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

On 8/7/19 3:36 AM, Laura Smith wrote:
> 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
>
>

If the target field is really an integer type and you have lots of rows
you might be better off with arithmetic functions.

create table short as select id/10 as base, array_agg(mod(id,10)) as
odds from head group by base;

select * from short;
  base   |        odds
---------+---------------------
  879740 | {0}
 8797389 | {1}
 8797397 | {0,1,2,3,5,6,7,8,9}
(3 rows)

 select case when array_length(odds,1) = 1 then 10*base + odds[1] else
base end from short;
   base
----------
  8797400
 87973891
  8797397
(3 rows)