Question

我收到一条错误消息： org.springframework.web.util.NestedServletException：请求处理失败;嵌套异常是java.lang.ClassCastException：java.lang.Object无法强制转换为com.crimetrack.business.Login

Login.java

public class Login {

    private String userName;
    private String password;
    private boolean loggedin;

    public Login(){};

    /**
     * @return the loggedin
     */
    public boolean isLoggedin() {
        return loggedin;
    }

    /**
     * @param loggedin the loggedin to set
     */
    public void setLoggedin(boolean loggedin) {
        this.loggedin = loggedin;
    }

    /**
     * @param userName
     * @param password
     */
    public Login(String userName, String password) {
        this.userName = userName;
        this.password = password;
    }

    /**
     * @return the userName
     */
    public String getUserName() {
        return userName;
    }

    /**
     * @param userName the userName to set
     */
    public void setUserName(String userName) {
        this.userName = userName;
    }

    /**
     * @return the password
     */
    public String getPassword() {
        return password;
    }

    /**
     * @param password the password to set
     */
    public void setPassword(String password) {
        this.password = password;
    }

}

@Controller
public class AuthenticationController {

    private final Logger logger = Logger.getLogger(getClass());

    private AuthenticationManager authenticationManager;
    private Login login = new Login();

    String message = "Congrulations You Have Sucessfully Login";
    String errorMsg = "Login Unsucessful";

    @RequestMapping(value="login.htm")
    public ModelAndView onSubmit(Object command) throws ServletException {

        String userName = ((Login)command).getUserName();
        String password = ((Login)command).getPassword();

        login.setUserName(userName);
        login.setPassword(password);

        logger.info("Login was set");

        logger.info("the username was set to " + login.getUserName());
        logger.info("the password was set to " + login.getPassword());

        if (authenticationManager.Authenticate(login) == true){
            return new ModelAndView("main","welcomeMessage", message);
        }

        //return new ModelAndView("main","welcomeMessage", message);
        return new ModelAndView("login","errorMsg", "Error!!!");
    }

}

Answer 1

试试这个解决方案：

<强> JSP

<form:form action="yourUrl" modelAttribute="login" method="POST">
<% ... %>
</form:form>

<强>控制器

// your method that prints the form
public ModelAndView onGet(@ModelAttribute Login login) {
    // return ...
}

@RequestMapping(value="login.htm")
public ModelAndView onSubmit(@ModelAttribute Login login) {
    String userName = login.getUserName();
    String password = login.getPassword();
    // ...
}

<强>解释

注释@ModelAttribute与model.addAttribute(String name, Object value)完全相同。例如，@ModelAttribute Login login与model.addAttribute("login", new Login());相同。

也就是说，使用onGet方法，将这样的对象传递给视图。由于属性modelAttribute="login"，标记<form:form>将查看模型的属性列表，以查找名称为login的属性。如果找不到，则抛出异常。

然后，这是神奇的部分：使用标记<form:input path="userName" />，Spring MVC将自动设置userName属性中bean的属性modelAttribute="login"，即在你的情况下，login。如果你输入<form:input path="wtf" />之类的东西，它会引发异常，因为bean Login没有这样的属性。

最后，在您的onSubmit方法上（再次感谢注释@ModelAttribute），您可以访问以前由Spring MVC自动提取的login bean。

注意

我个人（几乎）从不使用ModelAndView实例，但请按以下步骤操作：

// the methods can have the name you want // not only onGet, onPost, etc. as in servlets @RequestMapping("url1.htm") public String loadAnyJsp(@ModelAttribute Login login) { return "path/to/my/views/login"; } @RequestMapping("url2.htm") public String redirectToAnotherController(@ModelAttribute Login login) { return "redirect:url1.htm"; }

JSP的路径在web.xml文件中指定，例如：

... <bean class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver" p:favorPathExtension="true" p:favorParameter="true" p:ignoreAcceptHeader="true" p:defaultContentType="text/html"> <description>Depending on extension, return html with no decoration (.html), json (.json) or xml (.xml), remaining pages are decoracted</description> <property name="mediaTypes"> <map> <entry key="xml" value="application/xml" /> <entry key="json" value="application/json" /> <entry key="html" value="text/html" /> <entry key="action" value="text/html" /> </map> </property> <property name="defaultViews"> <list> <bean class="org.springframework.web.servlet.view.xml.MarshallingView" p:marshaller-ref="xstreamMarshaller" /> <bean class="org.springframework.web.servlet.view.json.MappingJacksonJsonView" /> </list> </property> <property name="viewResolvers"> <list> <bean id="nameViewResolver" class="org.springframework.web.servlet.view.BeanNameViewResolver"> <description>Maps a logical view name to a View instance configured as a Spring bean</description> </bean> <bean class="org.springframework.web.servlet.view.InternalResourceViewResolver" p:prefix="/WEB-INF/views/" p:suffix=".jsp" /> </list> </property> </bean> ...

您应阅读the doc以获取更多信息（参见16.5解析视图）。

Answer 2

在“onSubmit”方法中，您将command投射到Login。

显然不是。

在用户使用Spring MVC提交表单后，如何获取表单的数据？

2 个答案: