Sql 如何在select子句中使用多个LISTAGG（）函数时选择不同的值？

我正在尝试使用以下查询检索race_代码、chara_代码和reason_代码作为列表：

SELECT a.pid,
       LISTAGG(a.rc, ',') WITHIN GROUP (ORDER BY a.rc) AS race,
       LISTAGG(a.cc, ',') WITHIN GROUP (ORDER BY a.cc) as chara_codes,
       LISTAGG(a.rrc, ',') WITHIN GROUP (ORDER BY a.rrc) AS removal_reason
FROM (
   SELECT UNIQUE
          p.person_id pid,
          r.race_code rc,
          c.characteristic_code cc,
          rr.removal_reason_code rrc
     FROM person p left outer join race r on p.person_id = r.person_id
          left outer join characteristic c on p.person_id = c.person_id
          left outer join placement_episode pe on p.person_id = pe.child_id
          left outer join removal_reason rr on pe.placement_episode_id = rr.placement_episode_id
     ) a
GROUP BY a.pid

我在引用了一些链接（如和）后尝试了此查询。但在这样做之后，我也无法获得所有字段的唯一值

我的o/p如下所示：

pid      race_code     chara_code      reason_code
 1        a,b,b,c     c1,c1,c2,c3     r1,r2,r3,r3
 2       a,c,d,d,d      c1,c2,c2        r3,r3

and so on.

如果我试图一次只检索一个字段，并保留所需的联接操作符，那么它会给出正确的结果。但对于多个LISTAGG（）函数，其值是重复的。

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我没有办法做到这一点。是否有其他方法可以获得不同的值？

不幸的是，这比需要的更复杂。但是，你可以做到。其思想是枚举每个值，然后使用

case

将

NULL

参数传递给

listag（）

查询将基于每个人员和字段组合枚举行。第一次看到该值时，其值为“1”，随后的值将递增。

LISTAGG（）

仅选择第一个值

---

你应该学习分析函数。它们非常有用。

您可以使用标量子查询，如本例所示：

select p.person_id
     , (select LISTAGG(rc, ',') WITHIN GROUP (ORDER BY rc) 
          from race r where r.person_id = p.person_id
         group by r.person_id) race
     , (select LISTAGG(cc, ',') WITHIN GROUP (ORDER BY cc) 
          from characteristic r where r.person_id = p.person_id
         group by r.person_id) chara_codes
     , (select LISTAGG(rrc, ',') WITHIN GROUP (ORDER BY rrc) 
          from removal_reason r where r.person_id = p.person_id
         group by r.person_id) removal_reason
  from person p;

with race_list as (
  select person_id
       , LISTAGG(rc, ',') WITHIN GROUP (ORDER BY rc) race
    from race
   group by person_id
), characteristic_list as (
  select person_id
       , LISTAGG(cc, ',') WITHIN GROUP (ORDER BY cc) chara_codes
    from characteristic
   group by person_id
), removal_reason_list as (
  select person_id
       , LISTAGG(rrc, ',') WITHIN GROUP (ORDER BY rrc) removal_reason
    from removal_reason
   group by person_id
)
select p.person_id
     , r.race
     , c.chara_codes
     , rr.removal_reason
  from person p
  join race_list r
    on r.person_id = p.person_id
  join characteristic_list c
    on c.person_id = p.person_id
  join removal_reason_list rr
    on rr.person_id = p.person_id;

或者，您可以如本例所示预计算列表：

select p.person_id
     , (select LISTAGG(rc, ',') WITHIN GROUP (ORDER BY rc) 
          from race r where r.person_id = p.person_id
         group by r.person_id) race
     , (select LISTAGG(cc, ',') WITHIN GROUP (ORDER BY cc) 
          from characteristic r where r.person_id = p.person_id
         group by r.person_id) chara_codes
     , (select LISTAGG(rrc, ',') WITHIN GROUP (ORDER BY rrc) 
          from removal_reason r where r.person_id = p.person_id
         group by r.person_id) removal_reason
  from person p;

with race_list as (
  select person_id
       , LISTAGG(rc, ',') WITHIN GROUP (ORDER BY rc) race
    from race
   group by person_id
), characteristic_list as (
  select person_id
       , LISTAGG(cc, ',') WITHIN GROUP (ORDER BY cc) chara_codes
    from characteristic
   group by person_id
), removal_reason_list as (
  select person_id
       , LISTAGG(rrc, ',') WITHIN GROUP (ORDER BY rrc) removal_reason
    from removal_reason
   group by person_id
)
select p.person_id
     , r.race
     , c.chara_codes
     , rr.removal_reason
  from person p
  join race_list r
    on r.person_id = p.person_id
  join characteristic_list c
    on c.person_id = p.person_id
  join removal_reason_list rr
    on rr.person_id = p.person_id;

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谢谢你的回复。在查询的第二行中有额外的“'）。请把它拿走。它的投掷错误。你能给我一些解释或者一些你使用的逻辑链接吗？我不知道您使用的分区概念。这个查询很好：）