How to configure one implementation based on a properties file using Spring configuration?

Question

I have an application that can use two implementations of a service:

package test;

public interface MyService {
    void doSomething();
}


package test;

public class MyServiceA implements MyService {

    private final String whatever;

    public MyServiceA(final String whatever) {
        this.whatever = whatever;
    }

    @Override
    public void doSomething() {
        System.out.println("MyServiceA did "+whatever);
    }
}


package test;

public class MyServiceB implements MyService {

    private final String what;

    public MyServiceB(final String what) {
       this.what = what;
    }

    @Override
    public void doSomething() {
        System.out.println("MyServiceB did "+what);
    }
}

with different configurations.

I want to select what implementation to use with a system property.

I want to have the configuration for each implementation in its own property file and also its own spring configuration. So I can remove all the non used configuration altogether when not in use.

How may I configure any of the two implementations without requiring the property file of the non configured implementation?

Other solutions to this problem welcome.

score 1 · Answer 1 · answered Sep 19 '16 at 07:24

1

You may use profiles and different @Configuration classes for different profiles. http://docs.spring.io/spring/docs/current/spring-framework-reference/htmlsingle/#beans-definition-profiles-java

answered Sep 19 '16 at 07:24

Yuri Plevako

311
1
6

Actually you may configure it in your web.xml or WebApplicationInitializer, not system property. Changing one row in web.xml is not really a hardcore way. And if you don't want to provide another configuration for another service why do you need property? Actually when you switch profile spring will not even create a @Configuration class bean for non-active profile, so you don't need any properties for it and it will just stay calm in it's package) – Yuri Plevako Sep 20 '16 at 06:35
And since you want different property files for different beans you may declare alternative PropertyPlaceHolderConfigurer bean and set a prefix to it's properties. And then you may use prefixed properties for your alternative bean – Yuri Plevako Sep 20 '16 at 06:35
I am not supposed to modify the war application, so I cannot change the web.xml file. I can only configure it externally. – John Smith Sep 23 '16 at 20:28

score 0 · Answer 2 · edited May 23 '17 at 12:26

0

This can be achieved via @Qualifier. Here is the link that can provide you the implementation approach you are looking for. Spring Qualifier and property placeholder

edited May 23 '17 at 12:26

Community

1
1

answered Sep 19 '16 at 06:32

mhasan

3,703
1
18
37

Note that the first answer isn't the best, OP should read the second one. – Asoub Sep 19 '16 at 07:06
1

@Asoub You should explain what is the problem with an answer instead of saying that the other answer is better – John Smith Sep 19 '16 at 18:45

score 0 · Accepted Answer · edited May 23 '17 at 10:27

I decided to implement it as: https://stackoverflow.com/a/3036044/5759622

These are the required spring XML files:

applicationContext.xml

<?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">

    <import resource="MyService-${MyService}.xml"/>

</beans>

MyService-MyServiceA.xml

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

    <context:property-placeholder location="classpath:test/MyServiceA.properties" />

    <bean class="test.MyServiceA">
        <constructor-arg value="${MyServiceA.whatever}"/>
    </bean>

</beans>

MyService-MyServiceB.xml

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

    <context:property-placeholder location="classpath:test/MyServiceB.properties" />

    <bean class="test.MyServiceB">
        <constructor-arg value="${MyServiceB.what}"/>
    </bean>

</beans>

How to configure one implementation based on a properties file using Spring configuration?

3 Answers3