Question

我是Spring annotation和MVC的新手。我的第一页是home.jsp，现在它没有在Tomcat服务器中出现我在控制台的末尾显示错误。我试图创建一个非常简单的注释应用程序。这是控制器。

import javax.validation.Valid;

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.ui.ModelMap;
import org.springframework.validation.BindingResult;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;

@Controller

    //@RequestMapping()
    public class UserLoginController {

    public UserLoginController(){

    }
        //@RequestMapping(value="/get" , method = RequestMethod.GET)
    //@ModelAttribute("user")
        public String get(final ModelMap model) {

            User userForm = new User();
      model.addAttribute("userLogin", userForm);
       return "form";
        }

        @RequestMapping(value="/home.jsp", method = RequestMethod.POST)
        public String post(Model model) {

           String test=" inside here";

            return "success";
        }
    }

这是JSP

<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form" %>
<%@ page session="false" %>
<%@ include file="/WEB-INF/views/header.jsp" %>

 <html>
<head>
<meta http-equiv="Content-Type" content="text/html">
<title>iBank - Home-version 2.0</title>
</head>
<body>
<h1 align="center">Welcome to iBank-Dhiren</h1>
<h2 align="center">Your Online Bank Portal</h2>
<p align="center">   
Today is ${today}.<br/>
<a href="<%=request.getContextPath()%>/admin.htm">Modified Administration Site-version-1.0 </a>
</p>
</body>
</html>

<p>
    <form:form  method="post" commandName="user">
        <div>
            <form:label path="name">Name:</form:label>
            <form:input path="name"/>
            <form:errors path="name" />
        </div>
        <div>
            <form:label path="email">Email:</form:label>
            <form:input path="email" />
            <form:errors path="email" />
        </div>
        <div>
            <input type="submit" value="  OK  "/>
        </div>
    </form:form>
</p>

</html>

现在如何提交JSP应该是UserLoginController。我无法在注释值之间看到任何连接，并且我的Tomcat服务器也会返回此错误。

SEVERE: Servlet.service() for servlet jsp threw exception
java.lang.IllegalStateException: Neither BindingResult nor plain target object for bean name 'user' available as request attribute
    at org.springframework.web.servlet.support.BindStatus.<init>(BindStatus.java:141)
    at org.springframework.web.servlet.tags.form.AbstractDataBoundFormElementTag.getBindStatus(AbstractDataBoundFormElementTag.java:174)

由于 Dhiren

好的，我无法再进一步了。我尝试了所有但仍然完全陷入困境。

我的JSP home.jsp

<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form" %>
<%@ page session="false" %>
<%@ include file="/WEB-INF/views/header.jsp" %>

 <html>
<head>
<meta http-equiv="Content-Type" content="text/html">
<title>iBank - Home-version 2.0</title>
</head>
<body>
<h1 align="center">Welcome to iBank-Dhiren</h1>
<h2 align="center">Your Online Bank Portal</h2>
<p align="center">   
Today is ${today}.<br/>
<a href="<%=request.getContextPath()%>/admin.htm">Modified Administration Site-version-1.0 </a>
</p>
</body>
</html>

<p>
    <form:form  method="post" action="/user.jsp" modelAttribute="user">

        <div>
            <form:label path="firstName">Name:</form:label>
            <form:input path="firstName"/>
            <form:errors path="firstName" />
        </div>
        <div>
            <form:label path="password">Password:</form:label>
            <form:input path="password" />
            <form:errors path="password" />
        </div>
        <div>
            <form:label path="middleName">Middle name:</form:label>
            <form:input path="middleName" />
            <form:errors path="middleName" />
        </div>
        <div>
            <form:label path="lastName">LastName:</form:label>
            <form:input path="lastName" />
            <form:errors path="lastName" />
        </div>


        <div>
            <input type="submit" value="  OK  "/>
        </div>
    </form:form>
</p>

</html>

我的UserLoginController

import javax.validation.Valid;

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.ui.ModelMap;
import org.springframework.validation.BindingResult;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;

@Controller

    //@RequestMapping()
    public class UserLoginController {

    public UserLoginController(){

    }
        //@RequestMapping(value="/get" , method = RequestMethod.GET)
    //@ModelAttribute("user")
        public String get(final ModelMap model) {

            User userForm = new User();
      model.addAttribute("userLogin", userForm);
       return "form";
        }

        @RequestMapping(value="/user.jsp", method = RequestMethod.POST)
        public String post(final User user, final BindingResult result, Model mv) {

