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：

　　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);

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
             )

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

　　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