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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T03:29:09+00:00 2026-05-14T03:29:09+00:00

Here is part of my web.xml: <context-param> <param-name>contextConfigLocation</param-name> <param-value> classpath:application-config.xml </param-value> </context-param> application-config.xml uses

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Here is part of my web.xml:

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

application-config.xml uses property placeholder:

<context:property-placeholder location="classpath:properties/db.properties"/>

Is it possible somehow to define which properties file to use in web.xml rather than in application-config.xml?

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  1. Editorial Team
    Editorial Team
    2026-05-14T03:29:09+00:00Added an answer on May 14, 2026 at 3:29 am

    Yes, you can use ServletContextParameterFactoryBean to expose context-param value (it also requires full form of PropertyPlaceholderConfigurer instead of simple context:property-placeholder):

    <bean id = "myLocation" class = 
        "org.springframework.web.context.support.ServletContextParameterFactoryBean">
        <property name="initParamName" value = "myParameter" />
    </bean>
    
    <bean class = 
        "org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
        <property name="location" ref = "myLocation" />
    </bean>
    

    Or use Spring 3.0’s EL:

    <bean class = 
        "org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
        <property name="location" value = "#{contextParameters.myParameter}" />
    </bean>
    
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