Question

My application can run on different environments. I need configure MY data model per environment. The data model is build using Spring beans.

I use Spring 3.0.5, so I cannot conditionally load resources.
I have this:

<bean id="Template1" class="...
..............
</bean>

<bean id="Template2" class="...
..............
</bean>

<bean id="Template3" class="...
..............
</bean>
................

<bean id="Factory" ...>
<propety name="type"><value>${app.type}</value></property>
<property>
 <map>
   <entry key="Temlate1" value-ref="Template1">
   <entry key="Temlate2" value-ref="Template1">
   <entry key="Temlate3" value-ref="Template1">

..................

Real bean I create by factory:

<bean id="real" factory="Factory" factory-method="getInstance"
 <constructor-arg>Factory</.....
 .............
 </bean>

Java code:

class Factory {
private Map<String, Object> templateBeans;

 Object getInstance(String name) {
  return templateBeans.get(name);
 ...........

Is it possible in some way to declare abstract template beans? Because I have very big problem with memory. Does another way exist to instantiate different beans conditionally in Spring before version 3.1? It would be good to use only EL because I don’t have access to the Java code of the beans as they are from a third-party library.

Answer 1

I haven’t tried this but I fairly sure you could use bean aliases for this.

Firstly declare all your template beans to lazy so they aren’t instantiated on startup.

<bean id="lazy" class="com.foo.ExpensiveToCreateBean" lazy-init="true"/>

Then use a bean alias with a variable to point to the real one:

<alias name="real" alias="${beanForEnvironment}"/>

See my answer here for how to nicely load properties per environment:

property-placeholder location from another property