我尝试了很多但没有达到。我想从控制器返回有效的Json对象。我在数据库中放了一些数据并希望列出它们。这是我的代码
的web.xml
<web-app version="2.5" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xsi:schemalocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<servlet>
<servlet-name>sdnext</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name><param-value>/WEB-INF/config/sdnext-servlet.xml</param-value></init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>sdnext</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
</web-app>
sdnext-servlet.xml中
<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:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="
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.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context.xsd
http://www.springframework.org/schema/mvc
http://www.springframework.org/schema/mvc/spring-mvc.xsd">
<mvc:annotation-driven/>
<context:annotation-config />
<context:property-placeholder location="classpath:resources/database.properties"/>
<context:component-scan base-package="com.dineshonjava"/>
<tx:annotation-driven transaction-manager="hibernateTransactionManager"/>
<bean
class="org.springframework.web.servlet.view.InternalResourceViewResolver"
id="jspViewResolver">
<property name="prefix" value="/WEB-INF/views/"></property>
<property name="suffix" value=".jsp"></property>
</bean>
<bean class="org.springframework.jdbc.datasource.DriverManagerDataSource"
id="dataSource">
<property name="driverClassName" value="${database.driver}"></property>
<property name="url" value="${database.url}"></property>
<property name="username" value="${database.user}"></property>
<property name="password" value="${database.password}"></property>
</bean>
<bean
class="org.springframework.orm.hibernate4.LocalSessionFactoryBean"
id="sessionFactory">
<property name="dataSource" ref="dataSource"></property>
<property name="annotatedClasses">
<list>
<value>com.dineshonjava.model.Employee</value>
</list>
</property>
<property name="hibernateProperties">
<props>
<prop key="hibernate.dialect">${hibernate.dialect}</prop>
<prop key="hibernate.show_sql">${hibernate.show_sql}</prop>
<prop key="hibernate.hbm2ddl.auto">${hibernate.hbm2ddl.auto} </prop>
</props>
</property>
</bean>
<bean class="org.springframework.orm.hibernate4.HibernateTransactionManager"
id="hibernateTransactionManager">
<property name="sessionFactory" ref="sessionFactory"></property>
</bean>
</beans>
和 EmployeeController.java
import org.springframework.beans.factory.annotation.Autowired;
@Controller
public class EmployeeController {
@Autowired
private EmployeeService employeeService;
@RequestMapping(value = "/index", method = RequestMethod.GET)
public @ResponseBody List<Employee> listAllUsers() {
return employeeService.listEmployeess();
}
}
它在控制台中给了我这一行:
休眠:
select this_.EMPID as EMPID1_0_0_, this_.ADDRESS as ADDRESS2_0_0_, this_.EMPAGE as EMPAGE3_0_0_,
this_.EMPNAME as EMPNAME4_0_0_, this_.SALARY as SALARY5_0_0_ from employee this_
那么如何获取表的值并将其放入Json对象?