c# print barcode labels Creating and populating the OpenSchema table in Visual Studio .NET

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Listing 6-2. Creating and populating the OpenSchema table
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SET NOCOUNT ON; USE tempdb; GO IF OBJECT_ID('dbo.OpenSchema') IS NOT NULL DROP TABLE dbo.OpenSchema; GO CREATE TABLE dbo.OpenSchema ( objectid INT NOT NULL, attribute NVARCHAR(30) NOT NULL, value SQL_VARIANT NOT NULL, PRIMARY KEY (objectid, attribute) ); INSERT INTO VALUES(1, INSERT INTO VALUES(1, INSERT INTO VALUES(1, INSERT INTO VALUES(2, INSERT INTO VALUES(2, INSERT INTO VALUES(2, INSERT INTO VALUES(2, INSERT INTO VALUES(3, INSERT INTO VALUES(3, INSERT INTO VALUES(3, dbo.OpenSchema(objectid, attribute, value) N'attr1', CAST('ABC' AS VARCHAR(10)) ); dbo.OpenSchema(objectid, attribute, value) N'attr2', CAST(10 AS INT) ); dbo.OpenSchema(objectid, attribute, value) N'attr3', CAST('20040101' AS SMALLDATETIME)); dbo.OpenSchema(objectid, attribute, value) N'attr2', CAST(12 AS INT) ); dbo.OpenSchema(objectid, attribute, value) N'attr3', CAST('20060101' AS SMALLDATETIME)); dbo.OpenSchema(objectid, attribute, value) N'attr4', CAST('Y' AS CHAR(1)) ); dbo.OpenSchema(objectid, attribute, value) N'attr5', CAST(13.7 AS DECIMAL(9,3)) ); dbo.OpenSchema(objectid, attribute, value) N'attr1', CAST('XYZ' AS VARCHAR(10)) ); dbo.OpenSchema(objectid, attribute, value) N'attr2', CAST(20 AS INT) ); dbo.OpenSchema(objectid, attribute, value) N'attr3', CAST('20050101' AS SMALLDATETIME));
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The contents of the OpenSchema table are shown in Table 6-10.
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Table 6-10. Contents of OpenSchema Table
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objectid 1 1 1 2 attribute attr1 attr2 attr3 attr2 value ABC 10 2004-01-01 00:00:00.000 12
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objectid 2 2 2 3 3 3
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attribute attr3 attr4 attr5 attr1 attr2 attr3
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value 2006-01-01 00:00:00.000 Y 13.700 XYZ 20 2005-01-01 00:00:00.000
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Representing data this way allows logical schema changes to be implemented without adding, altering, or dropping tables and columns, but by using DML INSERTs, UPDATEs, and DELETEs instead. Of course, other aspects of working with the data (such as enforcing integrity, tuning, and querying) become more complex and expensive with such a representation. There are other approaches to deal with frequent data definition changesfor example, storing the data in XML format. However, when you weigh the advantages and disadvantages of each representation, you might find the open schema representation demonstrated here more favorable in some scenariosfor example, representing auction data. Keep in mind that this representation of the data requires very complex queries even for simple requests, because different attributes of the same entity instance are spread over multiple rows. Before you query such data, you might want to rotate it to a traditional form with one column for each attributeperhaps store the result in a temporary table, index it, query it, and then get rid of the temporary table. To rotate the data from its open schema form into a traditional form, you need to use a pivoting technique. In the following section, I'll describe the steps involved in solving pivoting problems. I'd like to point out that to understand the steps of the solution, it can be very helpful if you think about query logical processing phases, which I described in detail in 1. I discussed the query processing phases involved with the PIVOT table operator in SQL Server 2005, but those phases apply just as well to the solution in SQL Server 2000. Moreover, in SQL 2000 the phases are more apparent in the code, while in SQL Server 2005 they are implicit. The first step you might want to try when solving pivoting problems is to figure out how the number of rows in the result correlates to the number of rows in the source data. Here, you need to create a single result row out of the multiple base rows for each object. This can mean creating a GROUP BY objectid. As the next step in a pivoting problem, you can think in terms of the result columns. You need a result column for each unique attribute. Because the data contains five unique attributes (attr1, attr2, attr3, attr4, and attr5), you need five expressions in the SELECT list. Each expression is supposed to extract, out of the rows belonging to the grouped object, the value corresponding to a specific attribute. This can be done with the following MAX(CASE...) expression, which in this example is applied to the attribute attr2: MAX(CASE WHEN attribute = 'attr2' THEN value END) AS attr2
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Remember that with no ELSE clause CASE assumes an implicit ELSE NULL. The CASE expression just shown will yield NULL for rows where attribute does not equal 'attr2' and yield value when attribute does equal 'attr2'. This means that among the rows with a given value of objectid (say, 1), the CASE expression would yield several NULLs and, at most, one known value (10 in our example), which
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represents the value of the target attribute (attr2 in our example) for the given objectid. The trick to extracting the one known value is to use MAX or MIN. Both ignore NULLs and will return the one nonNULL value present, because both the minimum and the maximum of a set containing one value is that value. Here's the complete query that pivots the attributes from OpenSchema, yielding the output shown in Table 6-11: SELECT objectid, MAX(CASE WHEN attribute MAX(CASE WHEN attribute MAX(CASE WHEN attribute MAX(CASE WHEN attribute MAX(CASE WHEN attribute FROM dbo.OpenSchema GROUP BY objectid;
= = = = =
'attr1' 'attr2' 'attr3' 'attr4' 'attr5'
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