On 8/6/19 6:25 PM, Laura Smith wrote:
Hi,
I've seen various Postgres examples here and elsewhere that deal with the old
common-prefix problem (i.e. "given 1234 show me the longest match").
I'm in need of a bit of guidance on how best to implement an alternative take.
Frankly I don't quite know where to start but I'm guessing it will probably
involve CTEs, which is an area I'm very weak on.
So, without further ado, here's the scenario:
Given an SQL filtering query output that includes the following column:
87973891
87973970
87973971
87973972
87973973
87973975
87973976
87973977
87973978
87973979
8797400
The final output should be further filtered down to:
87973891
8797397
8797400
i.e. if $last_digit is present 0–9 inclusive, recursively filter until the
remaining string is all the same (i.e. in this case, when $last_digit[0-9] is
removed, 8797397 is the same).
So, coming back to the example above:
8797397[0-9] is present
so the "nearest common" I would be looking for is 8797397 because once [0-9] is
removed, the 7 is the same on the preceeding digit.
The other two rows ( 87973891 and 8797400) are left untouched because
$last_digit is not present in [0-9].
Hope this question makes sense !
Laura
Hows this?
select distinct
case cc
when 1 then num
else left(num,-1)
end
from (
select
num,
(select count(*) as cc from numbers n2 where left(n2.num, -1) =
left(numbers.num, -1))
from numbers
) as tmpx ;
-Andy
