在简单的Spring 4 REST服务上获得404

Question

我正在尝试访问我编写的RESTful Web服务:

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

但是使用浏览器中的地址栏获取404并且不知道原因.我想恢复JSON.我把杰克逊2放在我的课堂上:

<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'

这是我的控制器:

@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>

任何帮助将不胜感激.

Answer 1

有些事情在这里不正确.

首先,在您的请求映射中,映射应该是一致的.您的类应该映射到, "/ws"并且您的方法应该生成结果"/family/{familyId}"

在你的web.xml中,你已经配置了servlet来响应/ws/*,你的控制器ws再次被请求映射.这不会工作.

一旦"/ws/*"被servlet截获,就不应该在请求映射中重复它.Controller仅响应其上下文中的URL模式.无论是后"/ws"您的网址只在控制器的情况下.

我通常更喜欢将servlet映射到"/"控制器内的所有其他分辨率.不过只是我的偏好.

所以正确的配置是

web.xml中

    <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>

和控制器

   @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;
       }
   }