value="false" /&
&/property&
&bean class="org.springframework.jdbc.core.JdbcTemplate"& --&
&property name="dataSource" ref="dataSource" /& --&
&/bean& --&
&bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager"&
&property name="entityManagerFactory" ref="emf" /&
&tx:annotation-driven
transaction-manager="transactionManager" proxy-target-class="true"/& --&
&tx:annotation-driven mode="aspectj"
transaction-manager="transactionManager" /&
@Table(name="STUDENT")
public class Student implements Serializable{
private static final long serialVersionUID = -7067776L;
@Column(name="STU_SEQ", nullable=false)
private Integer stuS
@Column(name="STUDENT_NAME", nullable=false, length=21)
private String studentN
@Column(name="STU_AGE", nullable=false, length=4)
private Integer stuA
public Integer getStuSeq() {
return stuS
public void setStuSeq(Integer stuSeq) {
this.stuSeq = stuS
public String getStudentName() {
return studentN
public void setStudentName(String studentName) {
this.studentName = studentN
public Integer getStuAge() {
return stuA
public void setStuAge(Integer stuAge) {
this.stuAge = stuA
public String toString() {
return "Student [stuSeq=" + stuSeq + ", studentName=" + studentName
+ ", stuAge=" + stuAge + "]";
测试代码：
@ContextConfiguration({ "/test-context.xml"})
public class PersistTest extends AbstractTest{
@PersistenceContext
private EntityM
public void test(){
persist();
System.out.println("Over!");
@Transactional
private void persist(){
Student s = new Student();
s.setStuAge(30);
s.setStudentName("张三");
s.setStuSeq(1);
em.persist(s);
@Transactional
private void query(){
QStudent q = QStudent.
Student student = new JPAQuery(em).from(q).where(q.stuSeq.eq(1)).uniqueResult(q);
System.out.println(student);
按赞数排序
问题解决了！原来还是因为事务的问题，spring配置了事务，代码里也有@Transactional注解，可还是事务不能提交。这是为什么呢？原来在同一个
类中，方法a调用加了@Transactional注解的方法b，方法b的事务是不起作用的。（另一个类调用b方法会有事务）
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