Question

我需要一次获得16列的最小值。

即。 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16

minimum = 1

我需要这个为一个新的16组滑过1，并获得那个的最小值

即。 2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17

最低2

有2571列，11,000行。

到目前为止，我从以下行获取行：

DECLARE @Start INT =1 
DECLARE @End INT=16 
Begin Transaction
SELECT * 
FROM   (SELECT Row=Row_number() 
                     OVER ( 
                       ORDER BY column_name), 
               column_name 
        FROM   information_schema.columns 
        WHERE  table_name IN ( '100', '200', '300', '400' ) 
               AND column_name NOT IN ( 'time', 'scan' ))Row 
WHERE  row >= @start 
       AND row <= @end 
Commit

这给了我前16列，但是在一列中。我想打开它并将返回的结果放入游标中，我可以执行逐行分析。

像这样的东西，虽然如果我为每个我想要的行号添加1，它应该移动它。

DECLARE @Column01 NVARCHAR(50), 
        @Column02 NVARCHAR(50), 
        @Column03 NVARCHAR(50), 
        @Column04 NVARCHAR(50), 
        @Column05 NVARCHAR(50), 
        @Column06 NVARCHAR(50), 
        @Column07 NVARCHAR(50), 
        @Column08 NVARCHAR(50), 
        @Column09 NVARCHAR(50), 
        @Column10 NVARCHAR(50), 
        @Column11 NVARCHAR(50), 
        @Column12 NVARCHAR(50), 
        @Column13 NVARCHAR(50), 
        @Column14 NVARCHAR(50), 
        @Column15 NVARCHAR(50), 
        @Column16 NVARCHAR(50) 

DECLARE @Start INT =1 
DECLARE @End INT=16 


DECLARE db_cursor CURSOR FOR 
  SELECT column_name 
  FROM   (SELECT Row=Row_number() 
                       OVER ( 
                         ORDER BY column_name), 
                 column_name 
          FROM   information_schema.columns 
          WHERE  table_name IN ( '100', '200', '300', '400' ) 
                 AND column_name NOT IN ( 'time', 'scan' ))Row 
  WHERE  row >= @start
         AND row <= @end 

OPEN db_cursor  
FETCH NEXT FROM db_cursor INTO  @Column01 ,@Column02

WHILE @@FETCH_STATUS = 0  
BEGIN  

      --do work here    
      print @Column01 + ' ' + @Column02

    FETCH NEXT FROM db_cursor INTO  @Column01 , 
@Column02 
END  

CLOSE db_cursor  
DEALLOCATE db_cursor

我知道我可以转动它，但是因为我不需要聚合而能够满足我的要求吗？有什么建议吗？

Answer 1

您不需要动态SQL。只需使用UNPIVOT标准化您的数据。

WITH t AS (
  SELECT row, scan, col, val
  FROM MyTable
  UNPIVOT(val FOR col IN ([1],[2],[3],[4],...<snip>...,[2570],[2571]))p
)
SELECT
  row, scan, col AS range_start, col + 15 AS range_end, min_val
FROM t t1
CROSS APPLY (
  SELECT
    MIN(t2.val) AS min_val
  FROM t
  WHERE
    row = t1.row AND
    scan = t1.scan AND
    col BETWEEN t1.col AND t1.col + 15
) t2

列中的结果集不是行

1 个答案: