我是 Java Spring 框架的新手,我正在创建简单的重定向应用程序。但是我收到了 http 404 错误 代码如下
applicationContext.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:aop="http://www.springframework.org/schema/aop"
xmlns:c="http://www.springframework.org/schema/c"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:flow="http://www.springframework.org/schema/webflow-config"
xmlns:jee="http://www.springframework.org/schema/jee"
xmlns:jms="http://www.springframework.org/schema/jms"
xmlns:lang="http://www.springframework.org/schema/lang"
xmlns:osgi="http://www.springframework.org/schema/osgi"
xmlns:p="http://www.springframework.org/schema/p"
xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:util="http://www.springframework.org/schema/util"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-4.0.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-4.0.xsd
http://www.springframework.org/schema/webflow-config http://www.springframework.org/schema/webflow-config/spring-webflow-config-2.0.xsd
http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee-4.0.xsd
http://www.springframework.org/schema/jms http://www.springframework.org/schema/jms/spring-jms-4.0.xsd
http://www.springframework.org/schema/lang http://www.springframework.org/schema/lang/spring-lang-4.0.xsd
http://www.springframework.org/schema/osgi http://www.springframework.org/schema/osgi/spring-osgi-1.2.xsd
http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-4.0.xsd
http://www.springframework.org/schema/util http://www.springframework.org/schema/util/spring-util-4.0.xsd">
<context:component-scan base-package="com.webservlet"/>
</beans>
调度器-servlet.xml
<?xml version='1.0' encoding='UTF-8' ?>
<!-- was: <?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: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-4.0.xsd
http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-4.0.xsd
http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-4.0.xsd" >
<bean class="org.springframework.web.servlet.mvc.support.ControllerClassNameHandlerMapping"/>
<bean class="com.webservlet.LoginServlet">
</bean>
<!--
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="indexMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="index.htm">indexController</prop>
</props>
</property>
</bean>
<bean id="homeMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="home.htm">homeController</prop>
</props>
</property>
</bean>
<bean id="loginMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="login.htm">loginController</prop>
</props>
</property>
</bean>
<bean id="about_usMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props>
<prop key="about_us.htm">about_usController</prop>
</props>
</property>
</bean>
<bean id="contact_usMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="mappings">
<props> <prop key="contact_us.htm">contact_usController</prop>
</props>
</property>
</bean>
<!--
The index controller.
-->
<bean name="indexController"
class="org.springframework.web.servlet.mvc.ParameterizableViewController"
p:viewName="index" />
<bean name="homeController"
class="org.springframework.web.servlet.mvc.ParameterizableViewController"
p:viewName="home" />
<bean name="loginController"
class="org.springframework.web.servlet.mvc.ParameterizableViewController"
p:viewName="login" />
<bean name="about_usController"
class="org.springframework.web.servlet.mvc.ParameterizableViewController"
p:viewName="about_us" />
<bean name="contact_usController"
class="org.springframework.web.servlet.mvc.ParameterizableViewController"
p:viewName="contact_us" />
<!--
The viewResolver
-->
<bean id="viewResolver"
class="org.springframework.web.servlet.view.InternalResourceViewResolver"
p:prefix="/WEB-INF/jsp/"
p:suffix=".jsp" />
</beans>
LoginServlet
。
/*
* 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 com.webservlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
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.config.annotation.EnableWebMvc;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.servlet.mvc.ParameterizableViewController;
/**
*
* @author sharathsind
*/
@Controller
@EnableWebMvc
public class LoginServlet {
@RequestMapping(value="/login.do",method = RequestMethod.POST)
public ModelAndView handleRequestInternal() {
return new ModelAndView("home");
}
}
登录.jsp
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Login Page</title>
</head>
<body>
<script>
function display(string) {
document.getElementById("form").action = string;
}
</script>
<form method="post" id="form" onsubmit="display('login.do')">
UserName: <input type="text" name="userName"><br>
PassWord: <input type="password" name="passWord">
<input type="submit" value="submit"/>
</form>
</body>
</html>
当我重定向到 login.do 时它没有显示。任何人都可以帮助我 提前致谢
最佳答案
您的配置有缺陷,更不用说它配置太多了。对于初学者,删除 @EnableWebMvc
来自你的 LoginServlet
因为它什么也没做。它应该放在 @Configuration
上基于类并在您使用 XML 的基于 Java Config 的应用程序中使用。
applicationContext.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"
xsi:schemaLocation="
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd">
<context:component-scan base-package="com.webservlet">
<context:exclude-filter type="annotation" expression="org.springframework.stereotype.Controller" />
</context:component-scan>
</beans>
应该扫描 Controller 以外的所有内容(注意 exclude-filter
)。
dispatcherServlet.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:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context.xsd
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd">
<context:component-scan base-package="com.webservlet" use-default-filters="false">
<context:include-filter type="annotation" expression="org.springframework.stereotype.Controller" />
</context:component-scan>
<mvc:annotation-driven />
<mvc:view-controller path="/index.html" view-name="index" />
<mvc:view-controller path="/home.html" view-name="home" />
<mvc:view-controller path="/login.html" view-name="login />
<mvc:view-controller path="/about_us.html" view-name="about_us" />
<mvc:view-controller path="/contact_us.html" view-name="contact_us" />
<bean id="viewResolver"
class="org.springframework.web.servlet.view.InternalResourceViewResolver"
p:prefix="/WEB-INF/jsp/"
p:suffix=".jsp" />
</beans>
只扫描 Controller ,注意 include-filter
和 use-default-filters="false"
属性。启用 @RequestMapping
用 <mvc:annotation-driven />
解析对于那些讨厌的 View Controller ,请使用 <mvc:view-controller />
.
有关 mvc
的更多信息命名空间我建议阅读 this section引用指南。
另一个技巧是使用无版本的 xsd 文件而不是版本化的文件,这将负责使用类路径上可用的最新版本的 xsd。
关于java - Spring MVC 404错误http请求,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24603981/