如何比較SQL Server中的前兩行

Question

我有以下表格作為示例：

MTRL表

|MTRL|  CODE  |
---------------
| 1  | 080109 |
| 2  | 085145 |
| 3  | 084141 |

地鐵表

|MTRL|  PRICE  | FINDOC |
-------------------------
| 1  |  4.95   | 12345  |
| 1  |  4.50   | 23421  |
| 1  |  3.90   | 23499  |


|MTRL|  PRICE  | FINDOC |
-------------------------
| 2  |  2.95   | 45345  |

我當前正在使用這兩個查詢，然后將輸出存儲到變量中。 我正在與編程進行比較。 但是與其運行兩個查詢，不如運行一個查詢然后獲取輸出，這會更好。

所以我想比較最后兩個價格。 如果最后一個價格已發生重大變化，或者只有一個價格（如mtrl2），則輸出價格1

SELECT 
      , SUB.PRICE
FROM (
        SELECT 
              , ML.PRICE
              , rn = row_number() over (PARTITION BY M.CODE ORDER BY FINDOC DESC)
        FROM  MTRLINES ML
              INNER JOIN MTRL M
                    ON M.MTRL = ML.MTRL  AND FINDOC IN (SELECT FINDOC FROM FINDOC WHERE SOSOURCE=1251 AND FPRMS IN (1,2)) 
              WHERE  M.SODTYPE=51 AND M.COMPANY=1 AND M.CODE=:kod_an
      ) sub
WHERE rn IN (1)

這存儲在變量price1

SELECT 
      , SUB.PRICE
FROM (
        SELECT 
              , ML.PRICE
              , rn = row_number() over (PARTITION BY M.CODE ORDER BY FINDOC DESC)
        FROM  MTRLINES ML
              INNER JOIN MTRL M
                    ON M.MTRL = ML.MTRL  AND FINDOC IN (SELECT FINDOC FROM FINDOC WHERE SOSOURCE=1251 AND FPRMS IN (1,2)) 
              WHERE  M.SODTYPE=51 AND M.COMPANY=1 AND M.CODE=:kod_an
      ) sub
WHERE rn IN (2)

這存儲在可變的price2

我正在像編程一樣做（這是vbscript）：

If ABS(price1-price2)>0.02 Then
    Result=Price1
Else
    Result=0
End If

如何僅使用SQL來實現？

Answer 1

我可能會建議：

with prices as (
      select code,
             max(case when seqnum = 1 then price end) as price_last,  
             max(case when seqnum = 2 then price end) as price_second,
             count(*) as num_prices 
      from (select ml.price, m.code,
                   row_number() over (partition by M.CODE order by FINDOC desc) as seqnum
            from MTRLINES ML inner join
                 MTRL M
                 ON M.MTRL = ML.MTRL and
                    FINDOC in (select FINDOC from FINDOC where SOSOURCE = 1251 and FPRMS in (1, 2)
                              ) 
            where M.SODTYPE = 51 and M.COMPANY = 1 and
                  M.CODE = :kod_an
           ) m
      where seqnum in (1, 2)
      group by code
     )
select (case when num_prices = 1 then price_last
             when abs(price_last - price_second) > 0.02 then price_last
             else 0
        end)
from prices;

請注意，您的代碼一次只能執行一個代碼。 您可以刪除M.CODE = :kod_an並在多個代碼上運行該代碼。

Answer 2

嘗試這樣的事情。 在沒有樣本數據的情況下沒有對其進行測試，但是其想法是首先在CTE中獲得兩行（rn = 1和rn = 2）。 然后，使用case語句查找abs值，並為rn=返回0或PRICE 。

我也假設您的其余查詢是正確的

;with t as (
SELECT 
      SUB.PRICE
FROM (
        SELECT 
               ML.PRICE
              , rn = row_number() over (PARTITION BY M.CODE ORDER BY FINDOC DESC)
        FROM  MTRLINES ML
              INNER JOIN MTRL M
                    ON M.MTRL = ML.MTRL  AND FINDOC IN (SELECT FINDOC FROM FINDOC WHERE SOSOURCE=1251 AND FPRMS IN (1,2)) 
              WHERE  M.SODTYPE=51 AND M.COMPANY=1 AND M.CODE=:kod_an
      ) sub
WHERE rn IN (1,2)
)
select case 
            when abs(sum(case when rn=1 
                            then price 
                            else (price * -1 ) 
                        end)
                    )
                    > 0.02  
        then (select price1 from t where rn=1) 
        else 0 end
         as result
from t
;