Arrays 连接联接表上的数组元素PostgreSQL
如果我有这样一个查询,是否可以进行1对1元素数组连接:Arrays 连接联接表上的数组元素PostgreSQL,arrays,postgresql,concatenation,unnest,Arrays,Postgresql,Concatenation,Unnest,如果我有这样一个查询,是否可以进行1对1元素数组连接: 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 ... 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,arra
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 ...
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,array2有时有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
这里我想得到的是数组1和数组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unnest,但无法使其工作:
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
;
编辑:
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);
++多亏了
编辑:
我找到了一个非常接近的解决方案,但一旦数组之间的连接完成,它就混合了一些中间值,尽管如此,我仍然需要一个完美的解决方案
我现在采用的方法是:
1基于两个单独的表创建一个表
2使用横向聚合:
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;
我觉得你想得太远了,试试这个: 像这样的
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);
使用..的联接运算符。。将自动从联接的表中获取非空值。这样就不需要使用例如Coalesca.id、b.id.n
对于大型桌子来说,它并不漂亮,也绝对不高效,但似乎可以满足您的所有需求
对于元素数量不相同的数组,结果将只包含其中一个数组中的元素
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']);
它返回以下结果:
id |元素
--+--------
1{a-x,b-y,c-z,d}
2{d-m,e-n,f-o,p}
旁注:正确的JSON带有not with'。在PostgreSQL中,您还可以将包含数组的字符串定义为:选择数组['a','b','c']作为数组1;我编辑了这个问题,问题是我在每个数组中的元素数量并不总是相同的。是否只加入元素数量相等的数组?这很难看,但是你可以对4个和8个元素进行上述操作,如果它是['a','b','c','d']和['x','y','z'],那么结果应该是['a-x','b-y','c-z','d-']或['a-x','b-y','c-z','d-null']