Question

如果我有这样的查询，是否可以进行1对1元素数组连接：

编辑：数组并不总是具有相同数量的元素。可能是array1有时有4个元素和array2 8个元素。

drop table if exists a;
drop table if exists b;
create temporary table a as (select 1 as id,array['a','b','c'] as array1);
create temporary table b as (select 1 as id,array['x','y','z'] as array2);

select
a.id,
a.array1,
b.array2,
array_concat--This has to be a 1 to 1 ordered concatenation (see                   
            --example below)
from a
left join b on b.id=a.id

我想在这里得到的是数组1和2的成对串联，如下所示：

id       array11          array2         array_concat
 1    ['a','b','c']   ['d','e','f']   ['a-d','b-e','c-f']
 2    ['x','y','z']   ['i','j','k']   ['x-i','y-j','z-k']
 3    ...

我尝试使用unfst但我无法使其工作：

    select
    a.id,
    a.array1,
    b.array2,
    array_concat
    from table a
    left join b on b.id=a.id
    left join (select a.array1,b.array2, array_agg(a1||b2)
                      FROM unnest(a.array1, b.array2) 
                           ab (a1, b2)
              ) ag on ag.array1=a.array1 and  ag.array2=b.array2
;

修改

这仅适用于一个表：

SELECT array_agg(el1||el2)
FROM unnest(ARRAY['a','b','c'], ARRAY['d','e','f']) el (el1, el2);

++感谢https://stackoverflow.com/users/1463595/%D0%9D%D0%9B%D0%9E

修改

我找到了一个非常接近的解决方案，但是一旦数组之间的连接完成，它会混合一些中间值，从来没有我需要一个完美的解决方案......

我现在使用的方法是：

1）根据2个独立的表创建一个表 2）使用Lateral汇总：

create temporary table new_table as
SELECT
    id,
    a.a,
    b.b
    FROM a a
    LEFT JOIN b b on a.id=b.id;

SELECT id,
       ab_unified
       FROM pair_sources_mediums_campaigns,
       LATERAL (SELECT ARRAY_AGG(a||'[-]'||b order by grp1) as ab_unified
               FROM (SELECT DISTINCT case when a null
                                   then 'not tracked'
                                   else a
                                   end as a
                         ,case when b is null
                                   then 'none'
                                   else b
                                   end as b
                            ,rn - ROW_NUMBER() OVER(PARTITION BY a,b ORDER BY rn) AS grp1

                    FROM unnest(a,b) with ordinality as el (a,b,rn)
                ) AS sub
           ) AS lat1
           order by 1;

Answer 1

像这样。

with a_elements (id, element, idx) as (
  select a.id,
         u.element, 
         u.idx
  from a 
    cross join lateral unnest(a.array1) with ordinality as u(element, idx)
), b_elements (id, element, idx) as (
  select b.id,
         u.element, 
         u.idx
  from b
    cross join lateral unnest(b.array2) with ordinality as u(element, idx)
)
select id,
       array_agg(concat_ws('-', a.element, b.element) order by idx) as elements
from a_elements a
  full outer join b_elements b using (id, idx)
group by coalesce(a.id, b.id);

连接运算符using (..)将自动从连接表中获取非空值。这消除了使用例如coalesce(a.id, b.id)。1.

对于大型桌子来说，它并不漂亮，绝对没有效率，但似乎可以随心所欲。

对于没有相同数量元素的数组，结果将只包含其中一个数组的元素。

对于此数据集：

insert into a 
  (id, array1)
values
  (1, array['a','b','c','d']),
  (2, array['d','e','f']);

insert into b 
  (id, array2)
values 
  (1, array['x','y','z']), 
  (2, array['m','n','o','p']);

返回此结果：

id | elements       
---+----------------
 1 | {a-x,b-y,c-z,d}
 2 | {d-m,e-n,f-o,p}

Answer 2

我认为你想得太远了，试试这个（SQLFiddle）：

select
   a.id,
   a.array1,
   b.array2,
   array[a.array1[1] || '-' || b.array2[1],
         a.array1[2] || '-' || b.array2[2],
         a.array1[3] || '-' || b.array2[3]] array_concat
from
   a inner join
   b on b.id = a.id
;

连接表PostgreSQL上的连接数组元素

2 个答案: