Re: how to use aggregate functions in this case
| От | BladeOfLight16 |
|---|---|
| Тема | Re: how to use aggregate functions in this case |
| Дата | |
| Msg-id | CA+=1U=VgJ-Mb_cNe6mANCT71ZTkXtUxm4Sr=AiKViHKbhAYN3Q@mail.gmail.com обсуждение исходный текст |
| Ответ на | Re: how to use aggregate functions in this case (BladeOfLight16 <bladeoflight16@gmail.com>) |
| Ответы |
Re: how to use aggregate functions in this case
|
| Список | pgsql-general |
On Sun, Aug 25, 2013 at 8:36 PM, BladeOfLight16 <bladeoflight16@gmail.com> wrote:
WIDTH_BUCKET(v_rec_fts.lev, 0, 100, 4) AS bucketThis appears to be some kind of equal interval problem.SELECT v_rec1.user,COUNT(*) as count,
FROM v_rec2
GROUP BY user, bucket;(Untested, but this should be the gist.)Bucket 1 would be 0 to 25, bucket 2 is 25 to 50, 3 is 50 to 75, 4 is 75 to 100. If you really need to change the bucket number to some kind of text, you can probably nest this query inside another that uses a CASE to pick the text based on on the bucket number.Good luck.
Then again, I guess you don't need a nested query.
SELECT v_rec1.user,
CASE WIDTH_BUCKET(v_rec_fts.lev, 0, 100, 4)
SELECT v_rec1.user,
CASE WIDTH_BUCKET(v_rec_fts.lev, 0, 100, 4)
WHEN 1 THEN '0 to 25'
WHEN 2 THEN '25 to 50'
WHEN 3 THEN '50 to 75'
WHEN 4 THEN '75 to 100'
ELSE 'But how?'
END CASE AS quarter_percentage
END CASE AS quarter_percentage
COUNT(*) as count,
FROM v_rec2
GROUP BY user, quarter_percentage;
FROM v_rec2
GROUP BY user, quarter_percentage;
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