是否可以从表中连接和选择字段,其中表和字段名称以文本形式存储在另一个表中?
假设我有表TableofValues、TableA、TableB和TableC,如下所示。
是否可以在TableofValues中根据TableName在相关表中连接,然后选择基于FieldName的字段,其中TableRowID =连接表中的ID?
这是SQL Server 2008。
示例表格集:
预期结果:
发布于 2015-07-09 12:39:21
您也可以在没有动态查询的情况下实现这一点。请参阅以下代码:
-- Create Demo Data
CREATE TABLE #TableOfValues(id int identity(1,1), value float, TableName nvarchar(30), FieldName nvarchar(30), TableRowId int)
INSERT INTO #TableOfValues(value, TableName, FieldName, TableRowId)
VALUES (1234.2,N'TableA',N'Customer',1),
(245.5,N'TableB',N'Cust',3),
(248.59,N'TableA',N'Customer',2),
(8526.36,N'TableC',N'Cstmer',1),
(224.15,N'TableB',N'Cust',5),
(148.98,N'TableC',N'Cstmer',2)
CREATE TABLE #TableA(id int identity(1,1), Customer nvarchar(30))
INSERT INTO #TableA(Customer)
VALUES(N'Bloggs'),(N'Smith'),(N'Jones')
CREATE TABLE #TableB(id int identity(1,1), Cust nvarchar(30))
INSERT INTO #TableB(Cust)
VALUES (N'Aother'),(N'NJONES'), (N'FBLOGGS'), (N'SMITH'), (N'BARTHUR')
CREATE TABLE #TableC(id int identity(1,1), Cstmer nvarchar(30))
INSERT INTO #TableC(Cstmer)
VALUES(N'Mr Fred Bloggs'),(N'Tony Smith')
-- Your Part
SELECT tov.id, tov.value, tov.TableName, tov.FieldName, tov.TableRowId, tabs.tabVal
FROM #TableOfValues as tov
INNER JOIN (
SELECT N'TableA' as tabName, a.id, a.Customer as tabVal
FROM #TableA as a
UNION ALL
SELECT N'TableB' as tabName, b.id, b.Cust as tabVal
FROM #TableB as b
UNION ALL
SELECT N'TableC' as tabName, c.id, c.Cstmer as tabVal
FROM #TableC as c
) as tabs
ON tov.TableName = tabs.tabName
AND tov.TableRowId = tabs.id
-- Cleanup
DROP TABLE #TableOfValues
DROP TABLE #TableA
DROP TABLE #TableB
DROP TABLE #TableC
如果需要,您可以向表中添加(如果它们不存在)索引。如果您寻找特定值,这些将进一步提高查询性能。
CREATE CLUSTERED INDEX [NCI_TableOfValues_id]
ON #TableOfValues ([id])
CREATE CLUSTERED INDEX [NCI_TableA_id]
ON #TableA ([id])
CREATE CLUSTERED INDEX [NCI_TableB_id]
ON #TableB ([id])
CREATE CLUSTERED INDEX [NCI_TableC_id]
ON #TableC ([id])
CREATE NONCLUSTERED INDEX [NCI_TableOfValues_TableName]
ON #TableOfValues([TableName])
INCLUDE ([id],[value],[FieldName],[TableRowId])
除了解决方案之外,表的设计也有些糟糕。它将进一步提高您的SELECT
性能。
发布于 2015-07-09 12:37:39
除此之外,你应该有一个很好的理由这样做,因为它是非常糟糕的设计,否则,你可以考虑使用SchemaBinding视图。
您必须为任何给定的TableName创建一个视图。这可以在JOIN中使用
发布于 2015-07-09 12:53:41
您也可以这样做。您希望组合#TableA
、#TableB
和#TableC
中的所有行,并添加一列以确定FieldName
。
SELECT
Customer = FieldValue,
Value = v.value
FROM #TableOfValues v
LEFT JOIN (
SELECT id, Customer AS FieldValue, 'Customer' AS FieldName FROM #TableA UNION ALL
SELECT id, Cust, 'Cust' FROM #TableB UNION ALL
SELECT id, Cstmer, 'Cst' FROM #TableC
)t
ON v.FieldName = t.FieldName
AND v.TableRowId = t.id
https://stackoverflow.com/questions/31317400
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