쓰기 SQL Server 쿼리

Question

는 쿼리 제발 도움말 :

select * 
from sc84 as nom 
join sc319 as p on p.PARENTEXT = nom.id 
join sc219 as pt on p.sp327 = pt.id 
join _1SCONST as c on c.objid=p.id 

을 결과적으로 내가 원하는

Car/price_base/08-08-2016:13-40/100/ 
Car/price_base/08-08-2016:14-40/150/ 
Car/price_base/08-09-2016:13-40/190/ 
Car/price_super/08-09-2016:18-40/210/ 
Car/price_super/08-10-2016:13-40/290/  

약 같은 테이블 좋은

Car/price_base/08-09-2016:13-40/190/
Car/price_super/08-10-2016:13-40/290/

을 반환, 날짜의 가격 유형과 그 가치. (가격과 각 상품의 각 유형에 대한 현재 가격) 마지막 옵션을 얻는 방법을 알려주십시오. 그룹과 함께 옵션을 시도했지만 분명히 충분한 기술이 없습니다.

nom.id - PK SKU  
pt.id - PK price type 
p.id -PK price  
p.parentext - parent price (sku)  
p.sp327 - FK to price type 
date = date column 

나는 테이블과 열 구조를 제공하십시오 SQL 서버를 2008

표 구조

T=1SCONST | 

----Columns------- 
Name     |Descr    |Type|Length|Precision 
F=ROW_ID    |Row ID    |I |0  |0   
F=OBJID     |ID obj(0-cons)  |C |9  |0   
F=ID     |ID parameter  |I |0  |0   
F=DATE     |Fix date   |D |0  |0   
F=VALUE     |valume    |V |255 |0   
F=DOCID     |ID Document   |C |9  |0   
F=TIME     |Time    |I |0  |0   
F=ACTNO     |Action No   |I |0  |0   
F=LINENO_    |LineNo    |S |0  |0   
F=TVALUE    |     |C |3  |0  

T=SC319 | 

----Columns------- 
Name     |Descr    |Type|Length|Precision 
F=ROW_ID    |Row ID    |I |0  |0   
F=ID     |ID object   |C |9  |0   
F=PARENTEXT    |Parent in other tabl|C |9  |0   
F=ISMARK    |Object is Marked for|L |0  |0   
F=VERSTAMP    |Version stamp  |I |0  |0   
F=SP6681    |(P)Âàëþòà   |C |9  |0   
F=SP6682    |(P)Ïðîöåíò   |N |5  |2   
F=SP327     |(P)ÒèïÖåí   |C |9  |0  

T=SC319 |Ñïðàâî÷íèê Öåíû    |SC319  |R   

# ----Columns------- 
# Name     |Descr    |Type|Length|Precision 
F=ROW_ID    |Row ID    |I |0  |0   
F=ID     |ID object   |C |9  |0   
F=PARENTEXT    |Parent in other tabl|C |9  |0   
F=ISMARK    |Object is Marked for|L |0  |0   
F=VERSTAMP    |Version stamp  |I |0  |0   
F=SP6681    |(P)Âàëþòà   |C |9  |0   
F=SP6682    |(P)Ïðîöåíò   |N |5  |2   
F=SP327     |(P)ÒèïÖåí   |C |9  |0   

#==TABLE no 24  : Ñïðàâî÷íèê Íîìåíêëàòóðà 
# Name |Descr       |SQLTableNam|RecordLock 
T=SC84 |Ñïðàâî÷íèê Íîìåíêëàòóðà  |SC84  |R   

# ----Columns------- 
# Name     |Descr    |Type|Length|Precision 
F=ROW_ID    |Row ID    |I |0  |0   
F=ID     |ID object   |C |9  |0   
F=PARENTID    |ID parent obj  |C |9  |0   
F=CODE     |object code   |C |8  |0   
F=DESCR     |object description |C |70 |0   
F=ISFOLDER    |Is Line - Folder |Y |0  |0   
F=ISMARK    |Object is Marked for|L |0  |0   
F=VERSTAMP    |Version stamp  |I |0  |0   
F=SP85     |(P)Àðòèêóë   |C |25 |0   
F=SP86     |(P)ÁàçîâàÿÅäèíèöà |C |9  |0   
F=SP208     |(P)Âåñîâîé   |N |1  |0   
F=SP2417    |(P)ÂèäÍîìåíêëàòóðû |C |9  |0   
F=SP97     |(P)ÌèíÎñòàòîê  |N |13 |3   
F=SP5066    |(P)ÍåÂêëþ÷àòüÂïðàéñ |N |1  |0   
F=SP5013    |(P)ÍîìåðÃÒÄ   |C |9  |0   
F=SP94     |(P)ÎñíîâíàÿÅäèíèöà |C |9  |0   
F=SP4427    |(P)ÎñíîâíîåÑâîéñòâî |C |9  |0   
F=SP103     |(P)ÑòàâêàÍÄÑ  |C |9  |0   
F=SP104     |(P)ÑòàâêàÍÏ   |C |9  |0   
F=SP5012    |(P)ÑòðàíàÏðîèñõîæäåí|C |9  |0   
F=SP8574    |(P)ÍèçÎòêëîíåíèÿÑåáå|N |4  |2   
F=SP8575    |(P)ÂåðõÎòêëîíåíèÿÑåá|N |4  |2   
F=SP8576    |(P)ÍèçÎòêëîíåíèÿÑïåö|N |4  |2   
F=SP8577    |(P)ÂåðõÎòêëîíåíèÿÑïå|N |4  |2   
F=SP8578    |(P)ÍèçÎòêëîíåíèÿÇàêó|N |4  |2   
F=SP8579    |(P)ÂåðõÎòêëîíåíèÿÇàê|N |4  |2   
F=SP8580    |(P)ÍèçÎòêëîíåíèÿÐîçí|N |4  |2   
F=SP8581    |(P)ÂåðõÎòêëîíåíèÿÐîç|N |4  |2   
F=SP8599    |(P)Äëèíà   |N |6  |2   
F=SP8600    |(P)Øèðèíà   |N |6  |2   
F=SP8601    |(P)Âûñîòà   |N |6  |2   
F=SP8602    |(P)Îáúåì   |N |14 |5   
F=SP8606    |(P)ÌàêñèìàëüíûéÏðîöå|N |4  |2   
F=SP8607    |(P)Àäðåñ   |C |9  |0   
F=SP95     |(P)Êîììåíòàðèé  |M |0  |0   
F=SP101     |(P)ÏîëíÍàèìåíîâàíèå |M |0  |0 

Answer 1

select * from (
    select *, ROW_NUMBER() over(partition by nom.id,pt.id order by c.date desc) NUM 
    from sc84 as nom 
    join sc319 as p 
     on p.PARENTEXT = nom.id 
    join sc219 as pt 
     on p.sp327 = pt.id 
    join _1SCONST as c 
     on c.objid=p.id 
) A 
    where NUM=1