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モンスタークエリを作成しました。最適化できると確信しており、クエリ自体に関するコメントやガイダンスをいただければ幸いです。ただし、具体的な質問があります。

私が返すデータは、複数の列に複製されることがあります。

+-------+------+----------+------+-------+--------+----------+-------+------+
| first | last |  deaID   | cert | count |  npi   | clientid | month | year |
+-------+------+----------+------+-------+--------+----------+-------+------+
| Alex  | Jue  | UNKNOWN  | MD   |    11 | 123123 |   102889 |     7 | 2012 |
| Alex  | Jue  | BJ123123 | MD   |    11 | 123123 |   102889 |     7 | 2012 |
+-------+------+----------+------+-------+--------+----------+-------+------+

ご覧のとおり、を除いてすべてのフィールドが等しいです。deaID

この場合、私は戻りたいだけです：

+------+-----+----------+----+----+--------+--------+---+------+
|      |     |          |    |    |        |        |   |      |
+------+-----+----------+----+----+--------+--------+---+------+
| Alex | Jue | BJ123123 | MD | 11 | 123123 | 102889 | 7 | 2012 |
+------+-----+----------+----+----+--------+--------+---+------+

ただし、重複がない場合：

+-------+------+---------+------+-------+--------+----------+-------+------+
| first | last |  deaID  | cert | count |  npi   | clientid | month | year |
+-------+------+---------+------+-------+--------+----------+-------+------+
| Alex  | Jue  | UNKNOWN | MD   |    11 | 123123 |   102889 |     7 | 2012 |
+-------+------+---------+------+-------+--------+----------+-------+------+

それなら私はそれを保ちたいです！

重複'deaID=unknown'がある場合は、 ; を使用してすべてのレコードを削除します。ただし、一致するものが1つしかない場合は、その一致を返します

unknown1つの一致がある場合にレコードIFFを 返すにはどうすればよいですか？

誰かが興味を持っている場合のモンスタークエリはここにあります:)

with ctebiggie  as (

select distinct
p.[IMS_PRESCRIBER_ID],
p.PHYSICIAN_NPI as MLISNPI,
a.CLIENT_ID,
p.MLIS_FIRSTNAME,
p.MLIS_LASTNAME,
p_address.IMS_DEA_NBR,
p.IMS_PROFESSIONAL_ID_NBR,
p.IMS_PROFESSIONAL_ID_NBR_src,
p.IMS_CERTIFICATION_CODE,
datepart(mm,a.RECEIVED_DATE) as [Month],
datepart(yyyy,a.RECEIVED_DATE) as [Year]

from

MILLENNIUM_DW_dev..D_PHYSICIAN p
left outer join
MILLENNIUM_DW_dev..F_ACCESSION_DAILY a
on a.REQUESTOR_NPI=p.PHYSICIAN_NPI
left outer join MILLENNIUM_DW_dev..D_PHYSICIAN_ADDRESS p_address
on p.PHYSICIAN_NPI=p_address.PHYSICIAN_NPI

where 
a.RECEIVED_DATE is not null
--and p.IMS_PRESCRIBER_ID is not null
--and p_address.IMS_DEA_NBR !='UNKNOWN'
and p.REC_ACTIVE_FLG=1
and p_address.REC_ACTIVE_FLG=1
and DATEPART(yyyy,received_date)=2012
  and DATEPART(mm,received_date)=7


group by 
p.[IMS_PRESCRIBER_ID],
p.PHYSICIAN_NPI,
p.IMS_PROFESSIONAL_ID_NBR,
p.MLIS_FIRSTNAME,
p.MLIS_LASTNAME,
p_address.IMS_DEA_NBR,
p.IMS_PROFESSIONAL_ID_NBR,
p.IMS_PROFESSIONAL_ID_NBR_src,
p.IMS_CERTIFICATION_CODE,
datepart(mm,a.RECEIVED_DATE),
datepart(yyyy,a.RECEIVED_DATE),
a.CLIENT_ID

)
,
ctecount as 
(select
 COUNT (Distinct f.ACCESSION_ID) [count],
 f.REQUESTOR_NPI,f.CLIENT_ID,
 datepart(mm,f.RECEIVED_DATE) mm,
datepart(yyyy,f.RECEIVED_DATE)yyyy
from MILLENNIUM_DW_dev..F_ACCESSION_DAILY f

where 
 f.CLIENT_ID not in (select * from SalesDWH..TestPractices)

 and DATEPART(yyyy,f.received_date)=2012
  and DATEPART(mm,f.received_date)=7


group by f.REQUESTOR_NPI,
f.CLIENT_ID,
datepart(mm,f.RECEIVED_DATE),
datepart(yyyy,f.RECEIVED_DATE)
)

select ctebiggie.*,c.* from
ctebiggie
full outer join
ctecount c
on c.REQUESTOR_NPI=ctebiggie.MLISNPI
and c.mm=ctebiggie.[Month]
and c.yyyy=ctebiggie.[Year]
and c.CLIENT_ID=ctebiggie.CLIENT_ID
4

2 に答える 2

3

基本クエリがあると仮定して、この結果セットにrow_numberとcountbypartition関数を割り当てます。次に、外側の選択で、カウントが1の場合は不明が選択され、それ以外の場合は選択されません。

SELECT first,
       last,
       deaID,
       cert,
       count,
       npi,
       clientid,
       month,
       year
  FROM (
         SELECT first,
                last,
                deaID,
                cert,
                count,
                npi,
                clientid,
                month,
                year,
                ROW_NUMBER() OVER (PARTITION BY
                                     first,last,cert,count,npi,clientid,month,year 
                                    ORDER BY CASE WHEN deaID = 'Unkown' THEN 0 ELSE 1 END,
                                       deaID) AS RowNumberInGroup,
                COUNT() OVER (PARTITION BY first,last,cert,count,npi,clientid,month,year)
                    AS CountPerGroup,
                 SUM(CASE WHEN deaID = 'Unkown' THEN 1 ELSE 0 END) 
                     OVER (PARTITION BY first,last,cert,count,npi,clientid,month,year)
                     AS UnknownCountPerGroup
           FROM BaseQuery
      ) T
 WHERE (T.CountPerGroup = T.UnknownCountPerGroup AND T.RowNumberInGroup = 1) OR T.RowNumberInGroup > T.UnknownCountPerGroup
于 2012-09-07T17:30:46.313 に答える
2

これが役立つかどうかを参照してください

select distinct main.col1,main.col2  ,
       isnull(( select col3 from table1 where table1.col1=main.col1
       and table1.col2=main.col2 and col3 <>'UNKNOWN'),'UNKNOWN')
from   table1 main

SQLフィドルのサンプル

またはあなたの公正なバージョンは

SELECT distinct first,
       last,
       cert,
       count,
       npi,
       clientid,
       month,
       year,
      isnull(
      select top 1 dealid from table1 intable where 
      intable.first=maintable.first and
      intable.last=maintable.last and
      intable.cert=maintable.cert and
      intable.npi=maintable.npi and
      intable.clientid=outtable.clientid and
      intable.month=outtable.month and
      intable.year=outtable.year
      where dealid<>'UNKNOWN'),'UNKNOWN') as dealId
FROM  table1 maintable
于 2012-09-07T17:30:08.683 に答える