Passing maven properties to spring

Question

I know this is probably a dumb question but I can't figure it out for the life of me. Basically I am using maven to set my dataSource username, password, and driver class name. When I look in the effective Pom.xml it all appears fine as follows

<dataSource.driverClassName>oracle.jdbc.driver.OracleDriver</dataSource.driverClassName>
<dataSource.username>someUsername</dataSource.username>
<dataSource.password>somePassword</dataSource.password>

I am trying to use this information when declaring a spring datasource. The code appears as follows.

<bean id="dataSource" class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close">
    <property name="driverClassName" value="${dataSource.driverClassName}"/>
    <property name="url" value="${dataSource.url}"/>
    <property name="username" value="${dataSource.username}"/>
    <property name="password" value="${dataSource.password}"/>
</bean>

I then pass the datasource into a jdbcTemplate but when I use the template to run sql statements in my code I get an error saying that no driver with the name ${dataSource.driverClassName} can be found. This is obviously because the string constant is being passed rather than the variable. What am I missing?

Thanks

Thanks for all the help. Looks like PropertyConfigurationPlaceholder was what I was looking for. — Mike Baglio Jr., Feb 15 '12 at 22:02

score 11 · Answer 1 · edited Oct 02 '18 at 18:58

Overview

A simple way to configure the Maven Resources plugin without adding Java classes is to define a resource filter. For the purposes of an example, we'll configure spring-config.xml using properties defined in pom.xml.

pom.xml

The pom.xml file defines a property as follows:

<project>
  <properties>
    <environment.deploy>local</environment.deploy>
  </properties>
  ...

Then, in the build section, define the location of resources (files) that will have variables replaced. In the following example, only the file spring-config.xml will be updated:

  <build>
    <resources>
        <resource>
            <directory>src/main/resources</directory>
            <filtering>true</filtering>
            <includes>
                <include>spring-config.xml</include>
            </includes>
        </resource>
    </resources>
  </build>
</project>

The syntax for pom.xml is powerful, and can be used to substitute values within a number of files, such as:

<resources>
  <resource>
    <directory>src/main/resources</directory>
    <filtering>true</filtering>
    <includes>
      <include>**/*included_filename.xml</include>
    </includes>
  </resource>
</resources>

Exclusions are also possible:

<excludes>**/*excluded_filename.xml</excludes>

If only one file needs to have variables applied, set up different resources, such as:

<resources>
    <resource>
        <directory>src/main/resources</directory>
        <filtering>true</filtering>
        <includes>
            <include>spring-config.xml</include>
        </includes>
    </resource>
    <resource>
        <directory>src/main/resources</directory>
        <filtering>false</filtering>
        <includes>
            <include>**/*</include>
        </includes>
    </resource>
</resources>

This will ensure that all files are bundled, but only spring-config.xml is changed.

spring-config.xml

Located in src/main/resources, relative to the project's root directory, the properties loaded by Maven can be used. For example:

<beans ...>
  <bean ...>
    <property name="deploymentEnvironment">
        <list>
            <value>${environment.deploy}</value>
        </list>
    </property>
</bean>

During the build process, Maven replaces the value above inside the target's spring-config.xml file:

<beans ...>
  <bean ...>
    <property name="deploymentEnvironment">
        <list>
            <value>local</value>
        </list>
    </property>
</bean>

External Properties

For the truly adventurous, Maven properties can also be loaded from external files.

score 3 · Answer 2 · answered Feb 15 '12 at 21:51

I think that you cannot do it in that way, I mean, from pom.xml to spring application context xml.

Put your properties in a property file, something like:

dataSource.username=${dataSource.username}
dataSource.driverClassName=${dataSource.driverClassName}
dataSource.username=${dataSource.username}
dataSource.password=${dataSource.password}

Then, use PropertyConfigurationPlaceholder to load the property file and make properties availables on spring applicationt context file.

score 2 · Accepted Answer · answered Feb 15 '12 at 21:47

Spring is expecting to find those values in a .properties file.

Inject a PropertyConfigurationPlaceholder with those values and Spring will find them.

http://static.springsource.org/spring/docs/3.0.x/javadoc-api/org/springframework/beans/factory/config/PropertyPlaceholderConfigurer.html

Look for example 3.8.2.1 below:

http://static.springsource.org/spring/docs/3.0.x/spring-framework-reference/html/beans.html

score 2 · Answer 4 · answered Feb 15 '12 at 21:48

2

Maven can not transfer any data to the application, because maven works at compile time and the parameter a evaluated at runtime.

But you can use maven filters. To substiture some markers with values while maven is processing the resources.

answered Feb 15 '12 at 21:48

Ralph

118,862
56
287
383

I agree with this answer. Using maven filtering updates the values in the spring config files at build time, thus having the correct values at run time. See http://stackoverflow.com/questions/3886753/access-maven-project-version-in-spring-config-files – Noremac Jul 09 '13 at 20:11
But filtering must explicitely switched on in the POM, as shown in the answer by Carlos AG. – gsl Apr 10 '18 at 15:01

Passing maven properties to spring

4 Answers4

Overview

pom.xml

spring-config.xml

External Properties

Linked