1
我遇到了针对我们的MS SQL Server数据库之一的查询问题。为了简洁起见，下面的表格和视图被简化了，但应该用来描述问题。SQL Server递归查询问题
每个表格的得分都被编译为其直系子女的平均值。这些视图对于固定结构已经足够了，但是对于当前处于嵌套集合形式的位置层次结构来说，它变得更加复杂。由于用户定义了位置层次结构中没有固定数量的层。
我试着用递归CTE解决这个问题，但是他们不允许在递归部分进行聚合。
CREATE TABLE [dbo].[locations_main](
[id] [smallint] NOT NULL,
[name] [nchar](50) NOT NULL,
[lft] [smallint] NOT NULL,
[rgt] [smallint] NOT NULL,
[parent_id] [smallint] NULL,
CONSTRAINT [PK_locations_main] PRIMARY KEY CLUSTERED ([id] ASC)
)
GO
INSERT INTO [dbo].[locations_main] VALUES
(1, 'location 1', 1, 16, NULL),
(2, 'location 1-1', 2, 9, 1),
(3, 'location 1-1-1', 3, 4, 2),
(4, 'location 1-1-2', 5, 6, 2),
(5, 'location 1-1-3', 7, 8, 2),
(7, 'location 1-2', 10, 15, 1),
(8, 'location 1-2-1', 11, 12, 7),
(9, 'location 1-2-2', 13, 14, 7)
GO
CREATE TABLE [dbo].[outcomes](
[id] [smallint] NOT NULL,
[location_id] [smallint] NOT NULL,
[name] [nvarchar](50) NOT NULL,
CONSTRAINT [PK_outcomes] PRIMARY KEY CLUSTERED ([id] ASC)
)
GO
INSERT INTO [dbo].[outcomes] VALUES
(1, 3, 'outcome 1'),
(2, 4, 'outcome 2'),
(3, 5, 'outcome 3'),
(4, 8, 'outcome 4'),
(5, 9, 'outcome 5')
GO
CREATE TABLE [dbo].[prompts](
[id] [smallint] NOT NULL,
[outcome_id] [smallint] NOT NULL,
[name] [nvarchar](50) NOT NULL,
CONSTRAINT [PK_prompts] PRIMARY KEY CLUSTERED ([id] ASC)
)
GO
INSERT INTO [dbo].[prompts] VALUES
(1, 1, 'prompt 1'),
(2, 2, 'prompt 2'),
(3, 3, 'prompt 3'),
(4, 4, 'prompt 4'),
(5, 5, 'prompt 5')
GO
CREATE TABLE [dbo].[subprompts](
[id] [smallint] NOT NULL,
[prompt_id] [smallint] NOT NULL,
[name] [nvarchar](50) NOT NULL,
[score] [smallint] NOT NULL,
CONSTRAINT [PK_subprompts] PRIMARY KEY CLUSTERED ([id] ASC)
)
GO
INSERT INTO [dbo].[subprompts] VALUES
(1, 1, 'subprompt 1', 1),
(2, 1, 'subprompt 2', 1),
(3, 2, 'subprompt 3', 1),
(4, 2, 'subprompt 4', 3),
(5, 3, 'subprompt 5', 2),
(6, 3, 'subprompt 6', 4),
(7, 4, 'subprompt 7', 1),
(8, 4, 'subprompt 8', 5),
(9, 5, 'subprompt 9', 3),
(10, 5, 'subprompt 10', 3)
GO
CREATE VIEW [dbo].[vw_prompts]
AS
SELECT
dbo.prompts.id,
dbo.prompts.outcome_id,
dbo.prompts.name,
AVG(dbo.subprompts.score) AS score
FROM dbo.prompts
LEFT OUTER JOIN dbo.subprompts
ON dbo.prompts.id = dbo.subprompts.prompt_id
GROUP BY
dbo.prompts.id,
dbo.prompts.outcome_id,
dbo.prompts.name
GO
CREATE VIEW [dbo].[vw_outcomes]
AS
SELECT
dbo.outcomes.id,
dbo.outcomes.location_id,
dbo.outcomes.name,
AVG(dbo.vw_prompts.score) AS score
FROM dbo.outcomes
LEFT OUTER JOIN dbo.vw_prompts
ON dbo.outcomes.id = dbo.vw_prompts.id
GROUP BY
dbo.outcomes.id,
dbo.outcomes.location_id,
dbo.outcomes.name
GO
下面的查询检索所有的位置，但它的叶节点计算平均值不是位置的问题直接子 -
SELECT loc_main_ag.name, AVG(CAST(vw_outcomes.score AS FLOAT))
FROM locations_main loc_main_ag
LEFT JOIN locations_main loc_main
ON loc_main_ag.lft <= loc_main.lft
AND loc_main_ag.rgt >= loc_main.rgt
INNER JOIN vw_outcomes
