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?

Do you have a org.springframework.web.context.ContextLoaderListener registered as a element in your web.xml? — Rob Lockwood-Blake, Mar 12 '14 at 11:56
Right, well that is causing the issue then. By default the ContextLoaderListener will look for a Spring configuration at /WEB-INF/applicationContext.xml. Simply remove the element and things should work fine. — Rob Lockwood-Blake, Mar 12 '14 at 11:59
@RobBlake Good catch! Copy it to answer so I can accept it. Thanks a lot — Leos Literak, Mar 12 '14 at 12:01

score 72 · Accepted Answer · edited Dec 25 '14 at 11:26

72

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.

edited Dec 25 '14 at 11:26

AdrieanKhisbe

3,899
8
37
45

answered Mar 12 '14 at 12:01

gerrytan

40,313
9
84
99

It is not shared but becomes the parent context or root context. Servlet contexts become children and thus can see beans in the parent but not the other way around. Which might have been what you meant. – Adam Gent Mar 12 '14 at 12:13
This solved a completely different issue that has been plaguing me for the last day. Good on you. – DasBooten Mar 03 '16 at 17:33
What is "somewhere-else" supposed to be - sorry, im completely new to this? – Vic Torious Jan 05 '17 at 01:48

Aalkhodiry · Answer 2 · 2015-03-20T20:09:49.863

Update: This will create a second context same as in applicationContext.xml

or you can add this code snippet to your 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

<servlet>
    <servlet-name>dispatcher</servlet-name>
    <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
    <load-on-startup>1</load-on-startup>
</servlet>

Can you explain the difference? explanation? – Asif Mushtaq Sep 15 '18 at 14:05 — Asif Mushtaq, Sep 15 '18 at 14:05

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

2 Answers2

Linked