           String test=" inside here";

            return "success";
        }
    }

用户类

import javax.validation.constraints.Size;

import org.hibernate.validator.constraints.NotEmpty;

public class User {
/*  @NotEmpty
    @Size(max = 20)
    */
private String userId;
    /*@NotEmpty
    @Size(max = 20)
*/
private String password;
private String firstName;
private String middleName;
private String lastName;
//private int userAccessLevel;
/**
 * @return the userId
 */
public String getUserId() {
    return userId;
}
/**
 * @param userId the userId to set
 */
public void setUserId(String userId) {
    this.userId = userId;
}
/**
 * @return the password
 */
public String getPassword() {
    return password;
}
/**
 * @param password the password to set
 */
public void setPassword(String password) {
    this.password = password;
}
/**
 * @return the firstName
 */
public String getFirstName() {
    return firstName;
}
/**
 * @param firstName the firstName to set
 */
public void setFirstName(String firstName) {
    this.firstName = firstName;
}
/**
 * @return the middleName
 */
public String getMiddleName() {
    return middleName;
}
/**
 * @param middleName the middleName to set
 */
public void setMiddleName(String middleName) {
    this.middleName = middleName;
}
/**
 * @return the lastName
 */
public String getLastName() {
    return lastName;
}
/**
 * @param lastName the lastName to set
 */
public void setLastName(String lastName) {
    this.lastName = lastName;
}
/**
 * @return the userAccessLevel
 *
public int getUserAccessLevel() {
    return userAccessLevel;
}
*/
/**
 * @param userAccessLevel the userAccessLevel to set
 *
public void setUserAccessLevel(int userAccessLevel) {
    this.userAccessLevel = userAccessLevel;
}*/

}

这些是context.xml文件 servlet-context.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:beans="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
        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">

    <!-- DispatcherServlet Context: defines this servlet's request-processing infrastructure -->

    <!-- Enables the Spring MVC @Controller programming model -->
    <annotation-driven />

    <!-- Handles HTTP GET requests for /resources/** by efficiently serving up static resources in the ${webappRoot}/resources directory -->
    <resources mapping="/resources/**" location="/resources/" />

    <!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
    <beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <beans:property name="prefix" value="/WEB-INF/views/" />
        <beans:property name="suffix" value=".jsp" />
    </beans:bean>

        <!-- Imports user-defined @Controller beans that process client requests -->

    <context:component-scan base-package="mytest.apps" />

</beans:beans>

有人可以告诉我为什么home.jsp没有出现

web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
    <display-name>appServlet</display-name>
    <!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
    <context-param>  
  <param-name>log4jConfigLocation</param-name>  
  <param-value>/WEB-INF/log4j.xml</param-value>  
</context-param>  
<listener>  
  <listener-class>org.springframework.web.util.Log4jConfigListener</listener-class>  
</listener> 

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring/appServlet/root-context.xml</param-value>
    </context-param>

    <!-- Creates the Spring Container shared by all Servlets and Filters -->
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <!-- Processes application requests -->
    <servlet>
        <servlet-name>appServlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>appServlet</servlet-name>
        <url-pattern>/</url-pattern>

    </servlet-mapping>
</web-app>

当我尝试访问Web应用程序的第一页时出现所有这些错误，我在Tomcat中收到此错误。

INFO: At least one JAR was scanned for TLDs yet contained no TLDs. Enable debug logging for this logger for a complete list of JARs that were scanned where no TLDs were found. Skipping JAR scanning can improve startup time and JSP compilation time.
log4j:ERROR Attempted to append to closed appender named [console].
Sep 14, 2011 10:02:39 PM org.apache.catalina.core.ApplicationDispatcher invoke
SEVERE: Servlet.service() for servlet jsp threw exception
java.lang.IllegalStateException: Neither BindingResult nor plain target object for bean name 'user' available as request attribute
    at org.springframework.web.servlet.support.BindStatus.<init>(BindStatus.java:141)
    at org.springframework.web.servlet.tags.form.AbstractDataBoundFormElementTag.getBindStatus(AbstractDataBoundFormElementTag.java:174)
    at org.springframework.web.servlet.tags.form.AbstractDataBoundFormElementTag.getPropertyPath(AbstractDataBoundFormElementTag.java:194)
    at org.springframework.web.servlet.tags.form.LabelTag.autogenerateFor(LabelTag.java:129)
    at org.springframework.web.servlet.tags.form.LabelTag.resolveFor(LabelTag.java:119)
    at org.springframework.web.servlet.tags.form.LabelTag.writeTagContent(LabelTag.java:89)
    at org.springframework.web.servlet.tags.form.AbstractFormTag.doStartTagInternal(AbstractFormTag.java:102)
.
.
Sep 14, 2011 10:02:39 PM org.apache.catalina.core.StandardWrapperValve invoke
SEVERE: Servlet.service() for servlet [appServlet] in context with path [/AdministrativeApplication] threw exception [An exception occurred processing JSP page /WEB-INF/views/home.jsp at line 25

