Spring :: @ Transactional not working

Question

저는 Spring에 익숙하지 않고 트랜잭션 개념을 배우고 있습니다. @Transactional을 사용할 수 없습니다.

사용 케이스 : GetEmployee라는이()의 RuntimeException를 throw 할 때 직원 및 직원 세부
데이터 삽입 롤백합니다. 그러나 롤백은 일어나지 않습니다. Oracle 데이터베이스 11g 및 Spring 4.3.1.RELEASE를 사용하고 있습니다. 아래는 독립 실행 형 자바 코드입니다.

코드

공공 정적 무효 메인 (문자열 []에 args) { AbstractApplicationContext의 CTX = 새로운 ClassPathXmlApplicationContext ("스프링 bean.xml"); ctx.registerShutdownHook(); 에는 트랜잭션되지

Employee emp = new Employee("149518", "Mickey", "Mouse", "15 years", "tennis"); 

IEmployee empIntfc = (IEmployee)ctx.getBean("empService"); 
try { 
     empIntfc.createEmployee(emp); 
     empIntfc.createEmployeeDetails(emp); 

     //Below will throw RunTime Exception 
     empIntfc.getEmployee(2); 

    }catch (Exception e) { 
     e.printStackTrace(); 

    } finally { 
     ctx.close();  
    } 

}

EmployeeService.java

public class EmployeeService implements IEmployee { 

private DataSource dataSource; 
private JdbcTemplate jdbcTemplate; 

public void setDataSource(DataSource dataSource) { 
    this.dataSource = dataSource; 
    jdbcTemplate = new JdbcTemplate(this.dataSource); 
} 
public JdbcTemplate getJdbcTemplate() { 
    return jdbcTemplate; 
} 

@Override 
@Transactional 
public int createEmployee(Employee emp) {  

    String sql1 = "INSERT INTO TEST_T1(EMP_ID, EMP_FNAME, EMP_LNAME) values 
    (?,?,?)"; 
    return getJdbcTemplate().update(sql1, emp.getEmpId(), 
    emp.getEmpFirstName(), emp.getEmpLastName()); 
} 

@Override 
@Transactional 
public int createEmployeeDetails(Employee emp) { 

    String sql = "INSERT INTO TEST_T2(EMP_ID, EXP, SKILLS) values (?,?,?)";  
    return getJdbcTemplate().update(sql, emp.getEmpId(), emp.getExp(), 
    emp.getSkills()); 
} 

@Override 
@Transactional(readOnly = true, noRollbackFor=RuntimeException.class) 
public Employee getEmployee(int empId) { 
    throw new RuntimeException("Intentional runtime exception"); 
} 

}

봄 bean.xml

<beans xmlns="http://www.springframework.org/schema/beans"> 

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

<bean id="transactionManager" 
class="org.springframework.jdbc.datasource.DataSourceTransactionManager"> 
<property name="dataSource" ref="dataSource"/> 
</bean> 
<bean id="dataSource" 
    class="org.springframework.jdbc.datasource.DriverManagerDataSource" > 
    <property name="driverClassName" value="oracle.jdbc.driver.OracleDriver"/> 
    <property name="url" value="jdbc:oracle:thin:@//xxxx:1521/xxxx"/> 
    <property name="username" value="user"/> 
    <property name="password" value="user"/> 
    </bean> 
    <bean id="empService" class="com.service.EmployeeService"> 
    <property name="dataSource" ref="dataSource"/> 
    </bean> 
    </beans> 

Answer 1

귀하의 주요 방법은 ... 의미 'createEmployee'를 입력하면 새 트랜잭션을 커밋하고 커밋하면 'createEmployeeDetails'가 새 트랜잭션을 만들고 커밋합니다.