ON loc_main.id = vw_outcomes.location_id
GROUP BY loc_main_ag.name
回报
location 1 2.4
location 1-1 2
location 1-1-1 1
location 1-1-2 2
location 1-1-3 3
location 1-2 3
location 1-2-1 3
location 1-2-2 3
“位置1“具有”位置1-1-1“，”位置1-1-2“，”位置1-1-3“，”位置1-2-1“和”位置1-2-2“的平均值 - （1 + 2 + 3 + 3 + 3）/ 5 = 2.4，而不是“位置1-1”和“位置的平均值Ñ1-2" - （2 + 3）/ 2 = 2.5
我试图通过使用CTE来解决这个打了一个问题，使用CTE的递归部分内GROUP BY和聚合函数 -
WITH location_scores
AS
(
-- Anchor member definition
-- Get score for all leaf node locations
SELECT locations_main.id, locations_main.name, locations_main.parent_id, AVG(CAST(vw_outcomes.score AS FLOAT)) AS score
FROM locations_main
INNER JOIN vw_outcomes
ON locations_main.id = vw_outcomes.location_id
WHERE locations_main.rgt - locations_main.lft = 1
GROUP BY locations_main.id, locations_main.name, locations_main.parent_id
UNION ALL
-- Recursive member definition
-- Rollup through locations parents to build averages
SELECT locations_main.id, locations_main.name, locations_main.parent_id, AVG(CAST(location_scores.score AS FLOAT)) AS score
FROM locations_main
INNER JOIN vw_outcomes
ON locations_main.id = vw_outcomes.location_id
INNER JOIN location_scores
ON locations_main.id = location_scores.parent_id
GROUP BY locations_main.id, locations_main.name, locations_main.parent_id
)
-- Statement that executes the CTE
SELECT *
FROM location_scores
UPDATE：这是我对表值函数的尝试。它根据这里包含的简单示例返回正确的结果，但我担心这将如何扩展。它将在野外执行的层次可以在15^5个记录的区域中的某个地方。
CREATE FUNCTION scores() RETURNS
@result TABLE
(
id SMALLINT,
name NVARCHAR(50),
lft SMALLINT,
rgt SMALLINT,
parent_id SMALLINT,
score FLOAT,
[level] SMALLINT
) AS
BEGIN
DECLARE @level INT
SET @level = 1
INSERT INTO @result
SELECT
locations_main.id,
locations_main.name,
locations_main.lft,
locations_main.rgt,
locations_main.parent_id,
AVG(CAST(vw_outcomes.score AS FLOAT)) AS score,
@level AS [level]
FROM locations_main
INNER JOIN vw_outcomes
ON locations_main.id = vw_outcomes.location_id
WHERE locations_main.rgt - locations_main.lft = 1
GROUP BY
locations_main.id,
locations_main.name,
locations_main.lft,
locations_main.rgt,
locations_main.parent_id
WHILE (SELECT COUNT(*) FROM @result WHERE level = @level AND parent_id IS NOT NULL) > 0 BEGIN
INSERT INTO @result
SELECT
locations_main.id,
locations_main.name,
locations_main.lft,
locations_main.rgt,
locations_main.parent_id,
AVG(CAST(res.score AS FLOAT)) AS score,
(@level + 1) AS [level]
FROM locations_main
INNER JOIN @result res
ON locations_main.id = res.parent_id
AND res.level = @level
GROUP BY
locations_main.id,
locations_main.name,
locations_main.lft,
locations_main.rgt,
locations_main.parent_id
SET @level = @level + 1
END
RETURN
END
我真的很感谢一些意见，这是否是一种合适的方法。
您的问题并不十分清楚，因为您的数据看起来并不明确，您期望的结果如何。你能发布一个简单的测试用例，显示你想要达到的目标吗？测试用例不必使用真实的表格或数据，只需说明您的要求即可。 – Pondlife 2012-02-06 08:43:41