Re: Guidance needed on an alternative take on common prefix SQL

On Wednesday, August 7, 2019 2:01 AM, Andy Colson <andy(at)squeakycode(dot)net> wrote:

> 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

Hi Andy,

That looks supremely clever !

I have just done a quick test and looks like it works as intended. Will do some more thorough testing with a larger dataset in due course.

Thank you very much indeed

Laura