代码之家 › 专栏 › 技术社区 › Dennis G

分组分层数据(parentID+ID)和运行总和?

ssrs-2008 reporting-services sql

1

Dennis G · 技术社区 · 16 年前

我有以下数据:

ID    parentID    Text        Price
1                 Root
2     1           Flowers
3     1           Electro
4     2           Rose        10
5     2           Violet      5
6     4           Red Rose    12
7     3           Television  100
8     3           Radio       70
9     8           Webradio    90

我正试图将此数据与Reporting Services 2008进行分组,并获得级别1(Flowers/Electro)和级别0(Root)的每组价格的总和。

ID    Text        Price
1     Root
|2    Flowers
|-4   Rose        10
|-5   Violet      5
| |-6 Red Rose    12
|     Group Sum-->27
|3    Electro
|-7   Television  100
|-8   Radio       70
  |-9 Webradio    90
      Group Sum-->260
----------------------
Total             287

(刚刚添加的ID缩进用于级别澄清)

ID    parentID    Text        Price    level0    level1    level2    level3
1                 Root                 1
2     1           Flowers              1         1
3     1           Electro              1         2
4     2           Rose        10       1         1         1
5     2           Violet      5        1         1         2
6     4           Red Rose    12       1         1         1         1
7     3           Television  100      1         2         1
8     3           Radio       70       1         2         2
9     8           Webradio    90       1         2         2         1

当具有上述结构时,我可以创建level0的外部分组,并相应地创建level1、level2和level3的子分组。当现在有一个级别1上的“组总和”,以及组外的总总和时,我正好得到了我想要的。

我的问题如下: 如何使用当前数据结构实现所需的结果,或者如何将当前数据结构(外部左连接?)临时转换为“新数据结构”——以便在临时表中运行报表?

谢谢你抽出时间, 丹尼斯

2 回复 | 直到 16 年前

1

Quassnoi 16 年前

WITH    q AS
        (
        SELECT  id, parentId, price
        FROM    mytable
        UNION ALL
        SELECT  p.id, p.parentID, q.price
        FROM    q
        JOIN    mytable p
        ON      p.id = q.parentID
        )
SELECT  id, SUM(price)
FROM    q
GROUP BY
        id

更新:

要检查的测试脚本:

DECLARE @table TABLE (id INT NOT NULL PRIMARY KEY, parentID INT, txt VARCHAR(200) NOT NULL, price MONEY)

INSERT
INTO @table
SELECT 1, NULL, 'Root', NULL
UNION ALL
SELECT 2, 1, 'Flowers', NULL
UNION ALL
SELECT 3, 1, 'Electro', NULL
UNION ALL
SELECT 4, 2, 'Rose', 10
UNION ALL
SELECT 5, 2, 'Violet', 5
UNION ALL
SELECT 6, 4, 'Red Rose', 12
UNION ALL
SELECT 7, 3, 'Television', 100
UNION ALL
SELECT 8, 3, 'Radio', 70
UNION ALL
SELECT 9, 8, 'Webradio', 90;

WITH    q AS
        (
        SELECT  id, parentId, price
        FROM    @table
        UNION ALL
        SELECT  p.id, p.parentID, q.price
        FROM    q
        JOIN    @table p
        ON      p.id = q.parentID
        )
SELECT  t.*, psum
FROM    (        
        SELECT  id, SUM(price) AS psum
        FROM    q
        GROUP BY
                id
        ) qo
JOIN    @table t
ON      t.id = qo.id

结果如下:

1       NULL    Root            NULL    287,00
2       1       Flowers         NULL    27,00
3       1       Electro         NULL    260,00
4       2       Rose            10,00   22,00
5       2       Violet          5,00    5,00
6       4       Red Rose        12,00   12,00
7       3       Television      100,00  100,00
8       3       Radio           70,00   160,00
9       8       Webradio        90,00   90,00

2

0

Dennis G 16 年前

SELECT A.Text, A.Price,                   
  CASE
    WHEN D.Text IS NULL
    THEN
        CASE
            WHEN C.Text IS NULL
            THEN
                CASE
                    WHEN B.Text IS NULL
                    THEN
                        A.ID
                    ELSE B.ID
                END
            ELSE C.ID
            END
    ELSE D.ID
  END
  AS LEV0,

  CASE
    WHEN D.Text IS NULL
    THEN
        CASE
            WHEN C.Text IS NULL
            THEN
                CASE
                    WHEN B.Text IS NULL
                    THEN
                        NULL
                    ELSE A.ID
                END
            ELSE B.ID
            END
    ELSE C.ID
  END
  AS LEV1,

  CASE
    WHEN D.Text IS NULL
    THEN
        CASE
            WHEN C.Text IS NULL
            THEN
                NULL
            ELSE A.ID
            END
    ELSE B.ID
  END
  AS LEV2,

  CASE
    WHEN D.Text IS NULL
    THEN NULL
    ELSE A.ID
  END
  AS LEV3

FROM         dbo.testOld AS A LEFT OUTER JOIN
                  dbo.testOld AS B ON A.parentID = B.ID LEFT OUTER JOIN
                  dbo.testOld AS C ON B.parentID = C.ID LEFT OUTER JOIN
                  dbo.testOld AS D ON C.parentID = D.ID

Text       Price       LEV0        LEV1        LEV2        LEV3
---------- ----------- ----------- ----------- ----------- -----------
Root       NULL        1           NULL        NULL        NULL
Flowers    NULL        1           3           NULL        NULL
Electro    NULL        1           4           NULL        NULL
Television 100         1           4           5           NULL
Radio      70          1           4           6           NULL
Rose       10          1           3           7           NULL
Violet     5           1           3           8           NULL
Webradio   90          1           4           5           14
Red Rose   12          1           3           7           15

有了这个结构,我可以继续在LEV0-3列上创建4个嵌套组,包括每个组的小计(如上面我想要的结果所示)。