Question

我有一个名称值对表，我在那里存储标签：

TAGID | NAME | VALUE

我有一个表格，这个标签适用于

THINGID | TAGID

我需要一个查询来生成一个resultSet，如果列/字段是给定THINGID的所有可能的TAG-NAMES（TAG表中的NAME字段），则值是对应的标记值。

THINGID | TAGNAME1 | TAGNAME2 | ... |etc.

我可以找到固定列的例子但不是这样的。

Answer 1

类似略有不同架构的东西（thing_id，tag_name，tag_value） - 没有标签表：

CREATE PROCEDURE dbo.PivotData(@table VARCHAR(128), @basefield
VARCHAR(128), @namefield VARCHAR(128), @valuefield VARCHAR(128))
AS
DECLARE
 @sql NVARCHAR(MAX)
SET @sql =
 'DECLARE
 @colName VARCHAR(128),
 @sqlBegin NVARCHAR(MAX),
 @sqlMiddle NVARCHAR(MAX),
 @sqlEnd NVARCHAR(MAX),
 @counter INT
 SET @counter = 1
 SET @sqlBegin = N''SELECT DISTINCT t0.'' + ''' + QUOTENAME(@basefield) + '''
 SET @sqlMiddle = N'' FROM '' + ''' + QUOTENAME(@table) + ''' + '' AS t0 ''
 DECLARE cols CURSOR FOR
 SELECT DISTINCT TOP 100 PERCENT ' + QUOTENAME(@namefield) + '
 FROM ' + QUOTENAME(@table) + '
 WHERE ' + QUOTENAME(@basefield) + ' IS NOT NULL
 ORDER BY ' + QUOTENAME(@namefield) + '
 OPEN cols
 FETCH NEXT FROM cols INTO @colName
 WHILE @@FETCH_STATUS = 0 BEGIN
 SET @sqlBegin = @sqlBegin + '', t'' + CAST(@counter AS VARCHAR) +
''.'' + ''' + QUOTENAME(@valuefield) + ''' + '' AS '' +
QUOTENAME(@colName) + ''''
 SET @sqlMiddle = @sqlMiddle + '' LEFT OUTER JOIN '' + ''' +
  QUOTENAME(@table) + ''' + '' AS t'' + CAST(@counter AS VARCHAR) +
'' ON t0.'' + ''' +
  QUOTENAME(@basefield) + ''' + '' = t'' + CAST(@counter AS VARCHAR)
+ ''.'' + ''' +
  QUOTENAME(@basefield) + ''' + '' AND t'' + CAST(@counter AS
VARCHAR) + ''.'' + ''' + QUOTENAME(@namefield) + ''' + ''='' +
QUOTENAME(@colName, '''''''') + ''''
 SET @counter = @counter + 1
 FETCH NEXT FROM cols INTO @colName
 END
 CLOSE cols
 DEALLOCATE cols
 SET @sqlEnd = '' WHERE t0.'' + ''' + QUOTENAME(@basefield) + ''' + ''
IS NOT NULL''
 DECLARE @sql NVARCHAR(MAX)
 SET @sql = @sqlBegin + @sqlMiddle + @sqlEnd
 EXEC sp_executesql @sql'
EXEC sp_executesql @sql
RETURN 0
GO
CREATE TABLE test_data (
id int identity primary key,
person_id int,
person_data_field VARCHAR (128),
person_data_value VARCHAR (128)
)
GO
INSERT INTO dbo.test_data (person_id, person_data_field, person_data_value)
VALUES (1, 'Name', 'John')
INSERT INTO dbo.test_data (person_id, person_data_field, person_data_value)
VALUES (1, 'Surname', 'Smith')
INSERT INTO dbo.test_data (person_id, person_data_field, person_data_value)
VALUES (1, 'Email', 'John@Smith.com')
INSERT INTO dbo.test_data (person_id, person_data_field, person_data_value)
VALUES (2, 'Name', 'Sarah')
INSERT INTO dbo.test_data (person_id, person_data_field, person_data_value)
VALUES (2, 'Surname', 'Lee')
INSERT INTO dbo.test_data (person_id, person_data_field, person_data_value)
VALUES (2, 'Phone', '012345678')
GO
EXEC [dbo].[PivotData] @table='test_data',
@basefield='person_id', @namefield='person_data_field', @valuefield='person_data_value'
GO

我发现这种方式太慢而且停止了进一步调整，但它可以满足您的要求。

在SQL 2005中使用Dynamic Pivoting需要帮助

1 个答案: