Java Lambda 函数接口提取到常用方法

如何解决Java Lambda 函数接口提取到常用方法

我有 2 个类并包含以下属性 languageMap

class Person {
    Map<String,String> languageMap;
}

class Employee {
    Map<String,String> languageMap;
}

有两个方法addressMap(List<Person> persons)和employeeMap(List<Employee> employee)调用Function接口，

public Map<String,String addressMap(List<Person> persons){
        Function<Collection<Person>,Map<String,String>> personFunc = CommonUtils::buildPersonMap;

               return personFunc.apply(persons);
  }

public Map<String,String employeeMap(List<Employee> employee){
    Function<Collection<Employee>,String>> addressFunc = CommonUtils::buildEmployeeMap;

         return addressFunc.apply(employee);
  }

private static Map<String,String> buildPersonMap(Collection<Person> personItem) {
    return personItem.stream()
            .filter(element -> element.getLanguageMap() != null)
            .flatMap(element -> element.getLanguageMap()
                    .entrySet()
                    .stream())
            .filter(map -> map.getKey() != null && map.getValue() != null)
            .collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue,(a,b) -> a));
}

private static Map<String,String> buildEmployeeMap(Collection<Employee> employeeItem) {
    return employeeItem.stream()
            .filter(element -> element.getLanguageMap() != null)
            .flatMap(element -> element.getLanguageMap()
                    .entrySet()
                    .stream())
            .filter(map -> map.getKey() != null && map.getValue() != null)
            .collect(Collectors.toMap(Map.Entry::getKey,b) -> a));
}

我想让 2 个 buildXXX() 方法通用，并尝试使用泛型如下，

private static Map<String,String> buildMap(Collection<?> input) {

    return input.stream()
      .filter(element -> element.getLanguageMap() != null). // ERROR
      .flatMap(element -> element.getLanguageMap().entrySet().stream()) // ERROR
      .filter(map -> map.getKey() != null && map.getValue() != null)
      .collect(Collectors.toMap(Map.Entry::getKey,b) -> a));
  }

有什么泛型或流技术可以解决这个问题？

解决方法

您可以让调用者为每种类型发送一个 getter，并以这种方式实现您的方法：

private static <T> Map<String,String> buildMap(Collection<T> input,Function<T,Map<String,String>> languageMapGetter) {

    return input.stream()
            .map(languageMapGetter) //you can also call it yourself
            .filter(Objects::nonNull)
            .flatMap(element -> element.entrySet().stream())
            .filter(map -> map.getKey() != null && map.getValue() != null)
            .collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue,(a,b) -> a));
}

这将使您的特定于类型的方法如下所示：

public Map<String,String> addressMap(List<Person> persons) {
    return CommonUtils.buildMap(persons,Person::getLanguageMap);
}

public Map<String,String> employeeMap(List<Employee> employee) {
    return CommonUtils.buildMap(employee,Employee::getLanguageMap);
}