java - 如何使用Stream过滤嵌套对象？

Question

我想使用 Stream API 过滤嵌套对象。问题是有太多的嵌套类，并且使用下面的方法我编写了太多重复的代码。

有没有办法在不重复代码的情况下处理这个流？

public class Country{
    Map<String, City> cities;
}

public class City{
    Map<String, School> schools;
}

public class School{
    String name;
    String address;
    Model model;
}

public class Model{
    String name;
    Teacher teacher;
}

public class Teacher{
    String name;
    String id;
}

我的直播；

country.getCities().values().stream().foreach(
     (City city) ->
         city.getSchools()
             .entrySet()      
             .stream()
             .filter(schoolEntry -> schoolEntry.getValue().getName().equals("test")) 
             .filter(schoolEntry -> schoolEntry.getValue().getModel().getName().equals("test2"))
             .filter(schoolEntry -> schoolEntry.getValue().getModel().getTeacher().getName().equals("test2"))
             .foreach(schoolEntry -> {
                  String schoolKey = schoolEntry.getKey();
                  resultList.put(schoolKey, schoolEntry.getValue().getModel().getTeacher().getId());
              })    
);

Answer 1

您可以定义一个方法来将其用作 Predicate过滤学校。

public static boolean matches(School school, String schoolName, String modelName, String teacherId) {
    return school.getName().equals(schoolName) 
            && school.getModel().getName().equals(modelName) 
            && school.getModel().getTeacher().getId().equals(teacherId);
}

将其应用于流:

public static Map<String, String> getSchoolAndTeacherFrom(Country country, Predicate<School> schoolWithModelAndTeacher) {
    return country.getCities().values().stream()
            .flatMap(c -> c.getSchools().entrySet().stream())
            .filter(s -> schoolWithModelAndTeacher.test(s.getValue()))
            .collect(Collectors.toMap(Entry::getKey, schoolEntry -> schoolEntry.getValue().getModel().getTeacher().getId()));
}

像这样使用:

    Country country = <county>
    Predicate<School> schoolWithModelAndTeacher = school -> matches(school, "test1", "test2", "test2");
    getSchoolAndTeacherFrom(country, schoolWithModelAndTeacher);

一些进一步的想法:

如果 map schools使用School.getName()作为键，那么我们可以写:

public static Map<String, String> getSchoolAndTeacherFrom(Country country, Predicate<School> schoolWithModelAndTeacher) {
    return country.getCities().values().stream()
            .flatMap(city -> city.getSchools().values().stream())
            .filter(schoolWithModelAndTeacher::test)
            .collect(Collectors.toMap(School::getName, school -> school.getModel().getTeacher().getId()));
}

假设一个国家/地区的学校名称和教师 ID 是唯一的(而模型名称很常见)，则过滤结果将是单个值(如果有)。但那就不需要 Map作为结果类型。类型为 Entry<String String> 的结果会做到的。
如果谓词的参数仍然已知(学校、模型、教师)，那么整个问题只是一个问题:在某个国家的给定学校中是否存在给定模型的给定教师。那么我们可以把它写得更短:

public static boolean isMatchingSchoolInCountryPresent(Country country, Predicate<School> schoolWithModelAndTeacher) {
    return country.getCities().values().stream()
            .flatMap(c -> c.getSchools().values().stream())
            .anyMatch(schoolWithModelAndTeacher::test);
}