Question

当我尝试加载http://localhost:8080/people时，收到404找不到错误的页面。

这是我的weblet中的servlet映射：

<servlet>
    <servlet-name>spring</servlet-name>
    <servlet-class>
        org.springframework.web.servlet.DispatcherServlet
    </servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>classpath:applicationContext.xml</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>spring</servlet-name>
    <url-pattern>/people/*</url-pattern>
</servlet-mapping>

以下是我理解它的工作原理：

servlet“spring”将拦截对http://localhost:8080/people的url请求，并调用类org.springframework.web.servlet.DispatcherServlet这是正确的吗？

我是否需要一些额外的配置才能正确加载此类？

更新：

这是控制器：

@Controller
public class PersonController {

    @Autowired
    private PersonService personService;

    @RequestMapping("/")
    public String listPeople(Map<String, Object> map) {

        map.put("person", new Person());
        map.put("peopleList", personService.listPeople());

        return "people";
    }

    @RequestMapping(value = "/add", method = RequestMethod.POST)
    public String addPerson(@ModelAttribute("person") Person person, BindingResult result) {

        personService.addPerson(person);

        return "redirect:/people/";
    }

    @RequestMapping("/delete/{personId}")
    public String deletePerson(@PathVariable("personId") Integer personId) {

        personService.removePerson(personId);

        return "redirect:/people/";
    }
}

Answer 1

您是否有controller支持来支持您的GET回复？

像这样的东西

@Controller
@RequestMapping(value = "/people")
public class LoginController {

    @RequestMapping(value = "/i_am_here", method = RequestMethod.GET)
    public String firstForm() {
        return "SHOW_ME_THE_JSP_PAGE";
    }
}

根据以上示例，这将使您的获取URL请求如 - ＆gt; / people / i_am_here

将调用方法，并且可以在JSP中发回回复。

Checkout this example under Github

https://github.com/hth/StatusInvoke/blob/master/src/com/example/UserController.java

Spring servlet映射不起作用

1 个答案: