Question

我正在使用Spring框架运行NetBeans。网页显示但没有格式化。它没有找到css和js文件。我做错了什么？

的NetBeans

的web.xml

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/applicationContext.xml</param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>2</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>*.htm</url-pattern>
        <url-pattern>/</url-pattern>
    </servlet-mapping>

    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
    <welcome-file-list>
        <welcome-file>redirect.jsp</welcome-file>
    </welcome-file-list>
</web-app>

的index.jsp

<%@page contentType="text/html" pageEncoding="UTF-8"%>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
<%@ taglib prefix="form" uri="http://www.springframework.org/tags/form" %>

<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <title>MedAI - Medical Artifical Intelligence</title>
    <script type="text/javascript" src="resources/js/jquery.js"></script>     
    <script type="text/javascript" src="resources/js/jquery-ui.js"></script>  
    <script type="text/javascript" src="resources/js/jquery.dataTables.js"></script>        
    <link href="resources/css/jquery-ui.css" rel="stylesheet" type="text/css" />
    <link href="resources/css/medai.css" rel="stylesheet"  type="text/css" />
    <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
    </head>

    <body>
    <div>
            <h1>MedAI</h1>
            <span class=version>V 1.0.0</span>            
    </div>

    <div id="tabs">
    <ul>
            <li><a href="<c:url value='/jsp/homeTab' />">Home</a></li>
            <li><a href="<c:url value='/jsp/geneNetworkTab' />">Gene Network</a></li>
            <li><a href="<c:url value='/jsp/diseaseNetworkTab' />">Disease Network</a></li>
    </ul>  
    </div>
        <script type="text/javascript">
            $(document).ready(function() {
            $('#tabs')
                .tabs()
                //.addClass('ui-tabs-vertical ui-helper-clearfix');
            });    
        </script>    
    </body>
</html>

调度-servlet.xml中

<?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: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"/>

    <!--
    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>

redirect.jsp中

<%--
Views should be stored under the WEB-INF folder so that
they are not accessible except through controller process.

This JSP is here to provide a redirect to the dispatcher
servlet but should be the only JSP outside of WEB-INF.
--%>
<%@page contentType="text/html" pageEncoding="UTF-8"%>
<% response.sendRedirect("index.htm"); %>

Answer 1

您应该在dispatcher-servlet.xml中启用default servlet handler，例如：

<?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: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/mvc
       http://www.springframework.org/schema/mvc/spring-mvc.xsd">

    <mvc:default-servlet-handler/>

</beans>

使用resources mapping代替

或，例如：

<?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: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/mvc
       http://www.springframework.org/schema/mvc/spring-mvc.xsd">

    <mvc:resources mapping="/resources/**" location="/resources/"/>
</beans>

OR 注册默认的servlet请求处理程序，例如：

<?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: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/mvc
           http://www.springframework.org/schema/mvc/spring-mvc.xsd">

    <bean class="org.springframework.web.servlet.mvc.method.annotation.RequestMappingHandlerMapping">
        <property name="defaultHandler">
            <bean class="org.springframework.web.servlet.resource.DefaultServletHttpRequestHandler"/>
        </property>
    </bean>
</beans>

IF 您仍然不知道该怎么做，我建议您先阅读整个Spring WebMvc Documentation。

Answer 2

我有同样的问题，我解决了它为这样的静态内容创建另一个servlet：

在您的调度程序之后将此行添加到您的web.xml：

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>2</load-on-startup>
    </servlet>
    <servlet>
        <servlet-name>resources</servlet-name>
        <servlet-class>com.eproducts.servlets.DefaultServlet</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
    <servlet-mapping>
        <servlet-name>resources</servlet-name>
        <url-pattern>/resources/*</url-pattern>     
    </servlet-mapping>

声明类处理程序：

package com.eproducts.servlets;

import java.io.*;

import javax.servlet.*;
import javax.servlet.http.*;

public class DefaultServlet extends HttpServlet
{   
    private static final long serialVersionUID = 1L;

    // Tomcat, Jetty, JBoss, and GlassFish 
    private static final String COMMON_DEFAULT_SERVLET_NAME = "default";

    // Resin 
    private static final String RESIN_DEFAULT_SERVLET_NAME = "resin-file";

    // WebLogic 
    private static final String WEBLOGIC_DEFAULT_SERVLET_NAME = "FileServlet";

    // WebSphere 
    private static final String WEBSPHERE_DEFAULT_SERVLET_NAME = "SimpleFileServlet";


    public String scanDefaultServlet(){
        if(this.getServletContext().getNamedDispatcher(COMMON_DEFAULT_SERVLET_NAME) != null) {
            return COMMON_DEFAULT_SERVLET_NAME;
        } else if(this.getServletContext().getNamedDispatcher(RESIN_DEFAULT_SERVLET_NAME) != null) {
            return RESIN_DEFAULT_SERVLET_NAME;
        } else if(this.getServletContext().getNamedDispatcher(WEBLOGIC_DEFAULT_SERVLET_NAME) != null){
            return WEBLOGIC_DEFAULT_SERVLET_NAME;
        } else if(this.getServletContext().getNamedDispatcher(WEBSPHERE_DEFAULT_SERVLET_NAME) != null){
            return WEBSPHERE_DEFAULT_SERVLET_NAME;
        } else {
            throw new IllegalStateException("Cannot determine what Server you currently use");
        }       
    }

    public void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        doGet(req, resp);
    }

    public void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException
    {
        RequestDispatcher rd = getServletContext().getNamedDispatcher(this.scanDefaultServlet());
        HttpServletRequest wrapped = new HttpServletRequestWrapper(req) {
                public String getServletPath() {return "";}
        };
        rd.forward(wrapped, resp);
    }
}

完成后，您可以像这样添加资源：

<script type="text/javascript" src="<c:url value="/resources/public/js/jquery-2.1.1.min.js"/>"></script>
<script type="text/javascript" src="<c:url value="/resources/public/js/bootstrap.min.js"/>"></script>

[编辑]：

在您的资源之前添加以下库，以使标记<c:url/>

工作

<%@taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>

Answer 3

在index.jsp中，更改

<script type="text/javascript" src="resources/js/jquery.js"></script>     
    <script type="text/javascript" src="resources/js/jquery-ui.js"></script>  
    <script type="text/javascript" src="resources/js/jquery.dataTables.js"></script>        
    <link href="resources/css/jquery-ui.css" rel="stylesheet" type="text/css" />
    <link href="resources/css/medai.css" rel="stylesheet"  type="text/css" />

到

<script type="text/javascript" src="<c:url value="resources/js/jquery.js" />" ></script>     
    <script type="text/javascript" src="<c:url value="resources/js/jquery-ui.js" />" ></script>  
    <script type="text/javascript" src="<c:url value="resources/js/jquery.dataTables.js" />" ></script>        
    <link href="<c:url value="resources/css/jquery-ui.css" />" rel="stylesheet" type="text/css" />
    <link href="<c:url value="resources/css/medai.css" />" rel="stylesheet"  type="text/css" />

然后在dispatcher-servlet.xml中添加

<mvc:resources mapping="/resources/**" location="/resources/"/>

Answer 4

我终于从头开始想出来了。唯一不对的是这行不应该出现在web.xml中

<url-pattern>/</url-pattern>

为什么Spring不能找到资源文件？

4 个答案: