I have a following code:
import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
public class Test {
/**
* @param args
*/
public static void main(String[] args)
{
Test t=new Test();
t.test();
}
class A
{
int id;
B b;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public B getB() {
return b;
}
public void setB(B b) {
this.b = b;
}
}
class B
{
int id;
String unique;
public String getUnique() {
return unique;
}
public void setUnique(String unique) {
this.unique = unique;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
}
public void test()
{
B b1=new B();
b1.setId(9999);
b1.setUnique("123456789");
B b2=new B();
b2.setId(7777);
b1.setUnique("123456789");
B b3=new B();
b3.setId(8888);
b3.setUnique("987654321");
A a1=new A();
a1.setId(1);;
a1.setB(b1);
A a2=new A();
a2.setId(2);;
a2.setB(b1);
A a3=new A();
a3.setId(3);;
a3.setB(b2);
A a4=new A();
a4.setId(4);;
a4.setB(b2);
A a5=new A();
a5.setId(5);;
a5.setB(b3);
A a6=new A();
a6.setId(6);;
a6.setB(b3);
List<A> list=new ArrayList<A>();
list.add(a1);
list.add(a2);
list.add(a3);
list.add(a4);
list.add(a5);
list.add(a6);
Map<B,List<A>> map=list.stream()
.collect(Collectors.groupingBy(A::getB));
//required output
//Map<String, List<A>>
}
}
In this example I have a class A
which is having a reference of class B
. The class B has a field named unique
. As a source I have got the List (a1 to a6) which are referencing Object of Class B (b1, b2 and b3). The Object b1 and b2 contains same unique values. I want the output using Java 8 Stream API that will return me the Map> such that if unique value of object B which is referenced by A is same then the resulting list should contains the merge of objects of Class A in a List. e.g following should be the output
Map<String,List<A>> output = [
"123456789" : [a1,a2,a3,a4],
"987654321" : [a5,a6]
]
答案 0 :(得分:3)
修好了这一行(它是SufraceHolder
,但原来是b1
):
b2
正确的映射代码是这样的:
b2.setUnique("123456789");
输出(用Map<String,List<A>> map=list.stream()
.collect(Collectors.groupingBy(p -> p.getB().unique));
表示的A)是:
{ID}