我想创建一个网站，能够与其他工作同一个Android应用程序。 对于这些剩余的呼叫我想利用一个单独的servlet
在web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
id="WebApp_ID" version="2.5">
<display-name>Stage XT-i</display-name>
<welcome-file-list>
<welcome-file>/WEB-INF/jsp/index.jsp</welcome-file>
</welcome-file-list>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<servlet>
<servlet-name>spring</servlet-name>
<servlet-class>
org.springframework.web.servlet.DispatcherServlet
</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>*.html</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/spring-servlet.xml
/WEB-INF/spring-security.xml
</param-value>
</context-param>
<!-- Spring Security -->
<filter>
<filter-name>springSecurityFilterChain</filter-name>
<filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
</filter>
<filter-mapping>
<filter-name>springSecurityFilterChain</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<servlet>
<servlet-name>Jersey Web Application</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>be.kdg.teamf</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
<param-value>true</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
正如你所看到人的球衣信息使用的URL通过：/ REST / *当我想使用URL字符串做一个GET：本地主机：9999 / REST /用户/品尝它很好地工作。 然而，当我使用@Autowired与球衣我得到一个空指针异常。
Java类：
@Path("/user")
public class UserRest {
@Autowired
private UserService userService;
@Context
UriInfo uriInfo;
// Another "injected" object. This allows us to use the information that's
// part of any incoming request.
// We could, for example, get header information, or the requestor's address.
@Context
Request request;
// Basic "is the service running" test
@GET
@Produces(MediaType.TEXT_PLAIN)
public String respondAsReady() {
return "Demo service is ready!";
}
@GET
@Path("sample")
@Produces(MediaType.APPLICATION_JSON)
public User getUserJson() {
User u = userService.listUsers().get(0);
System.out.println("Returning person");
return u;
}
}
有没有人看到这个问题？
提前致谢。
