Auto-answer: WITH uniques AS ( SELECT group_key, COUNT(DISTINCT reference) AS unique_references FROM record_data GROUP BY group_key ) SELECT group_key, record_data.*, uniques.unique_references FROM record_data JOIN uniques USING (group_key) ORDER BY group_key;
Super-PostgreSQL is my hero... 2009/7/21 Daniel Cristian Cruz <[email protected]> > Hi list, > > Does someone knows why this is not a supported feature? > > I was happily SQL querying when I woke up from my dreams with this message: > > ERROR: DISTINCT is not implemented for window functions > LINE 6: COUNT(DISTINCT handle) OVER (PARTITION BY pk_pessoas) AS ha... > > It could be very useful to find unique references within groups of clone > records. Or there is another way to write this kind of query? > > Example (reference is a column of record_data): > > SELECT > group_key, > record_data.*, > COUNT(DISTINCT reference) OVER (PARTITION BY group_key) AS > unique_references > FROM record_data > ORDER BY > group_key; > > Regards, > -- > Daniel Cristian Cruz > クルズ クリスチアン ダニエル > -- Daniel Cristian Cruz クルズ クリスチアン ダニエル
