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java - Spring MVC 表单错误

转载 作者：行者123 更新时间：2023-12-01 18:08:05 26 4
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我正在尝试在 Spring MVC 中运行一个项目。这是代码

index.jsp

<%@page contentType="text/html" 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>Welcome to Spring Web MVC project</title>
    </head>

    <body>
        <h1>Spring 3 Register!</h1>
        <a href="register.htm">click</a>
        <form:form action="${pageContext.request.contextPath}/register" method="POST" modelAttribute="userForm">
            <table>
                <tr>
                    <td colspan="2" align="center">Spring MVC Form Demo - Registration</td>
                </tr>
                <tr>
                    <td>User Name</td>
                    <td><form:input path="username" /></td>
                </tr>
                <tr>
                    <td>Password</td>
                    <td><form:password path="password" /></td>
                </tr>
                <tr>
                    <td>Email</td>
                    <td><form:input path="email" /></td>
                </tr>
                <tr>
                    <td>BirthDate (mm/dd/yyyy)</td>
                    <td><form:input path="birthDate" /></td>
                </tr>
                <tr>
                    <td>Profession</td>
                    <td><form:select path="profession" items="${professionList}" /></td>
                </tr>
                <tr>
                    <td colspan="2" align="center"><input type="submit" value="Register" /></td>
                </tr>

            </table>
        </form:form>
    </body>
</html>

RegistrationController.java

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

package RegisterInfo;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;

/**
 *
 * @author Harshit Shrivastava
 */
import RegisterInfo.model.User;

import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.ui.Model;

@Controller
@RequestMapping(value = "/register")
public class RegistrationController {

    @RequestMapping(method = RequestMethod.GET)
    public String viewRegistration(Model model)
    {
        User userForm = new User();
                model.addAttribute("userForm", new User());

                /*List<String> professionList = new ArrayList();
                professionList.add("Developer");
                professionList.add("Designer");
                professionList.add("IT Manager");
                model.put("professionList", professionList);*/
        return "index";
    }

    @RequestMapping(method = RequestMethod.POST)
    public String processRegistration(@ModelAttribute("userForm") User user, Map<String, Object> model)
    {       
        System.out.println("Username : " + user.getUserName());
        model.put("userForm", new User());
        return "index";
    }
}

用户.java

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

package RegisterInfo.model;

/**
 *
 * @author Harshit Shrivastava
 */
import java.util.Date;

public class User {
    private String username;
    private String password;
    private String email;
    private Date birthDate;
    private String profession;

    public String getUserName()
    {
        return username;
    }
    public void setUserName(String username)
    {
        this.username = username;
    }
        public String getPassword()
    {
        return password;
    }
    public void setPassword(String password)
    {
        this.password = password;
    }
        public String getEmail()
    {
        return email;
    }
    public void setEmail(String email)
    {
        this.email = email;
    }
        public Date getBirthDate()
    {
        return birthDate;
    }
    public void setBirthDate(Date birthDate)
    {
        this.birthDate = birthDate;
    }
        public String getProfession()
    {
        return profession;
    }
    public void setProfession(String profession)
    {
        this.profession = profession;
    }
}

调度程序-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"
       xmlns:context="http://www.springframework.org/schema/context"
       xmlns:mvc="http://www.springframework.org/schema/mvc"
       xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
       http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.1.xsd
       http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.1.xsd 
       http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.1.xsd
       http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.1.xsd ">

    <bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>

    <context:component-scan base-package="SpringRegister" />
    <mvc:annotation-driven />
    <!--
    Most controllers will use the ControllerClassNameHandlerMapping above, but
    for the index controller we are using ParameterizableViewController, so we must
    define an explicit mapping for it.
    -->
    <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" />

    <!--
    The index controller.
    -->
    <bean name="indexController"
          class="org.springframework.web.servlet.mvc.ParameterizableViewController"
          p:viewName="index" />

</beans>

在上面的程序中，每当我访问http://localhost:8080/SpringRegister/index.htm时，我总是会收到此错误

错误:

neither bindingresult nor plain target object for bean 'userForm' available as request attribute

代码有什么问题？

最佳答案

当您点击http://localhost:8080/SpringRegister/index.htm时从浏览器，您的请求由 ParameterizedViewController 处理，它不设置模型属性 (userForm) 并转发到 index.jsp，index.jsp 的表单标记需要模型属性 userForm <form:form action="${pageContext.request.contextPath}/register" method="POST" modelAttribute="userForm"> ，这是不可用的。这就是错误的原因。要解决此问题，您可以创建自己的 Controller 来处理您的 http://localhost:8080/SpringRegister/index.htm网址，

ModelAndView modelAndView  = new ModelAndView("your view name");
model.addAttribute("userForm", new User());

在 servlet xml 中配置该 Controller 而不是 class="org.springframework.web.servlet.mvc.ParameterizedViewController"。希望这会有所帮助。

关于java - Spring MVC 表单错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34758089/

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