SelVazis的答案是好的和干净的。然而，如果在任何表中都没有找到值，它将不会返回结果行。如果值在一个或多个表中不止一次，则会返回多个结果行。因此，最好先制作一个响应表并在该响应表中查找答案：

select 
    req.code, 
    resp."tableA", 
    resp."tableB", 
    resp."tableC", 
    resp."valueA", 
    resp."valueB", 
    resp."valueC" 
from 
    (select 'a' as code) req 
left join
    (Select 
        coalesce(codeA, codeB, codeC) as code,
        max(case when codeA is not null then 1 else 0 end) as "tableA",
        max(case when codeB is not null then 1 else 0 end) as "tableB",
        max(case when codeC is not null then 1 else 0 end) as "tableC",
        string_agg(distinct valueA, ', ') as "valueA",
        string_agg(distinct valueB, ', ') as "valueB",
        string_agg(distinct valueC, ', ') as "valueC"
    from 
        (select 
            * 
        from 
            (select code as codeA, value as valueA from tableA) a
        full outer join 
            (select code as codeB, value as valueB from tableB) b 
        on a.codeA = b.codeB
        full outer join 
            (select code as codeC, value as valueC from tableC) c 
        on c.codeC = b.codeB) 
        as subresp 
    group by 
        coalesce(codeA, codeB, codeC)
   ) resp 
   on req.code = resp.code

在本例中，我假设您希望从表中返回所有值，因此我对它们进行了字符串聚合。请随意使用min，max或任何浮动的值来替换它们。
DB fiddle for completeness here

我可能只是连接你正在寻找的三个数据集。你可以使用伪全外连接（因为没有真实的的连接条件）或聚合的交叉连接来完成这一点：

with a as (select min(value) as val from tablea where code = 'a')
   , b as (select min(value) as val from tableb where code = 'a')
   , c as (select min(value) as val from tablec where code = 'a')
select
  a.val is not null as code_exists_in_a,
  b.val is not null as code_exists_in_b,
  c.val is not null as code_exists_in_c,
  a.val as value_in_a,
  b.val as value_in_b,
  c.val as value_in_c
from a cross join b cross join c;

(For完整的外部连接可以用select code, value as val替换select min(value) as val，用from a full outer join b using (code) full outer join c using (code)替换from a cross join b cross join c）。
免责声明：如果表中的值可以为null，并且您希望将其显示为value exists，但value为null，则必须为此调整上述查询。

假设code在每个表上都是唯一的。
你可以使用full outer join来返回所有在left（tableA）或right（tableB）中匹配的记录：

Select coalesce(a.code, b.code, c.code) as code,
  MAX(case when a.code is not null then 1 else 0 end)  as tableA,
  MAX(case when b.code is not null then 1 else 0 end)  as tableB,
  MAX(case when c.code is not null then 1 else 0 end)  as tableC,
  MAX(coalesce(a.value,0))  as valueA,
  MAX(coalesce(b.value,0))  as valueB,
  MAX(coalesce(c.value,0)) as valueC
from tableA a
full outer join TableB b on a.code = b.code
full outer join TableC c on c.code = b.code
where a.code = 'a' or b.code = 'a' or c.code = 'a'
group by coalesce(a.code, b.code, c.code)

或者在full outer join上使用USING而不是ON：

Select coalesce(a.code, b.code, c.code) as code,
  case when a.code is not null then 1 else 0 end as tableA,
  case when b.code is not null then 1 else 0 end as tableB,
  case when c.code is not null then 1 else 0 end as tableC,
  coalesce(a.value,0) as valueA,
  coalesce(b.value,0) as valueB,
  coalesce(c.value,0) as valueC
from tableA a
full outer join TableB b using (code)
full outer join TableC c using (code)
where a.code = 'a' or b.code = 'a' or c.code = 'a'

结果：

code    tablea  tableb  tablec  valuea  valueb  valuec
a       1       1       0       1       4       null

Demo here