Could not open ServletContext resource

kuncajs picture kuncajs · May 17, 2011 · Viewed 107.9k times · Source

This is quite similar question to one older but the solution did not work for me.

I have a WAR package.

In web.xml

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>classpath:application-context.xml</param-value>
</context-param>

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

In application-context.xml

<bean id="placeholderConfig" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
    <property name="location">
        <value>classpath:social.properties</value>
    </property>
</bean>

But getting this:

org.springframework.beans.factory.BeanInitializationException: Could not load properties; nested exception is java.io.FileNotFoundException: Could not open ServletContext resource [/social.properties]

I checked the WAR package - .xml and .properties files are both in /WEB-INF/classes

.properties file is in src/main/resources and .xml in src/main/java (in default package both) and maven transports them (I think) correctly in the default package of WEB-INF/classes

Does anyone know why i could get this exception? Thank you.

EDIT: I just want to add that JUnit tests goes correctly (i mean they load what they should from social.properties) but when running the app it ignores my classpath: prefix

Answer

Roman K picture Roman K · Feb 7, 2012

Do not use classpath. This may cause problems with different ClassLoaders (container vs. application). WEB-INF is always the better choice.

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/spring-config.xml</param-value>
</context-param>

and

<bean id="placeholderConfig" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
     <property name="location">
         <value>/WEB-INF/social.properties</value>
     </property>
</bean>