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SQL:selezionare il valore distinto in sequenza più recente con raggruppamento

Hmmm. . . Un metodo consiste nell'ottenere l'ultimo valore. Quindi scegli tutte le ultime righe con quel valore e aggrega:

select min(rownum), colA, colB
from (select t.*,
             first_value(colA) over (partition by colB order by rownum desc) as last_colA
      from t
     ) t
where rownum > all (select t2.rownum
                    from t t2
                    where t2.colB = t.colB and t2.colA <> t.last_colA
                   )
group by colA, colB;

Oppure, senza l'aggregazione:

select t.*
from (select t.*,
             first_value(colA) over (partition by colB order by rownum desc) as last_colA,
             lag(colA) over (partition by colB order by rownum) as prev_clA
      from t
     ) t
where rownum > all (select t2.rownum
                    from t t2
                    where t2.colB = t.colB and t2.colA <> t.last_colA
                   ) and
      (prev_colA is null or prev_colA <> colA);

Ma in SQL Server 2008, trattiamo questo come un problema di lacune e isole:

select t.*
from (select t.*,
             min(rownum) over (partition by colB, colA, (seqnum_b - seqnum_ab) ) as min_rownum_group,
             max(rownum) over (partition by colB, colA, (seqnum_b - seqnum_ab) ) as max_rownum_group
      from (select t.*,
                   row_number() over (partition by colB order by rownum) as seqnum_b,
                   row_number() over (partition by colB, colA order by rownum) as seqnum_ab,
                   max(rownum) over (partition by colB order by rownum) as max_rownum
            from t
           ) t
     ) t
where rownum = min_rownum_group and  -- first row in the group defined by adjacent colA, colB
      max_rownum_group = max_rownum  -- last group for each colB;

Questo identifica ciascuno dei gruppi utilizzando una differenza di numeri di riga. Calcola il numero massimo di righe per il gruppo e complessivamente nei dati. Questi sono gli stessi per l'ultimo gruppo.