我尝试构建spring restfull服务教程,这是我的代码:
的web.xml
<servlet>
<servlet-name>accounts</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/mvc-config.xml
/WEB-INF/app-config.xml
</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>accounts</servlet-name>
<url-pattern>/app/*</url-pattern>
</servlet-mapping>
这是我的控制者:
@Controller
public class AccountController {
@RequestMapping(value = "/accounts", method = RequestMethod.GET, produces = {"application/json"})
@ResponseStatus(HttpStatus.OK)
public List<Account> accountSummary() {
return accountManager.getAllAccounts();
}
}
当我使用网址打开浏览器时:http://localhost:8080/rest-ws/app/accounts
我收到了这个错误:
javax.servlet.ServletException: Circular view path [accounts]: would dispatch back to the current handler URL [/rest-ws/app/accounts] again. Check your ViewResolver setup! (Hint: This may be the result of an unspecified view, due to default view name generation.)
我想念什么?
答案 0 :(得分:1)
试试这个:
@Controller
@RequestMapping(value = "/accounts")
public class AccountController {
@RequestMapping(value = "/list",
method = RequestMethod.GET, produces = {"application/json"})
@ResponseStatus(HttpStatus.OK)
public List<Account> accountSummary() {
return accountManager.getAllAccounts();
}
}
打开:
http://localhost:8080/rest-ws/app/accounts/list
有用吗?