spring - 在简单的 Spring 4 REST 服务上获取 404

Question

我正在尝试访问我编写的 RESTful Web 服务:

http://localhost:8080/dukegen/ws/family/1

但在浏览器中使用地址栏时出现 404 错误，不知道为什么。我正在尝试返回 JSON。我已将 Jackson 2 放在我的类路径中:

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.3.1</version>
</dependency>

这是服务器输出:

Jan 14, 2014 8:29:55 PM org.springframework.web.servlet.handler.AbstractUrlHandlerMapping registerHandler
INFO: Mapped URL path [/ws/family/{familyId}] onto handler 'familyResource'
Jan 14, 2014 8:29:55 PM org.springframework.web.servlet.handler.AbstractUrlHandlerMapping registerHandler
INFO: Mapped URL path [/ws/family/{familyId}.*] onto handler 'familyResource'
Jan 14, 2014 8:29:55 PM org.springframework.web.servlet.handler.AbstractUrlHandlerMapping registerHandler
INFO: Mapped URL path [/ws/family/{familyId}/] onto handler 'familyResource'
Jan 14, 2014 8:29:55 PM org.springframework.web.servlet.FrameworkServlet initServletBean
INFO: FrameworkServlet 'dispatcher': initialization completed in 360 ms
Jan 14, 2014 8:29:55 PM org.springframework.web.servlet.DispatcherServlet noHandlerFound
WARNING: No mapping found for HTTP request with URI [/dukegen/ws/family/1] in DispatcherServlet with name 'dispatcher'

这是我的 Controller :

@Controller
@RequestMapping("ws")
public class FamilyResource {

    @RequestMapping(value="family/{familyId}", method = RequestMethod.GET, produces="application/json")
    public @ResponseBody Family getFamily(@PathVariable long familyId) {
            .... builds Family object ....
             return family;
         }

}

这是我在 web.xml 中设置的调度程序:

 <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>classpath:/mvcContext.xml</param-value>
        </init-param>
  </servlet>

  <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/ws/*</url-pattern>
  </servlet-mapping>

我的 mvcContext.xml:

<?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:p="http://www.springframework.org/schema/p"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="
        http://www.springframework.org/schema/beans
        http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context
        http://www.springframework.org/schema/context/spring-context.xsd">

    <context:component-scan base-package="ws.hamacher.dukegen.resource"/>

</beans>

如有任何帮助，我们将不胜感激。

Answer 1

这里有几件事不正确。

首先，在您的请求映射中，映射应该是一致的。您的类应映射到 "/ws" 并且生成结果的方法应为 "/family/{familyId}"

在您的 web.xml 中，您已将 servlet 配置为响应 /ws/* 并且您的 Controller 再次请求映射到 ws。这将不起作用。

一旦 "/ws/*" 被您的 servlet 拦截，它就不应在请求映射中重复。 Controller 仅响应其上下文中的 URL 模式。 URL 中 "/ws" 之后的内容仅在 Controller 的上下文中。

我通常更喜欢将 servlet 映射到 "/" 以及在 Controller 内编码的所有进一步分辨率。不过，这只是我的偏好。

所以正确的配置是

web.xml

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>classpath:/mvcContext.xml</param-value>
        </init-param>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

和 Controller

   @Controller
   @RequestMapping("/ws")
   public class FamilyResource {
       @RequestMapping(value="/family/{familyId}", method = RequestMethod.GET, produces="application/json")
       public @ResponseBody Family getFamily(@PathVariable long familyId) {
          .... builds Family object ....
          return family;
       }
   }