[英]SQL Return different values on the same row, based on a value thats in the same table
我在尝试获取我的结果以显示他们需要它们来显示时遇到了非常困难的时间...这里和我的查询看起来像的示例,下面将是我需要的结果...
iv一直在尝试案例陈述和加入,但没有运气,如果有人可以帮助的话,我将非常感激。
table name dbo.DTLPAYMENTS
columns
PTNO
CD
AMT
DESCRIPTION
我的查询像这样向我显示结果...也永远不会有相等数量的贷方和借方行。 因此,一个人的借方(> 0)多于贷方(<0),反之亦然。...
PTNO / CD / AMT / DESCRIPTION
10007558931 30073 688.82 PAYMENT-ME
10007558931 30073 -704.44 PAYMENT-ME
10007558931 30073 704.44 PAYMENT-ME
10007558931 30073 -688.82 PAYMENT-ME
10007558931 30073 -698.82 PAYMENT-ME
我需要在单独的列中使用借方和贷方
如果有任何可能的话,我可以像这样...
PTNO / CD / AMT / DESCRIPTION / CD / AMT / DESCRIPTION
10007558931 30073 688.82 PAYMENT-ME 30073 -688.82 PAYMENT-ME
10007558931 30073 704.44 PAYMENT-ME 30073 -698.82 PAYMENT-ME
10007558931 30073 -704.44 PAYMENT-ME
如果有人能帮助我,谢谢你
首先,使用CTE将贷方和借方分离为单独的数据集,然后根据PTNO,CD和AMT的反值(* -1)进行完全联接。
编辑:由于基于AMT的加入不是目标,因此我使用BatchSeqID作为按时间顺序对数据进行排序的方法。 然后使用ROW_NUMBER()对两个数据集进行排序,然后在该值上加入。
;WITH
Credit AS
( SELECT PTNO,CD,AMT,DESCRIPTION,
ROW_NUMBER()OVER(PARTITION BY PTNO,CD ORDER BY BatchSeqID) AS Sort
FROM dbo.DTLPAYMENTS
WHERE AMT < 0 ),
Debit AS
( SELECT PTNO,CD,AMT,DESCRIPTION,
ROW_NUMBER()OVER(PARTITION BY PTNO,CD ORDER BY BatchSeqID) AS Sort
FROM dbo.DTLPAYMENTS
WHERE AMT > 0 )
SELECT ISNULL(c.PTNO,d.PTNO) AS PTNO,
ISNULL(c.CD,d.CD) AS CD,
--Credit data
c.AMT AS CRAMT,
c.DESCRIPTION AS CRDESCRIPTION,
--Debit data
d.AMT AS DBTAMT,
d.DESCRIPTION AS DBTDESCRIPTION
FROM Credit c
FULL JOIN Debit d
ON d.PTNO = c.PTNO
AND d.CD = c.CD
AND d.Sort= c.Sort
如果您不喜欢Matt的解决方案以及行重复的风险,则可以将每个事务放在不同的行上:
SELECT PTNO,
CASE WHEN AMT < 0 THEN cd ELSE null END DBTCD,
CASE WHEN AMT < 0 THEN amt ELSE null END AMTCD,
CASE WHEN AMT < 0 THEN DESCRIPTION ELSE null END DBTDESCRIPTION,
CASE WHEN AMT > 0 THEN cd ELSE null END CRCD,
CASE WHEN AMT > 0 THEN amt ELSE null END CRAMT,
CASE WHEN AMT > 0 THEN DESCRIPTION ELSE null END CRDESCRIPTION
FROM TABLE1
order by PTNO, AMT
在这里,您可以查看和播放结果。
PTNO DBTCD AMTCD DBTDESCRIPTION CRCD CRAMT CRDESCRIPTION
10007558931 30073 -704 PAYMENT-ME (null) (null) (null)
10007558931 30073 -699 PAYMENT-ME (null) (null) (null)
10007558931 30073 -689 PAYMENT-ME (null) (null) (null)
10007558931 (null) (null) (null) 30073 689 PAYMENT-ME
10007558931 (null) (null) (null) 30073 704 PAYMENT-ME
假设您要根据amt
列的大小对输出进行排序,请尝试:
select ptno,
max(case sign(amt) when 1 then cd end) dr_cd,
max(case sign(amt) when 1 then amt end) dr_amt,
max(case sign(amt) when 1 then description end) dr_desc,
max(case sign(amt) when -1 then cd end) cr_cd,
max(case sign(amt) when -1 then amt end) cr_amt,
max(case sign(amt) when -1 then description end) cr_desc
from (select d.*,
row_number() over (partition by ptno, cd, sign(amt)
order by abs(amt)) rn
from DTLPAYMENTS d) sq
group by ptno, cd, rn
order by ptno, cd, rn
SQLFiddle 在这里 。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.