如何重写 Java 8 Streaming 方法？

Question

我有以下entity 类：

@Entity
@Table(name = "KENNEL")
public class KENNEL{


    //other fields and vars

    @OneToMany(cascade = CascadeType.MERGE, orphanRemoval = true)
    @JoinColumn(name = "KENNEL_ID", referencedColumnName = "ID", updatable = true, insertable = true)
    private Set<Dog> dogs;

    public DateTime getOldestDogInKennel(){
            return dogList.stream().min(Comparator.comparing(Dog::getBirthday))
                .get().getBirthday();
        }


}

如何重新编写 getOldestDogInKennel 方法，使其无需使用 java 8 流式处理？

请注意，此实体不实现可比较的接口。

Answer 1

通过 Collections min 获取 Dog 对象的 min 元素

Collections.min(dogList, new Comparator<Dog>() {

    @Override
    public int compare(Dog o1, Dog o2) {
        return o1.getBirthday().compareTo(o2.getBirthday()));
    }

}).getBirthday()

Answer 2

我想这样可以；

Date oldest = new Date();
for (int i=0; i< doglist.size(); i++){
   if (doglist.get(i).getBirthday().compareTo(oldest) < 0)   
      oldest = doglist.get(i).getBirthday();
}

return oldest;

Answer 3

我不知道你想在节目后面做什么。所以这个方法在方法的末尾有最旧的dog。

public DateTime getOldestDogInKennel() {
    if (dogList.size() <= 0) { return null; }

    Dog dog = dogList.get(0);
    for (int i = 1; i < dogList.size(); i++) {
        Dog d = dogList.get(i);
        if (d.getBirthday().compareTo(dog.getBirthday()) < 0)) {
            dog = d;
        }
    }

    return dog.getBirthday();
}

没有Compareable接口使用：

public DateTime getOldestDogInKennel() {
    if (dogList.size() <= 0) { return null; }

    Dog dog = dogList.get(0);
    for (int i = 1; i < dogList.size(); i++) {
        Dog d = dogList.get(i);
        if (d.getBirthday().getTime() < dog.getBirthday().getTime()) {
            dog = d;
        }
    }

    return dog.getBirthday();
}