问题
given 1,2,3,6,7,8,11,12,15,18,19,20
, write a query to return the maximum of each group of the consecutive numbers are grouped by the query below, but I don't know how to obtain the maximum for each group of consecutive numbers with my current query
with trans as (
select c1,
case when lag(c1) over (order by c1) = c1 - 1 then 0 else 1 end as new
from table1
), groups as (
select c1, sum(new) over (order by c1) as grpnum
from trans
), ranges as (
select grpnum, min(c1) as low, max(c1) as high
from groups
group by grpnum
), texts as (
select grpnum,
case
when low = high then low::text
else low::text||'-'||high::text
end as txt
from ranges
)
select string_agg(txt, ',' order by grpnum) as number
from texts;
回答1:
Assuming you want 3, 8, 12, 15, and 20, you would use lead()
:
select c1
from (select t.*, lead(c1) over (order by c1) as next_c1
from table1 t
) t
where next_c1 is distinct from c1 + 1;
This uses the observation that you can find the end number just by comparing the "next number" to the current value plus 1.
If you want these in a string:
select string_agg(c1::text, ',' order by c1)
Here is a db<>fiddle.
来源:https://stackoverflow.com/questions/62606818/postgresql-given-1-2-3-6-7-8-11-12-15-18-19-20-a-query-to-return-the-maximum