Question

我正在尝试使用Spring和Hibernate构建REST API。

运行服务器时，它不会返回任何错误，但无法访问链接。

这是我的配置文件：

spring.xml

<mvc:annotation-driven />
<context:component-scan base-package="org.bloggy.spring" />
<bean id="sessionFactory" class="org.springframework.orm.hibernate5.LocalSessionFactoryBean">
    <property name="dataSource" ref="dataSource" />
    <property name="hibernateProperties">
        <props>
            <prop key="hbm2ddl.auto">create-drop</prop>
            <prop key="hibernate.dialect">org.hibernate.dialect.MySQL5Dialect</prop>
            <prop key="show_sql">true</prop>
        </props>
    </property>
</bean>

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

<bean id="txManager"
      class="org.springframework.transaction.jta.JtaTransactionManager" />

那是我的web.xml：

<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

<!-- Processes application requests -->

<servlet>
    <servlet-name>appServlet</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>WEB-INF/resources/spring.xml</param-value>
        <param-value>WEB-INF/resources/dao.xml</param-value>
        <param-value>WEB-INF/resources/service.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>

<servlet-mapping>
    <servlet-name>appServlet</servlet-name>
    <url-pattern>/api/*</url-pattern>
</servlet-mapping>

控制器：

@Controller
@RequestMapping(value = "users")
public class UserController {
  @Autowired
  private UserService userService;

  @RequestMapping(value = "/", method = RequestMethod.GET)
  public List<User> getUsers() {
    return userService.listUsers();
  }
}

Answer 1

如@Ken所述，为了将您的API公开为REST AP，您需要使用@RestController，或者如果您想使用@Controller，则使用@ResponseBody注释。

关于您的问题，我认为您需要将RequestMapping定义为

@Controller
@RequestMapping(value = "api")
public class UserController {
  @Autowired
  private UserService userService;

  @RequestMapping(value = "/users", method = RequestMethod.GET)
  public List<User> getUsers() {
    return userService.listUsers();
  }
}

或

@Controller
@RequestMapping(value = "api/users")
public class UserController {
  @Autowired
  private UserService userService;

  @RequestMapping(method = RequestMethod.GET)
  public List<User> getUsers() {
    return userService.listUsers();
  }
}

请查看是否有效。

Answer 2

在logs中可见，应用程序MyBloggy实际上并未初始化任何spring bean。 Tomcat上部署的缓存或war文件存在问题。理想情况下，在部署war时，除非已禁用，否则所有初始化的bean都可以在日志中看到。确保您在服务器上部署了正确的war。

Answer 3

您正在使用@Controller批注，但是如果要公开REST Web服务，则需要使用@RestController或@ResponseBody

@RestController
@RequestMapping(value = "users")
public class UserController

Spring无法获取映射

3 个答案: