Here the below code will show you how to inject the Set in Spring using the spring.xml file. By using that we will inject the Set object in spring.
Person.java
package com.evon;
public class Person {
String name;
String address;
int age;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getAddress() {
return address;
}
public void setAddress(String address) {
this.address = address;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
}
Customer.java
package com.evon;
import java.util.Set;
public class Customer {
private Set<Object> set;
public Set<Object> getSet() {
return set;
}
public void setSet(Set<Object> set) {
this.set = set;
}
}
spring.xml
<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-2.5.xsd">
<bean id="person" class="com.evon.Person">
<property name="name" value="Manish" />
<property name="address" value="New Delhi" />
<property name="age" value="25" />
</bean>
<bean id="customer" class="com.evon.Customer">
<!-- java.util.Set -->
<property name="set">
<set>
<value>1</value>
<ref bean="person" />
</set>
</property>
</bean>
</beans>
App.java
package com.evon;
import org.springframework.context.ApplicationContext;
import org.springframework.context.support.ClassPathXmlApplicationContext;
public class App {
public static void main(String[] args) {
ApplicationContext context = new ClassPathXmlApplicationContext(
"spring.xml");
Customer cust = (Customer) context.getBean("customer");
System.out.println(cust.getSet());
}
}
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