Could not open ServletContext resource

Question

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

I have a WAR package.

In web.xml

Answer 1

Put the things like /src/main/resources/foo/bar.properties and then reference them as classpath:/foo/bar.properties.

Answer 2

Try to use classpath*: prefix instead.

http://static.springsource.org/spring/docs/3.0.x/spring-framework-reference/html/resources.html#resources-classpath-wildcards

Also please try to deploy exploded war, to ensure that all files are there.

Answer 3

Mark sure propertie file is in "/WEB-INF/classes" try to use

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

Answer 4

I had the same error.

My filename was jpaContext.xml and it was placed in src/main/resources. I specified param value="classpath:/jpaContext.xml".

Finally I renamed the file to applicationContext.xml and moved it to the WEB-INF directory and changed param value to /WEB-INF/applicationContext.xml, then it worked!

Answer 5

Are you having Tomcat unpack the WAR file? It seems that the files cannot be found on the classpath when a WAR file is loaded and it is not being unpacked.

Answer 6

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>