22:     <form:form  method="post" action="/user.jsp" modelAttribute="user">
23:      
24:         <div>
25:             <form:label path="firstName">Name:</form:label>
26:             <form:input path="firstName"/>
27:             <form:errors path="firstName" />
28:         </div>


Stacktrace:] with root cause
java.lang.IllegalStateException: Neither BindingResult nor plain target object for bean name 'user' available as request attribute
    at org.springframework.web.servlet.support.BindStatus.<init>(BindStatus.java:141)

请帮忙

由于 Dhiren

Answer 1

class User() {
    String name;
    String email;

    //bean getters and setters
}

public String post(Model model, User user) {
   ....
}

我几天前刚刚开始使用弹簧绑定。

您设置的方式我相信您的表单会尝试使用“email”和“name”属性创建“User”对象。

修改的

我想我理解这个问题......

<form:form  method="post" commandName="user">

这告诉你，你想用来自用户对象的数据预先填充表单。如果这是你想要做的，那么你需要为jsp提供一个名为“user”的请求属性。

如果您尝试使用该表单提交数据，我认为是这种情况，我认为您要使用此代码

<form:form  method="post" modelAttribute="user">

Answer 2

model.addAttribute("**userLogin**", userForm); // in controller
return "form"; //in controller

在JSP中你应该给

form:form  method="post" action="/user.jsp" modelAttribute="userLogin"

（而不是modelAttribute/commandName="user"）

试试这个你不会

java.lang.IllegalStateException: Neither BindingResult nor plain target object for bean name 'user' available as request attribute" expection.

Answer 3

这很简单............

model.addAttribute("**user**",userForm );

这意味着，您的命令名称应与modelAttribute名称完全相同... 所以如果你的jsp文件中的命令名是＆＃34; user＆＃34;那么你的模型应该添加名为＆＃34; user＆＃34;。

的属性

Answer 4

您的post()方法应该将命令对象作为其参数之一：

public String post(final User user, final BindingResult result) {

在这种情况下，在发布数据后，表单中的字段name和email将分别绑定到user.name和user.email。

Answer 5

在get方法中，添加命令对象

public String get(final ModelMap model) {

  User userForm = new User();
  model.addAttribute("command", userForm);
  return "form";
}

Answer 6

“commandName”用于绑定表单中的模型和数据。如果你有类（MODEL）定义如下：

public class UserQuery {

    public String message;

    public String getMessage()
    {
        return message;
    }
    public void setMessage(String message) {
        this.message = message;
    } 
}

JSP文件（DATA）：

<prefix:form method="POST" commandName="UserQuery">
  <sf:input path="message" /><br/>
  <input type="submit"/>
</prefix:form>

和GET / POST控制器：

@Controller
public class ContactController {

    @RequestMapping(value="/Contact",method=RequestMethod.GET)
    ModelAndView Contact(Model model)
    {
        model.addAttribute("UserQuery",new Query());

        return new ModelAndView("contactView");
    }

    @RequestMapping(value="/Contact",method=RequestMethod.POST)
    ModelAndView Contact()
    {
        //Here you can get you're attribute and render it
        // in other view 
        return new ModelAndView("contactFormFiledView");
    }
}

GET方法 - ＆gt;用于创建属性“UserQuery”并将其与模型“查询”绑定，这是您的类。通过绑定我的意思是：如果用户单击表单中的提交按钮，表单中的数据将传递给“commandName”中指定的属性。

Answer 7

解决您的第一个问题。将home.jsp保存在webapp部分的resources文件夹下。对servlet-xml文件进行适当的更改，如下所示。

<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver"> <property name="prefix"> <value>resources/views/jsp/</value> </property> <property name="suffix"> <value>.jsp</value> </property> </bean>

在控制器中添加一个方法来处理主页的请求。

@RequestMapping(value = "/home", method = RequestMethod.GET) 
public String displayLogin(Model model) { 
    model.addAttribute("login", new Login()); 
    System.out.println("Returning index page");
    return "index"; 
}

您的新网址将是这样的http://localhost:8080/ProjectName/home

确保在上面提到的displayLogin方法中添加model.addAttribute("login", new Login());。

这应该有所帮助问候
BinoyC

Spring的注释关系

7 个答案: