How to configure Spring Bean if the instance was created inside if-else statement?

Question

I have class structure that look like below:

class Base {}

class ChildA extends Base {}

class ChildB extends Base {}

I got this in my application code:

Base theObj = null;
if( str.equalsIgnoreCase("object_A") == true ) {
   theObj = new ChildA();
} else {
   theObj = new ChildB();
}

May I know how to configure the bean in Spring?

If you're creating an object with "new" than it isn't a Spring-managed bean. — Jacob Mattison, Jun 22 '12 at 03:35
@JacobM: I think he is trying to say that how he can implement that in Spring. — Bhesh Gurung, Jun 22 '12 at 03:36
Looks like this relates to your situation: http://stackoverflow.com/q/3141130/738746 — Bhesh Gurung, Jun 22 '12 at 03:39

score 0 · Answer 1 · answered Jun 22 '12 at 08:24

0

One option is to create refs for two different beans (childA, childB) in spring xml context. And then in creation for main bean specify constructor with args. You can implement constructor with similar code as you mentioned.

answered Jun 22 '12 at 08:24

mishadoff

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What show in my mind is to use either AbstractFactoryBean because it contain a logic selection on which child bean to be create during runtime. – huahsin68 Jun 23 '12 at 07:21

How to configure Spring Bean if the instance was created inside if-else statement?

1 Answers1