Sql 仅 for each 唯一ID返回一个元素，并仅返回最新连接的记录

假设有以下情况:

create schema bv;
create table bv.user(id bigint primary key);
create table bv.user_photo (
  id bigint primary key,
  url varchar(255) not null,
  user_id bigint references bv.user(id)
);

insert into bv.user values (100), (101);
insert into bv.user_photo values
  (1, 'https://1.com', 100),
  (3, 'https://3.com', 100),
  (4, 'https://4.com', 101),
  (2, 'https://2.com', 100);

我想 for each 用户查询并构建一个对象，并在结果中只包含最新的图像.

这就是我所拥有的:

select
  json_build_object(
    'id', u.id,
    'latest_image', up.url
  ) user
from bv.user u
left join bv.user_photo up
  on u.id = up.user_id

然而，这将返回:

[
  {"id" : 100, "url" : "https://2.com"},
  {"id" : 100, "url" : "https://3.com"},
  {"id" : 100, "url" : "https://1.com"},
  {"id" : 101, "url" : "https://4.com"}
]

然而，预期的结果是:

[
  {"id" : 100, "url" : "https://3.com"},
  {"id" : 101, "url" : "https://4.com"}
]

我试过用distinct:

select distinct on(u.id)
  json_build_object(
    'id', u.id,
    'url', up.url
  ) user
from bv.user u
left join bv.user_photo up
  on u.id = up.user_id
order by u.id, up.id DESC

但我的问题是，这是不是正确的方法？我觉得在这种情况下我不应该用distinct.