Note: This is an answer that reflects the discussion in the comments.

给出示例 struct /数据:

create table people
(
    uuid            text,
    givenname       text,
    surname         text,
    bio_mother_uuid text,
    bio_father_uuid text
);

insert into people (uuid, givenname, surname, bio_mother_uuid, bio_father_uuid)
values ('aa', 'david', 'smith', 'ab', 'ac'),
       ('ab', 'mary', 'smith', 'ad', 'ae'),
       ('ac', 'john', 'smith', 'af', 'ag'),
       ('ad', 'gertude', 'snow', null, 'ah'),
       ('ae', 'Richard', 'snow', null, null),
       ('af', 'Katie', 'smith', null, null),
       ('ag', 'Mathew', 'smith', null, null),
       ('ah', 'Winston', 'Snow', null, null);   

您可以运行递归CTE，如下所示:

with recursive t as (
  select 
       p.uuid    as request_uuid,
       p.uuid                 as uuid,
       p.givenname            as givenname,
       p.surname              as surname,
       p.bio_mother_uuid      as bio_mother_uuid,
       p.bio_father_uuid      as bio_father_uuid,
       0 as generation
       from people p
  union all 
  select 
       child.request_uuid    as request_uuid,
       p.uuid                 as uuid,
       p.givenname            as givenname,
       p.surname              as surname,
       p.bio_mother_uuid      as bio_mother_uuid,
       p.bio_father_uuid      as bio_father_uuid,
       child.generation + 1 as generation       
       from 
       t child join 
       people p on p.uuid = child.bio_mother_uuid or p.uuid = child.bio_father_uuid
  )
  select * from t where request_uuid = 'aa'  and generation < 3  

结果是(fiddle):

然后，这将提供填充 struct 所需的所有信息(需要一些自定义代码来完成此操作).像这样在单个查询中检索所有数据可能比运行多个查询更快(但这取决于许多因素).

虽然可能有方法可以在一行中返回所有这些数据，但我觉得这将比上面的方法更难处理(您仍然需要自定义代码来解封，并且需要处理不同数量的列).另一种方法是在JSON中返回 struct (这会使SQL复杂化，但简化Go代码).