How to define a List bean in Spring?

Question

I'm using Spring to define stages in my application. It's configured that the necessary class (here called Configurator) is injected with the stages.
Now I need the List of Stages in another class, named LoginBean. The Configurator doesn't offer access to his List of Stages.

I cannot change the class Configurator.

My Idea:
Define a new bean called Stages and inject it to Configurator and LoginBean. My problem with this idea is that I don't know how to transform this property:

<property ...>
  <list>
    <bean ... >...</bean>
    <bean ... >...</bean>
    <bean ... >...</bean>
  </list>
</property>

into a bean.

Something like this does not work:

<bean id="stages" class="java.util.ArrayList">

Can anybody help me with this?

Answer 1

<bean id="someBean"
      class="com.somePackage.SomeClass">
    <property name="myList">
        <list value-type="com.somePackage.TypeForList">
            <ref bean="someBeanInTheList"/>
            <ref bean="someOtherBeanInTheList"/>
            <ref bean="someThirdBeanInTheList"/>
        </list>
    </property>
</bean>

And in SomeClass:

class SomeClass {

    private List<TypeForList> myList;

    @Required
    public void setMyList(List<TypeForList> myList) {
        this.myList = myList;
    }

}

Answer 2

I think you may be looking for org.springframework.beans.factory.config.ListFactoryBean.

You declare a ListFactoryBean instance, providing the list to be instantiated as a property withe a <list> element as its value, and give the bean an id attribute. Then, each time you use the declared id as a ref or similar in some other bean declaration, a new copy of the list is instantiated. You can also specify the List class to be used.

Answer 3

Import the spring util namespace. Then you can define a list bean as follows:

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:util="http://www.springframework.org/schema/util"
xsi:schemaLocation="http://www.springframework.org/schema/beans
                    http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
                    http://www.springframework.org/schema/util
                    http://www.springframework.org/schema/util/spring-util-2.5.xsd">


<util:list id="myList" value-type="java.lang.String">
    <value>foo</value>
    <value>bar</value>
</util:list>

The value-type is the generics type to be used, and is optional. You can also specify the list implementation class using the attribute list-class.

Answer 4

Here is one method:

<bean id="stage1" class="Stageclass"/>
<bean id="stage2" class="Stageclass"/>

<bean id="stages" class="java.util.ArrayList">
    <constructor-arg>
        <list>
            <ref bean="stage1" />
            <ref bean="stage2" />                
        </list>
    </constructor-arg>
</bean>

Answer 5

Use the util namespace, you will be able to register the list as a bean in your application context. You can then reuse the list to inject it in other bean definitions.

Answer 6

As an addition to Jakub's answer, if you plan to use JavaConfig, you can also autowire that way:

import com.google.common.collect.Lists;

import java.util.List;

import org.springframework.context.annotation.Configuration;
import org.springframework.context.annotation.Bean;

<...>

@Configuration
public class MyConfiguration {

    @Bean
    public List<Stage> stages(final Stage1 stage1, final Stage2 stage2) {
        return Lists.newArrayList(stage1, stage2);
    }
}