Question

我正在使用spring并且我坚持使用一个例子。如果我解决了一个错误，它会给我不同的错误......

我有一个名为UserController2.java的控制器程序，如下所示

package project2;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.stereotype.Controller;
import org.springframework.transaction.annotation.Transactional;
import org.springframework.ui.ModelMap;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.servlet.ModelAndView;
import org.springframework.web.servlet.mvc.multiaction.MultiActionController;

@Controller
public class UserController2 extends MultiActionController {

    private UserDAO userDAO;

    public void setUserDAO(UserDAO userDAO) {
        this.userDAO = userDAO;
    }
    @RequestMapping(params = "add", method = RequestMethod.POST)
    public ModelAndView add(HttpServletRequest request,
           HttpServletResponse response, User user) throws Exception {
        userDAO.saveUser(user);
        return new ModelAndView("redirect:list.htm");
    }
    @RequestMapping(params = "delete", method = RequestMethod.POST)
    @Transactional
    public ModelAndView delete(HttpServletRequest request,
        HttpServletResponse response, User user) throws Exception {
        userDAO.deleteUser(user);
        return new ModelAndView("redirect:list.htm");
    }

    public ModelAndView list(HttpServletRequest request,
        HttpServletResponse response) throws Exception {
        ModelMap modelMap = new ModelMap();
        modelMap.addAttribute("userList", userDAO.listUser());
        modelMap.addAttribute("user", new User());
        return new ModelAndView("userForm", modelMap);
    }
}

和jsp页面userForm.jsp如下

<body>
 <center> WELCOME TO CUSTOMER ACCESS SITE.PLEASE ENTER THE FOLLOWING 
  INFORMATION</center>
<form:form method="POST" action="add.htm"  modelAttribute="user">
   <table>
    <tr>
        <td><form:label path="name">Name</form:label></td>
        <td><form:input path="name" /></td>
    </tr>

    <tr>
        <td><form:label path="id">id</form:label></td>
        <td><form:input path="id" /></td>
    </tr>

    <tr>
        <td><form:label path="password">password</form:label></td>
        <td><form:input path="password" /></td>
    </tr>
    <tr>
        <td><form:label path="gender">Gender</form:label></td>
        <td><form:input path="gender" /></td>
    </tr>
    <tr>
        <td><form:label path="country">Country</form:label></td>
        <td><form:input path="country" /></td>
    </tr>
    <tr>
        <td colspan="2">
            <input type="submit" value="Submit"/>
        </td>
    </tr>
 </table>  
</form:form>

</body>

和UserController2-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:p="http://www.springframework.org/schema/p"
   xmlns:aop="http://www.springframework.org/schema/aop"
   xmlns:tx="http://www.springframework.org/schema/tx"
   xsi:schemaLocation="http://www.springframework.org/schema/beans  
    http://www.springframework.org/schema/beans/spring-beans-2.5.xsd
   http://www.springframework.org/schema/aop http://www.springframework.org/schema 
    /aop/spring-aop-2.5.xsd
   http://www.springframework.org/schema/tx http://www.springframework.org/schema 
    /tx/spring-tx-2.5.xsd">
<bean   
class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>


<bean id="urlMapping"  
class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
    <property name="mappings">
        <props>
            <prop key="index.htm">indexController</prop>
        </props>
    </property>
</bean>

<bean id="viewResolver"
      class="org.springframework.web.servlet.view.InternalResourceViewResolver"
      p:prefix="/WEB-INF/jsp/"
      p:suffix=".jsp" />

<bean id="myDataSource" class="org.apache.commons.dbcp.BasicDataSource" destroy- 
method="close">
    <property name="driverClassName" value="com.mysql.jdbc.Driver"/>
    <property name="url" value="jdbc:mysql://localhost:3306/indi"/>
    <property name="username" value="admin"/>
    <property name="password" value=""/>
</bean>

<bean id="mySessionFactory"   
class="org.springframework.orm.hibernate3.annotation.AnnotationSessionFactoryBean">
    <property name="dataSource" ref="myDataSource" />
    <property name="annotatedClasses">
        <list>
            <value>Spring.project2.User</value>

        </list>
    </property>
    <property name="hibernateProperties">
        <props>
            <prop key="hibernate.dialect">org.hibernate.dialect.MySQLDialect</prop>
            <prop key="hibernate.show_sql">true</prop>
            <prop key="hibernate.hbm2ddl.auto">create</prop>
        </props>
    </property>
</bean>

<bean id="myUserDAO" class="project2.UserDAOImpl">
     <property name="sessionFactory" ref="mySessionFactory"/>
</bean>

 <bean name="/user/*.htm" class="project2.UserController2" >
    <property name="userDAO" ref="myUserDAO" />
</bean>


<bean name="indexController"  
class="org.springframework.web.servlet.mvc.ParameterizableViewController"
      p:viewName="userForm" />

