Java 无法打开 ServletContext 资源 [/WEB-INF/applicationContext.xml]
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Could not open ServletContext resource [/WEB-INF/applicationContext.xml]
提问by Leos Literak
Ok, I am 500th user asking this question, I read many answersbut still having no luck.
好的，我是第 500 个问这个问题的用户，我阅读了很多答案，但仍然没有运气。
parent module pom contains:
父模块 pom 包含：
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-web</artifactId>
<version>${spring.framework.version}</version>
</dependency>
<dependency>
<groupId>org.springframework</groupId>
<artifactId>spring-webmvc</artifactId>
<version>${spring.framework.version}</version>
</dependency>
Child module has maven-jetty-pluginand I run my webapp module with jetty:run.
子模块有maven-jetty-plugin，我用jetty:run.
web.xmldefines standard dispatcher module:
web.xml定义标准调度模块：
<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>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
I have file dispatcher-servlet.xmlunder WEB-INF, though start fails with:
我在dispatcher-servlet.xml下有文件WEB-INF，但启动失败：
FileNotFoundException: Could not open ServletContext resource [/WEB-INF/applicationContext.xml]
What is wrong? Documentation and everybody says that Spring MVC will search for XX-servlet.xml, where XX is name of servlet. Why does it search for applicationContext.xml?
怎么了？文档和大家都说Spring MVC会搜索XX-servlet.xml，其中XX是servlet的名字。它为什么要搜索applicationContext.xml？
采纳答案by gerrytan
ContextLoaderListenerhas its own context which is shared by all servlets and filters. By default it will search /WEB-INF/applicationContext.xml
ContextLoaderListener有自己的上下文，由所有 servlet 和过滤器共享。默认情况下它会搜索/WEB-INF/applicationContext.xml
You can customize this by using
您可以使用
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/somewhere-else/root-context.xml</param-value>
</context-param>
on web.xml, or remove this listener if you don't need one.
on web.xml，或者如果您不需要，则删除此侦听器。
回答by Aalkhodiry
Update:This will create a second context same as in applicationContext.xml
更新：这将创建与 applicationContext.xml 中相同的第二个上下文
or you can add this code snippet to your web.xml
或者您可以将此代码片段添加到您的web.xml
<servlet>
<servlet-name>spring-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>classpath:applicationContext.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
instead of
代替
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
