java - Could not open ServletContext resource [/WEB-INF/applicationContext.xml]

Question

Ok, I am 500th user asking this question, I read many answers but still having no luck.

parent module pom contains:

<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-plugin and I run my webapp module with jetty:run.

web.xml defines standard dispatcher module:

<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.xml under WEB-INF, though start fails with:

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?

Answer 1

ContextLoaderListener has its own context which is shared by all servlets and filters. By default it will search /WEB-INF/applicationContext.xml

You can customize this by using

<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.