 </beans>

和web.xml文件如下

<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
    <servlet-name>UserController2</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>2</load-on-startup>
 </servlet>
<servlet-mapping>
     <servlet-name>UserController2</servlet-name>
    <url-pattern>*.htm</url-pattern>
</servlet-mapping>

现在我收到以下错误

org.apache.jasper.JasperException: java.lang.IllegalStateException: Neither  
BindingResult nor plain target object for bean name 'user' available as request 
attribute

我在jsp和控制器类中都只使用了“user”。有人告诉我要更改以下代码行

modelMap.addAttribute("user", new User());

到这个

modelMap.addAttribute("user", new user());

但我收到错误

  user cannot be resolved to a type

我已经彻底搜索了网，查看了各种示例，但我无法解析代码。任何建议？

Answer 1

如果我没有错，我认为你正在混合Spring新版本和旧版本的一些概念。在您的控制器中：

使用@Controller并移除extends MultiActionController（春季3路）

OR

从MultiActionController延伸并删除@Controller（春季2路）。您还应该在Spring XML配置文件中声明此bean。

也许这会有所帮助，或者说我错了，但是对我来说用这两种东西宣布一个控制器似乎很奇怪。

此外，您必须从UserController2中删除web.xml。如果您使用的是Spring 2方法，则必须在Spring XML配置文件中声明它。

Answer 2

我遇到了同样的问题。确保在控制器中返回页面userForm.jsp时，您将放入代表您的用户的Model对象。例如，在我的情况下：

@RequestMapping(value = "registration", method = RequestMethod.GET)
public String registration(Model model) {
    model.addAttribute("UserForm", new UserForm()); // !!!!
    return PATH_REGISTRATION;
}

 @RequestMapping(value = "/register", method = RequestMethod.POST)
public String register(
        @Valid @ModelAttribute("UserForm") UserForm userForm,
        BindingResult result, HttpServletRequest request, Model model,
        Principal principal) {
    if (!result.hasErrors()) {
        model.addAttribute("message", "Registration success");
    }
    try {
        userService.create(createUser(userForm));
    } catch (SQLException e) {
        model.addAttribute("message", "Can't create user");
        log.error("Can't create user " + userForm.getLogin(), e);
    }
    return PATH_REGISTRATION;
}

并在JSP中：

...
<form:form method="POST" action="registrate" commandName="UserForm">
....

Answer 3

我在项目中遇到了同样的问题。我通过修改控制器类修复了这个问题。这是代码： LoginForm.jsp页面

<%@ page language="java" contentType="text/html; charset=UTF-8"
    pageEncoding="UTF-8"%>
<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form"%>

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" 
    "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Login</title>
<style>
    .error { 
        color: red; font-weight: bold; 
    }
</style>
</head>
<body>
    <div align="center">
        <h2>Spring MVC Form Validation Demo - Login Form</h2>
        <table border="0" width="90%">
        <form:form action="login" commandName="userFormNew">
                <tr>
                    <td align="left" width="20%">Email: </td>
                    <td align="left" width="40%"><form:input path="email" size="30"/></td>
                    <td align="left"><form:errors path="email" cssClass="error"/></td>
                </tr>
                <tr>
                    <td>Password: </td>
                    <td><form:password path="password" size="30"/></td>
                    <td><form:errors path="password" cssClass="error"/></td>
                </tr>
                <tr>
                    <td></td>
                    <td align="center"><input type="submit" value="Login"/></td>
                    <td></td>
                </tr>
        </form:form>
        </table>
    </div>
</body>
</html>

在LoginForm.jsp页面中观察 commandName =“userFormNew” 在控制器中我有两个方法，一个是自动填充loginForm.jsp页面，当我们尝试使用 @RequestingMapping（value =“/”，method = RequestMethod.GET）运行应用程序时通过重定向LoginForm.jsp页面返回模型对象。

这是Controller类：

@Controller
public class LoginController {
    @RequestMapping(value = "/", method = RequestMethod.GET)
    public String viewLogin(Map<String, Object> model) {
        User user = new User();
        model.put("userFormNew", user);
        return "LoginForm";
    }

    @RequestMapping(value = "/login", method = RequestMethod.POST)
    public String doLogin(@Valid @ModelAttribute("userFormNew") User userForm,
            BindingResult result, Map<String, Object> model) {

        if (result.hasErrors()) {
            return "LoginForm";
        }

        return "LoginSuccess";
    }
}

我的模特课：

public class User {
    @NotEmpty
    @Email
    private String email;

    @NotEmpty(message = "Please enter your password.")
    @Size(min = 6, max = 15, message = "Your password must between 6 and 15 characters")
    private String password;

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }
}

BindingResult和bean名称'user'的普通目标对象都不可用作请求属性

3 个答案: