既不是BindingResault也不是普通目标对象…异常

Question

是的，我读到这是一个很常见的问题，但是阅读这些帖子并没有真正帮助我。

小故事是我想在showAllComments.jsp上提交一个表单

<form:form method="post" action="postNewComment.html">
        <table>
            <tr>
                <td><form:label path="comment">
                        COMMENT
                    </form:label></td>
                <td><form:input path="comment" /></td>
            </tr>
            <tr>
                <td colspan="2"><input type="submit"
                    value="WRITE" /></td>
            </tr>
        </table>
    </form:form>

以下是控制器：

@Controller
@SessionAttributes
public class CommentController {


    @Autowired
    private CommentService commentService;


    @RequestMapping(value = "/postNewComment", method = RequestMethod.POST)
    public ModelAndView showAllUsers(@ModelAttribute("command") Comment comment, BindingResult result) {


        System.out.println(comment.getComment());

        Map<String, Object> model = new HashMap<String, Object>();
        model.put("COMMENTS", commentService.getComments());

        return new ModelAndView("showAllComments", model);
    }
}

结果如下:Java . lang . illegalstateexception:bean名称“command”的BindingResult和plain target对象都不能作为请求属性

但您可能需要从一开始就了解整个故事：用户在索引上启动应用程序.jsp通过单击

<a href="toLoginPage.html">Log in</a> 

该链接将他带到登录控制器

@RequestMapping("/toLoginPage")
    public ModelAndView goToLoginPage() {
        return new ModelAndView("login", "command", new User());
}

然后他被带到login.jsp，在那里他提供了自己的用户名和密码。

<form:form method="post" action="log_in.html">

        <input type="text" name="uName" />
        <input type="password" name="pW" />

        <input type="submit" value="Log IN">
    </form:form> 

他提交了表格，然后被带回LoginController

@RequestMapping(value = "/log_in", method = RequestMethod.POST)
    public ModelAndView tryToLogin(@RequestParam("uName") String uName, @RequestParam("pW") String pW, HttpServletResponse response, HttpServletRequest request) {
        ModelAndView ret = new ModelAndView("login", "command", new User());
        User user = userService.existingUser(uName, pW);
        loggedInUser = new User();
        if (user != null) {
            Map<String, Object> model = new HashMap<String, Object>();
                model.put("COMMENTS", allComments);
                model.put("LOGGED_IN_USER", loggedInUser);
            ret = ModelAndView("showAllComments", model);
        }
        return ret;
    }

成功登录后，他在showAllComments页面上看到了所有的评论，他应该能够添加自己的评论，但提交上述表单会引发上述异常。我觉得少了点什么，但我不知道是什么。为了便于记录，我展示了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>Spring3MVC</display-name>

    <servlet>
        <servlet-name>spring</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>spring</servlet-name>
        <url-pattern>*.html</url-pattern>
    </servlet-mapping>

    <welcome-file-list>
        <welcome-file>index.jsp</welcome-file>
    </welcome-file-list>
</web-app>

还有spring-servlet.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:context="http://www.springframework.org/schema/context"
    xmlns:tx="http://www.springframework.org/schema/tx"
    xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">

    <context:property-placeholder location="classpath:jdbc.properties" />
    <context:component-scan base-package="net" />

    <tx:annotation-driven transaction-manager="hibernateTransactionManager" />

    <bean id="viewResolver"
        class="org.springframework.web.servlet.view.UrlBasedViewResolver">
        <property name="viewClass"
            value="org.springframework.web.servlet.view.JstlView" />
        <property name="prefix" value="/WEB-INF/jsp/" />
        <property name="suffix" value=".jsp" />
    </bean>

    <bean id="dataSource"
        class="org.springframework.jdbc.datasource.DriverManagerDataSource">
        <property name="driverClassName" value="${database.driver}" />
        <property name="url" value="${database.url}" />
        <property name="username" value="${database.user}" />
        <property name="password" value="${database.password}" />
    </bean>

    <bean id="sessionFactory"
        class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean">
        <property name="dataSource" ref="dataSource" />
        <property name="annotatedClasses">
            <list>
                <value>net.model.User</value>
                <value>net.model.Comment</value>
            </list>
        </property>
        <property name="hibernateProperties">
            <props>
                <prop key="hibernate.dialect">${hibernate.dialect}</prop>
                <prop key="hibernate.show_sql">${hibernate.show_sql}</prop>
            </props>
        </property>
    </bean>

    <bean id="hibernateTransactionManager"
        class="org.springframework.orm.hibernate3.HibernateTransactionManager">
        <property name="sessionFactory" ref="sessionFactory" />
    </bean>

    <bean id="messageSource"
        class="org.springframework.context.support.ReloadableResourceBundleMessageSource">
        <property name="basename" value="classpath:messages" />
        <property name="defaultEncoding" value="UTF-8" />
    </bean>

    <bean id="localeChangeInterceptor"
        class="org.springframework.web.servlet.i18n.LocaleChangeInterceptor">
        <property name="paramName" value="lang" />
    </bean>

    <bean id="localeResolver"
        class="org.springframework.web.servlet.i18n.CookieLocaleResolver">
        <property name="defaultLocale" value="en" />
    </bean>

    <bean id="handlerMapping"
        class="org.springframework.web.servlet.mvc.annotation.DefaultAnnotationHandlerMapping">
        <property name="interceptors">
            <ref bean="localeChangeInterceptor" />
        </property>
    </bean>

</beans>

Answer 1

在显示showAllComments时，需要添加表单bean类ie Comment作为模型的属性。jsp。

@RequestMapping(value = "/log_in", method = RequestMethod.POST)
public ModelAndView tryToLogin(@RequestParam("uName") String uName, @RequestParam("pW") String pW,      HttpServletResponse response, HttpServletRequest request) {
    ModelAndView ret = new ModelAndView("login", "command", new User());
    User user = userService.existingUser(uName, pW);
    loggedInUser = new User();
    model.addAttribute("command", new Comment());
    if (user != null) {
        Map<String, Object> model = new HashMap<String, Object>();
            model.put("COMMENTS", allComments);
            model.put("LOGGED_IN_USER", loggedInUser);
        ret = ModelAndView("showAllComments", model);
    }
    return ret;
}

这应该没问题。

更新

使用“command”作为命令对象名称不是一个好的做法。对于类注释，您可以使用“注释”或类似的内容。如果这样做，请使用以下代码更新表单。

<form:form method="post" action="postNewComment.html" commandName="comment">
    <table>
        <tr>
            <td><form:label path="comment">
                    COMMENT
                </form:label></td>
            <td><form:input path="comment" /></td>
        </tr>
        <tr>
            <td colspan="2"><input type="submit"
                value="WRITE" /></td>
        </tr>
    </table>
</form:form>

在所有其他地方做同样的改变，即

model.addAttribute("comment", new Comment());

和

@ModelAttribute("comment")

更新2

    @RequestMapping(value="userRegistration", method = RequestMethod.GET)
public ModelAndView showUserRegistrationForm(Model model){
    model.addAttribute("user", new AccountDetailsForm());
    return new ModelAndView("userRegistration");
}