人是我的根POJA,我有作为我的孩子对象的电话号码列表.
String firstName; String lastName; Long id; List<String> phoneNumber = new ArrayList<>(); int age; public Person(String firstName,String lastName,int age,Long id,List<String> phone) { super(); this.firstName = firstName; this.lastName = lastName; this.age = age; this.id = id; this.phoneNumber = phone; } List<Person> personList = Arrays.asList( new Person("Abdul","Razak",27,50L,Arrays.asList("100","101","102")),new Person("Udemy","tut",56,60L,Arrays.asList("200","201","202")),new Person("Coursera",78,20L,Arrays.asList("300","301","302")),new Person("linked",14,10L,Arrays.asList("400","401","402")),new Person("facebook",24,5L,Arrays.asList("500","501","502")),new Person("covila",34,22L,Arrays.asList("600","602","604")),new Person("twitter",64,32L,Arrays.asList("700","702","704")) ); List<String> list = personList.stream() .map(p -> p.getPhoneNumber().stream()) .flatMap(inputStream -> inputStream) .filter(p -> p.contains("502")) .collect(Collectors.toList());
我想检索其数字等于特定字符串的人.是否可以通过使用流来实现这一点?
List<String> list = personList.stream() .map(p -> p.getPhoneNumber().stream()) .flatMap(inputStream -> inputStream) .filter(p -> p.contains("502")) .collect(Collectors.toList());
简单来说,如何通过过滤子对象来检索父对象?
解决方法
personList.stream().filter((person)->person.getContacts().contains("100")) .collect(Collectors.toList());
会给你匹配的人.
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