HowToDoInJava

Spring version-less schema for latest version

By Lokesh Gupta | Filed Under: Spring Core

If you have worked on Spring framework projects, then you must have seen spring context configuration files (e.g. applicationContext.xml) where in header you specify the schema references (.xsd) for various spring modules. In schema references, we mention the xml namespace and as well as schema version number.

Well specifying schema version number is not mandatory at all, and you can omit it. If fact, you should omit it all the time. Consider it as a best practice to follow.

Spring automatically picks the highest version available from the project dependencies (jars). Also, as the project evolves and the Spring version will be updated, we won’t have to maintain all the XML config files to see the new features.

Example of version-less Spring schema

With Schema Version

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans" 
  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  xmlns:context="http://www.springframework.org/schema/context"
  xsi:schemaLocation="http://www.springframework.org/schema/beans 
  http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
  http://www.springframework.org/schema/context 
  http://www.springframework.org/schema/context/spring-context-3.0.xsd">

  
  <!-- Other bean definitions-->
  
</beans>

WITHOUT Schema Version

This can be written as:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans" 
  xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
  xmlns:context="http://www.springframework.org/schema/context"
  xsi:schemaLocation="http://www.springframework.org/schema/beans 
  http://www.springframework.org/schema/beans/spring-beans.xsd
  http://www.springframework.org/schema/context 
  http://www.springframework.org/schema/context/spring-context.xsd">

  
  <!-- Other bean definitions-->
  
</beans>

Drop me your questions in comments section.

Happy Learning !!

LinkedinRedditPocket

About Lokesh Gupta

Founded HowToDoInJava.com in late 2012. I love computers, programming and solving problems everyday. A family guy with fun loving nature. You can find me on Facebook, Twitter and Google Plus.

Feedback, Discussion and Comments

  1. Arpit Chinmay

    Hi,
    i’m new to Spring and while learning i encountered this error and could not resolve it.I did a little research and there is no article in stackOverflow that explains this.

    cvc-complex-type.2.4.c: The matching wildcard is strict, but no declaration can be found for element ‘context:property-placeholder’.

    <?xml version="1.0" encoding="UTF-8"?>
<beans xmlns = "http://www.springframework.org/schema/beans"
       xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance"
       xmlns:context = "http://www.springframework.org/schema/context"
       xsi:schemaLocation= "http://www.springframework.org/schema/beans
                 http://www.springframework.org/schema/beans/spring-beans.xsd
                 http://www.springframework.org.schema/context
                 http://www.springframework.org/schema/context/spring-context.xsd ">
                 
        <context:property-placeholder location="classpath:sport.properties"/>
                
        <bean id = "myFortune"
              class =  "com.love2code.springdemo.HappyFortuneService">  
        </bean>        
         
       <bean id = "myCoach"
             class = "com.love2code.springdemo.TrackCoach">
       <constructor-arg ref= "myFortune" />
       </bean>          
       
       <bean id="myCricketCoach"
            class = "com.love2code.springdemo.CricketCoach">
            <property name = "fortuneService" ref = "myFortune" />
            
            <!-- reference the loaded values from the properties file -->
            <property name = "emailAddress" value = "${foo.email}"/>
            <property name="name" value = "${foo.name}"/>
      </bean>
</beans>
    Reply

  2. Amol Gangawane

    Is it applicable to spring -security xsd also??

    Reply

  3. Gaurav Sachdeva

    How to remove version in this “”http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd”>””
    This is showing error in Web.xml
    Thanks

    Reply

  4. Kwen

    Hello,
    I’m using Spring 3.0.7.RELEASE but i have some problems with eclipse validation, the attribute local in “” cannot found in XSD if i remove the version in schemaLocation.

    How can I do ?

    My application-context :

    Thanx for help

    Reply

    • Lokesh

      Please put the code in code tag

      Reply

      • Kwen

        Sorry, i replace some characters by “-”

        -?xml version=”1.0″ encoding=”UTF-8″?-
        -beans xmlns=”http://www.springframework.org/schema/beans”
        xmlns:xsi=”http://www.w3.org/2001/XMLSchema-instance”
        xsi:schemaLocation=”http://www.springframework.org/schema/beans
        http://www.springframework.org/schema/beans/spring-beans-3.0.xsd“-
        -bean id=”myDaoTarget” class=”com.test.dao.MyDaoImpl”/-
        -bean id=”myDao” class=”org.springframework.aop.framework.ProxyFactoryBean”-
        -property name=”proxyInterfaces” value=”com.test.dao.MyDao” /-
        -property name=”target”–ref local=”myDaoTarget”/–/property-
        -/bean-
        -/beans-

        Reply

  5. Jayakumar Jayaraman

    That’s nice to know. Are there any impact because of this feature ?

    For instance we might be developing a project currently on 3.2.0 and at a later period if it refers to 5.0,…… any impact ?
    How how it affects related to the maven pom dependencies ?

    Thanks
    Jay

    Reply

    • Lokesh

      It will impact only when you use the config tags introduced in newer versions and you have old schema references.

      Reply

  6. goni

    nice

    Reply

Ask Questions & Share Feedback

Please do not submit a comment only to say "Thank you".

*Want to Post Code Snippets or XML content? Please use [java] ... [/java] tags otherwise code may not appear partially or even fully. e.g.
[java] 
public static void main (String[] args) {
...
} 
[/java]