家长班:
public class Person {
String firstName;
String lastName;
Long id;
List<Phone> phoneNumber = new ArrayList<>();
int age;
public Person(String firstName, String lastName, int age, Long id, List<Phone> phone) {
super();
this.firstName = firstName;
this.lastName = lastName;
this.age = age;
this.id = id;
this.phoneNumber = phone;
}
儿童电话对象(第1级):
public class Phone {
String number;
String type;
Long id;
public Phone(String number, String type, Long id) {
super();
this.number = number;
this.type = type;
this.id = id;
}
}
我正在尝试获取其电话对象类型为home且其编号应包含“888”的person对象.
List<Phone> list = personList.stream().map(p -> p.getPhoneNumber().stream()).
flatMap(inputStream -> inputStream).filter(p -> p.number.contains("888") && p.type.equals("HOME")).collect(Collectors.toList());
System.out.println(list.toString());
从上面的流代码,我能够获得电话对象.但是如何在同一个函数中获取该手机对象的父对象呢?
我试过这种方式,我为非匹配的对象获取null.
List<Person> finalList = personList.stream().map( per -> {
List<Phone> phones = per.getPhoneNumber();
Optional<Phone> ph = phones.stream().filter(p -> p.number.contains("888") && p.type.equals("HOME")).findAny();
if(ph.isPresent())
return per;
return null;
}).collect(Collectors.toList());
最佳答案
List<Person> result = personList.stream()
.filter( person -> person.phoneNumber.stream()
.anyMatch(phone -> phone.type.equals("HOME") &&
phone.number.contains("888"))
.collect(Collectors.toList());
您正在寻找的是手机流.它会做的伎俩.
让人们使用手机,您现在可以删除那些不符合条件的手机.
List<Person> result = personList.stream()
.filter( person -> person.phoneNumber.stream()
.anyMatch(phone -> phone.type.equals("HOME") &&
phone.number.contains("888"))
.map(person -> {
List<Phone> phones = person.phoneNumber.stream()
.filter(phone -> phone.type.equals("HOME") &&
phone.number.contains("888"))
.collect(Collectors.toList());
return new Person(person.firstName, person.lastName, person.age, person.id, phones);
})
.collect(Collectors.toList());