在spring mvc项目中缺少dispatcher-servlet.xml

Question

我在Spring工具套件中创建了一个Spring mvc项目。我正在关注YouTube教程视频，视频的作者是指将一些文本添加到dispatcher-servlet.xml中，但是在创建Spring项目时并未创建dispatcher-servlet.xml。这是为什么？它可以被称为别的东西吗？

Answer 1

我认为有几种方法可以创建简单的spring web项目。我知道的一些方式是：

1. 从eclipse（STS）创建动态Web项目/并把春天lirary它。

2. 形式STS文件/新建/ Maven项目（org.apache.maven.archetypes）/填写组ID，工件ID v.v ..

3. 从STS文件/新建/ Spring项目/简单的Spring Web Maven的。

4. Spring Roo的命令web mvc setup

不属于如何创建项目后，创建我们总是有文件的web.xml web.xml中的

内容类似这样

<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://java.sun.com/xml/ns/javaee" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd" id="WebApp_ID" version="3.0"> 

<!-- The most important thing to understand is that there are 2 layers of application contexts, 
and they each have default XML files to load beans from 
1.Servlet Application Context 
2.Root application context 
    --> 

    <display-name>Archetype Created Web Application</display-name> 
    <!-- 
    The Each Spring DispatcherServlet defined in the web.xml file gets its own application context. 
    If there is a Root application context (i.e. if a ContextLoaderListener is defined), then 
    that will be the parent of the servlet context and the beans will be available for use in 
    the servlet context(s). By default, DispatcherServlet loads beans from the file 
    /WEB-INF/<servlet-name>-servlet.xml --> 
    <servlet> 
     <servlet-name>dispatcher</servlet-name> 
     <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> 
     <init-param> 
      <param-name>contextConfigLocation</param-name> 
      <param-value>WEB-INF/dispatcher-servlet.xml</param-value> 
     </init-param> 
     <load-on-startup>1</load-on-startup> 
    </servlet> 
    <servlet-mapping> 
     <servlet-name>dispatcher</servlet-name> 
     <url-pattern>*.html</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> 

    <!--The Root application context is created 
     when you specify a ContextLoaderListener in your web.xml file. If you don't 
     specify a listener, then there will be no Root context (which is valid). 
     It allows you to define beans that will be available to all the servlet 
     contexts.REMEMBER: This context is OPTIONAL, and will only be created if 
     you specify a ContextLoaderListener --> 
    <listener> 
     <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> 
    </listener> 
    <context-param> 
     <param-name>contextConfigLocation</param-name> 
     <param-value> 
      /WEB-INF/spring-security.xml, 
      /WEB-INF/spring-database.xml 
     </param-value> 
    </context-param> 

    <filter> 
     <filter-name>springSecurityFilterChain</filter-name> 
     <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class> 
    </filter> 

    <filter-mapping> 
     <filter-name>springSecurityFilterChain</filter-name> 
     <url-pattern>/</url-pattern> 
    </filter-mapping> 
</web-app> 

声明

<init-param> 
     <param-name>contextConfigLocation</param-name> 
     <param-value>WEB-INF/dispatcher-servlet.xml</param-value> 
</init-param> 

是dispatcher-servlet xml配置文件的位置。如果缺少此声明，则缺省值将是direcrory webapp（或WebContent）中的文件名dispatcher-servlet.xml。如果缺少dispatcher-servlet.xml，不用担心，只需在那里创建文件即可。我们可以把这个declation到web.xml文件如上

<init-param> 
      <param-name>contextConfigLocation</param-name> 
      <param-value>WEB-INF/example/funny.xml</param-value> 
</init-param> 

认为它可以帮助

改变dispathcher servlet的XML配置文件的名称和位置