问

ArrayList获取对象属性的所有值

吴钧隆362 发布于 2023-02-13 10:15

list

api

java

假设我有一个User对象的ArrayList ArrayList.User对象具有属性userID.

是不是自己迭代列表并将userID添加到单独的列表中,是否有API调用,我可以在其中传递我想要的属性,并将这些属性的列表返回给我？看看API并没有什么突出的.

在Java 7或<中寻找解决方案.

1 个回答

您可以使用lambdas表达式(Java 8)执行此操作:

import java.util.*;
import java.util.function.*;
import java.util.stream.*;

public class Test {
  public static void main(String args[]){
    List<User> users = Arrays.asList(new User(1,"Alice"), new User(2,"Bob"), new User(3,"Charlie"), new User(4,"Dave"));
    List<Long> listUsersId = users.stream()
                                  .map(u -> u.id)
                                  .collect(Collectors.toList());
    System.out.println(listUsersId);
  }
}

class User {
  public long id;
  public String name;

  public User(long id, String name){
    this.id = id;
    this.name = name;
  }
}

输出:

[1, 2, 3, 4]

这里的代码片段.

使用反射的最丑的解决方案:

public class Test {
    public static void main (String[] args) throws NoSuchFieldException, SecurityException, IllegalArgumentException, IllegalAccessException{
         List<User> users = Arrays.asList(new User(1,"Alice"), new User(2,"Bob"), new User(3,"Charlie"), new User(4,"Dave"));
         List<Object> list = get(users,"id");
         System.out.println(list);
    }

    public static List<Object> get(List<User> l, String fieldName) throws NoSuchFieldException, SecurityException, IllegalArgumentException, IllegalAccessException{
        Field field = User.class.getDeclaredField(fieldName);
        field.setAccessible(true);
        List<Object> list = new ArrayList<>();
        for(User u : l){
            list.add(field.get(u));
        }
        field.setAccessible(false);
        return list;
    }
}
class User {
      private long id;
      private String name;

      public User(long id, String name){
        this.id = id;
        this.name = name;
      }
}

输出:

[1, 2, 3, 4]

2023-02-13 10:18 回答

于浮浅

撰写答案

今天，你开发时遇到什么问题呢？

立即提问

热门标签