Java 8 Stream() 具有复杂的 List 条件，它应该返回一个结果

Question

我有电话号码列表，我需要根据电话类型进行过滤。最终结果应该是单个对象 (PhoneUsageType)。示例场景 1. 如果列表有 2 种不同类型的电话号码，如 Cell_Phone 和 Home_Phone，它应该只返回 Cell_Phone 对象。2.如果它只有 Cell_Phone 那么它应该直接返回我那个对象 3.如果它没有 Cell_Phone 那么它应该返回我 Home_Phone如果 Cell_Phone 在列表中可用，则它应该是首选对象使用list.stream().filter代码PhoneUsageType phone1 = new PhoneUsageType();PhoneUsageKeyGrpType keyGroup1 = new PhoneUsageKeyGrpType();keyGroup1.setPurposeCode("CELL_PHONE");phone1.setPhoneUsageKeyGrp(keyGroup1);PhoneNumberType phoneNumber1 = new PhoneNumberType();phoneNumber1.setFullNumber("123");phone1.setPhoneNumber(phoneNumber1);PhoneUsageType phone2 = new PhoneUsageType();PhoneUsageKeyGrpType keyGroup2 = new PhoneUsageKeyGrpType();keyGroup2.setPurposeCode("CELL_PHONE");phone2.setPhoneUsageKeyGrp(keyGroup2);PhoneNumberType phoneNumber2 = new PhoneNumberType();phoneNumber2.setFullNumber("456");phone2.setPhoneNumber(phoneNumber2);List<PhoneUsageType> phoneUsageList = Lists.newArrayList();请帮我写下逻辑。提前致谢。如果您需要更多详细信息，请告诉我。

Answer 1

不要按类型使用filter、使用min或排序。max

像下面这样的东西：

enum PhoneType {

    CELL, WORK, HOME; // in order of preference

}

class Phone {

    public PhoneType getType() ...

    public String getNumber() ...

}

Phone best = Collections.min(phoneList, Comparator.comparing(Phone::getType));

Answer 2

更好地使用最小和最大排序类型

enum PhoneType {

    Mobile, Office, personal; // in order of preference

}

class Mobile {

    public PhoneType getType() 

    public String getNumber() 

}

Comparator<Mobile> preferredNumber = Comparator.comparing(Mobile::getType);

Mobile best = mobileList.stream().min(preferredNumber).orElseThrow();

希望你能明白。