Is it possible to read a combination of a System Property and a Customized Property in Spring config file? - spring

Question

I need to know if its possible in Spring to derive a value for a Property by combination of a System-Property and a Customized-Property defined over a PropertyFile.
What I've done:
1) Configured a -D argument in server as : -Dapp.Env="dev"
2) In my spring-config.xml, defined datasource details as below:
<!-- DataSource configurations -->
<bean id="myDataSource" class="com.atomikos.jdbc.AtomikosDataSourceBean"
init-method="init" destroy-method="close">
<property name="uniqueResourceName" value="myDataSource" />
<property name="xaDataSourceClassName" value="${my.database.xaDriver.class}" />
<property name="xaProperties">
<props>
<prop key="URL">#{systemProperties['appEnv']}.${my.database.jdbcUrl}</prop>
<prop key="user">#{systemProperties['appEnv']}.${my.database.user}</prop>
<prop key="password">#{systemProperties['appEnv']}.${my.database.password}</prop>
</props>
</property>
<property name="maxPoolSize" value="${my.database.maxPoolSize}" />
<property name="minPoolSize" value="${my.database.minPoolSize}" />
<property name="borrowConnectionTimeout" value="60" />
<property name="maintenanceInterval" value="120" />
</bean>
3) Defined the following properties in my .properties file:
my.database.xaDriver.class=oracle.jdbc.xa.client.OracleXADataSource
my.database.initialPoolSize=2
my.database.maxPoolSize=10
my.database.minPoolSize=2
dev.my.database.jdbcUrl=jdbc:oracle:thin:#test.com:1535:myDb
dev.my.database.user=myuserid
dev.my.database.password=mypwd
4) Defined .properties file location in the spring-config.xml as :
<context:property-placeholder location="/WEB-INF/spring/spring-config.properties" />
What I get on start of server :
Throwable occurred: org.springframework.beans.factory.BeanDefinitionStoreException: Invalid bean definition with name 'myDataSource' defined in ServletContext resource [/WEB-INF/spring/spring-config.xml]: Could not resolve placeholder 'my.database.jdbcUrl' in string value "#{systemProperties['appEnv']}.${my.database.jdbcUrl}"
What I also tried :
5) Defined a property in .properties file as below:
app.Env=dev
6) Modified the datasource bean configuration as below:
<prop key="URL">${app.Env}.${my.database.jdbcUrl}</prop>
<prop key="user">${app.Env}.${my.database.user}</prop>
<prop key="password">${app.Env}.${my.database.password}</prop>
However on start of server, I get same exception message :
Throwable occurred: org.springframework.beans.factory.BeanDefinitionStoreException: Invalid bean definition with name 'myDataSource' defined in ServletContext resource [/WEB-INF/spring/spring-config.xml]: Could not resolve placeholder 'my.database.jdbcUrl' in string value "${app.Env}.${my.database.jdbcUrl}"
Can someone help me out in pointing out what mistake I might be doing and what is/are the possible fix for this issue ?

Answer 1

1) You can read system properties into Spring bean:
<bean id="systemProperties" class="java.lang.System" factory-method="getProperties"/>
and then use multiple property resolvers
2) You can use flag systemPropertiesMode of PropertyPlaceholderConfigurer
3) You can write your custom bean factory that would produce single java.util.Properties object from system properties and file properties.
That bean would have method
public Properties merge() {
Properties merged = new Properties();
for (Properties p : properties) {
if (p != null) {
merged.putAll(p);
}
}
return merged;
}
Invoked in spring like that:
<bean id="factory" class="my.Factory">
<property name="properties">
<util:list>
<bean class="java.lang.System" factory-method="getProperties"/>
<bean ref="otherProperties">
</util:list>
</property>
</bean>