如何比较SQL Server中的前两行

Question

我有以下表格作为示例：

MTRL表

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

地铁表

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


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

我当前正在使用这两个查询，然后将输出存储到变量中。 我正在与编程进行比较。 但是与其运行两个查询，不如运行一个查询然后获取输出，这会更好。

所以我想比较最后两个价格。 如果最后一个价格已发生重大变化，或者只有一个价格（如mtrl2），则输出价格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)

这存储在变量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)

这存储在可变的price2

我正在像编程一样做（这是vbscript）：

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

如何仅使用SQL来实现？

Answer 1

我可能会建议：

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;

请注意，您的代码一次只能执行一个代码。 您可以删除M.CODE = :kod_an并在多个代码上运行该代码。

Answer 2

尝试这样的事情。 在没有样本数据的情况下没有对其进行测试，但是其想法是首先在CTE中获得两行（rn = 1和rn = 2）。 然后，使用case语句查找abs值，并为rn=返回0或PRICE 。

我也假设您的其余查询是正确的

;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
;