将重叠范围合并为数组元素相对容易。为简单起见,以下函数返回set of tstzrange
:
create or replace function merge_ranges(tstzrange[])
returns setof tstzrange language plpgsql as $$
declare
t tstzrange;
r tstzrange;
begin
foreach t in array $1 loop
if r && t then r:= r + t;
else
if r notnull then return next r;
end if;
r:= t;
end if;
end loop;
if r notnull then return next r;
end if;
end $$;
只需聚合用户的范围并使用该函数:
select fk_user_id, merge_ranges(array_agg(sessionrange))
from test
group by 1
order by 1, 2
fk_user_id | merge_ranges
------------+-----------------------------------------------------
1 | ["2016-01-14 17:57:01+01","2016-01-14 18:21:56+01"]
1 | ["2016-01-15 00:18:08+01","2016-01-15 00:18:15+01"]
1 | ["2016-01-15 00:18:16+01","2016-01-15 00:18:31+01"]
1 | ["2016-01-15 00:38:12+01","2016-01-15 00:48:20+01"]
2 | ["2016-01-15 00:18:00+01","2016-01-15 00:42:09+01"]
3 | ["2016-01-15 00:18:12+01","2016-01-15 00:18:20+01"]
3 | ["2016-01-15 01:32:12+01","2016-01-15 05:18:20+01"]
4 | ["2016-01-15 00:18:16+01","2016-01-15 00:18:26+01"]
(8 rows)
或者,该算法可以在一个函数循环中应用于整个表。我不确定,但对于大型数据集,这种方法应该更快。
create or replace function merge_ranges_in_test()
returns setof test language plpgsql as $$
declare
curr test;
prev test;
begin
for curr in
select *
from test
order by fk_user_id, sessionrange
loop
if prev notnull and prev.fk_user_id <> curr.fk_user_id then
return next prev;
prev:= null;
end if;
if prev.sessionrange && curr.sessionrange then
prev.sessionrange:= prev.sessionrange + curr.sessionrange;
else
if prev notnull then
return next prev;
end if;
prev:= curr;
end if;
end loop;
return next prev;
end $$;
结果:
select *
from merge_ranges_in_test();
id | sessionrange | fk_user_id
----+-----------------------------------------------------+------------
1 | ["2016-01-14 17:57:01+01","2016-01-14 18:21:56+01"] | 1
5 | ["2016-01-15 00:18:08+01","2016-01-15 00:18:15+01"] | 1
7 | ["2016-01-15 00:18:16+01","2016-01-15 00:18:31+01"] | 1
6 | ["2016-01-15 00:38:12+01","2016-01-15 00:48:20+01"] | 1
4 | ["2016-01-15 00:18:00+01","2016-01-15 00:42:09+01"] | 2
9 | ["2016-01-15 00:18:12+01","2016-01-15 00:18:20+01"] | 3
10 | ["2016-01-15 01:32:12+01","2016-01-15 05:18:20+01"] | 3
11 | ["2016-01-15 00:18:16+01","2016-01-15 00:18:26+01"] | 4
(8 rows)
这个问题非常有趣。我试图找到一个递归解决方案,但似乎程序尝试是最自然和最有效的。
我终于找到了一个递归解决方案。该查询删除重叠的行并插入它们的压缩等效项:
with recursive cte (user_id, ids, range) as (
select t1.fk_user_id, array[t1.id, t2.id], t1.sessionrange + t2.sessionrange
from test t1
join test t2
on t1.fk_user_id = t2.fk_user_id
and t1.id < t2.id
and t1.sessionrange && t2.sessionrange
union all
select user_id, ids || t.id, range + sessionrange
from cte
join test t
on user_id = t.fk_user_id
and ids[cardinality(ids)] < t.id
and range && t.sessionrange
),
list as (
select distinct on(id) id, range, user_id
from cte, unnest(ids) id
order by id, upper(range)- lower(range) desc
),
deleted as (
delete from test
where id in (select id from list)
)
insert into test
select distinct on (range) id, range, user_id
from list
order by range, id;
结果:
select *
from test
order by 3, 2;
id | sessionrange | fk_user_id
----+-----------------------------------------------------+------------
1 | ["2016-01-14 17:57:01+01","2016-01-14 18:21:56+01"] | 1
5 | ["2016-01-15 00:18:08+01","2016-01-15 00:18:15+01"] | 1
7 | ["2016-01-15 00:18:16+01","2016-01-15 00:18:31+01"] | 1
6 | ["2016-01-15 00:38:12+01","2016-01-15 00:48:20+01"] | 1
4 | ["2016-01-15 00:18:00+01","2016-01-15 00:42:09+01"] | 2
9 | ["2016-01-15 00:18:12+01","2016-01-15 00:18:20+01"] | 3
10 | ["2016-01-15 01:32:12+01","2016-01-15 05:18:20+01"] | 3
11 | ["2016-01-15 00:18:16+01","2016-01-15 00:18:26+01"] | 4
(8 rows)