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依据“坐标”生成趋势图

热度:93   发布时间:2016-04-24 08:57:01.0
根据“坐标”生成趋势图

数据库环境:SQL SERVER 2008R2

  有一“坐标”表t,表结构如下:

  id           int,

  num       int

  字段id是序号,递增且连续,字段num是数值类型。id可以看成是坐标轴的横轴,num则跟纵轴有关系,

连续的2行记录,如果后一行的num值比前一行的num值大,则是递增趋势,反之,是递减趋势。要实现的效果如下图1:

图1

  实现思路:

  将id=1的“坐标”所在纵轴设定为0,然后遍历后续所有的“坐标”,后面“坐标”的num值比前一个“坐标”的num大的,则纵坐标+1,

否则纵坐标-1。再对整理后的结果集进行行转列,根据纵坐标分组。

1.建表,插入测试数据

CREATE TABLE t(id INT,num INT)INSERT INTO t VALUES(1,1);INSERT INTO t VALUES(2,3);INSERT INTO t VALUES(3,4);INSERT INTO t VALUES(4,7);INSERT INTO t VALUES(5,5);INSERT INTO t VALUES(6,2);INSERT INTO t VALUES(7,6);INSERT INTO t VALUES(8,8);INSERT INTO t VALUES(9,4);INSERT INTO t VALUES(10,0);INSERT INTO t VALUES(11,9);INSERT INTO t VALUES(12,10);INSERT INTO t VALUES(13,12);INSERT INTO t VALUES(14,11);INSERT INTO t VALUES(15,17);INSERT INTO t VALUES(16,4);INSERT INTO t VALUES(17,2);INSERT INTO t VALUES(18,1);
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2.遍历所有“坐标”,生成分组依据

WITH x1 ( id, num, gp )          AS ( SELECT   id ,                        num ,                        0 AS gp               FROM     t x0               WHERE    id = 1               UNION ALL               SELECT   x0.id ,                        x0.num ,                        CASE WHEN x0.num > x1.num THEN x1.gp - 1                             ELSE x1.gp + 1                        END AS gp               FROM     t x0 ,                        x1               WHERE    x0.id = x1.id + 1             )
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3.行转列实现最终结果集

SELECT  ISNULL(CAST([1] AS VARCHAR(2)), '') [1] ,            ISNULL(CAST([2] AS VARCHAR(2)), '') [2] ,            ISNULL(CAST([3] AS VARCHAR(2)), '') [3] ,            ISNULL(CAST([4] AS VARCHAR(2)), '') [4] ,            ISNULL(CAST([5] AS VARCHAR(2)), '') [5] ,            ISNULL(CAST([6] AS VARCHAR(2)), '') [6] ,            ISNULL(CAST([7] AS VARCHAR(2)), '') [7] ,            ISNULL(CAST([8] AS VARCHAR(2)), '') [8] ,            ISNULL(CAST([9] AS VARCHAR(2)), '') [9] ,            ISNULL(CAST([10] AS VARCHAR(2)), '') [10] ,            ISNULL(CAST([11] AS VARCHAR(2)), '') [11] ,            ISNULL(CAST([12] AS VARCHAR(2)), '') [12] ,            ISNULL(CAST([13] AS VARCHAR(2)), '') [13] ,            ISNULL(CAST([14] AS VARCHAR(2)), '') [14] ,            ISNULL(CAST([15] AS VARCHAR(2)), '') [15] ,            ISNULL(CAST([16] AS VARCHAR(2)), '') [16] ,            ISNULL(CAST([17] AS VARCHAR(2)), '') [17] ,            ISNULL(CAST([18] AS VARCHAR(2)), '') [18]    FROM    ( SELECT    *              FROM      x1            ) AS t1 PIVOT( MAX(num) FOR id IN ( [1], [2], [3], [4], [5], [6],                                                [7], [8], [9], [10], [11],                                                [12], [13], [14], [15], [16],                                                [17], [18] ) )AS t2
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  如果要实现比较逼真的趋势图,用“/”、“\”替代数值,实现下图2的效果,只需在遍历坐标的时候再做些处理即可。

  实现的SQL脚本:

WITH    x1 ( id, num, cc, gp )          AS ( SELECT   id ,                        num ,                        '/' AS cc ,                        0 AS gp               FROM     t x0               WHERE    id = 1               UNION ALL               SELECT   x0.id ,                        x0.num ,                        CASE WHEN x0.num > x1.num THEN '/'                             ELSE '\'                        END AS cc ,                        CASE WHEN x0.num > x1.num THEN x1.gp - 1                             ELSE x1.gp + 1                        END AS gp               FROM     t x0 ,                        x1               WHERE    x0.id = x1.id + 1             )    SELECT  ISNULL([1], '') [1] ,            ISNULL([2], '') [2] ,            ISNULL([3], '') [3] ,            ISNULL([4], '') [4] ,            ISNULL([5], '') [5] ,            ISNULL([6], '') [6] ,            ISNULL([7], '') [7] ,            ISNULL([8], '') [8] ,            ISNULL([9], '') [9] ,            ISNULL([10], '') [10] ,            ISNULL([11], '') [11] ,            ISNULL([12], '') [12] ,            ISNULL([13], '') [13] ,            ISNULL([14], '') [14] ,            ISNULL([15], '') [15] ,            ISNULL([16], '') [16] ,            ISNULL([17], '') [17] ,            ISNULL([18], '') [18]    FROM    ( SELECT    id ,                        cc ,                        gp              FROM      x1            ) AS t1 PIVOT( MAX(cc) FOR id IN ( [1], [2], [3], [4], [5], [6],                                               [7], [8], [9], [10], [11], [12],                                               [13], [14], [15], [16], [17],                                               [18] ) )AS t2
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