http://www.pamaya.com/jquery-selectors-and-attribute-selectors-reference-and-examples/
Friday 17 July 2009
Callbacks vs Events
Callbacks vs Events.
A superb read which shows you a way around executing callbacks in the queue when one of them throws an exception. The dispatcher framework shouldn’t stop, but should continue to fire the callbacks which are registered for that event in the queue.
http://dean.edwards.name/weblog/2009/03/callbacks-vs-events/
Seems like a perfect solution..
Sunday 14 June 2009
Data Pivoting Techniques in SQL Server 2008 & 2005
The problem
A common problem in SQL database queries is the need to generate cross-tabulation reports to summarize data, By cross-tabulation or abbreviated as cross-tab, what i mean is converting rows to columns and columns to rows.
Lets take a look at the following example.
1: --Declare the table variable = @TblOrderDetails
2: 3: DECLARE @TblOrderDetails TABLE(
4: Id INT NOT NULL PRIMARY KEY, --1
5: UniqueOrderKey UNIQUEIDENTIFIER DEFAULT(NEWSEQUENTIALID()), --2,3
6: Employee NVARCHAR(50) NOT NULL,
7: ProductId INT NOT NULL,
8: Sales DECIMAL(8,3)
9: ) 10: 11: /*NOTE:--------------------------------------------------------------------------
12: 1. I've used an integer datatype to build the default index on the PK.
13: 2. Uniqueidentifier type for true uniqueness(useful for replication scenarios)
14: 3. You can use newid() or newsequentialid() for creating a new unique identifier,
15: for performance reasons, newsequentialid() is preferred.
16: ---------------------------------------------------------------------------------*/
17: 18: 19: --Insert Test Data into @TblOrderDetails
20: 21: INSERT INTO @TblOrderDetails(Id,Employee,ProductId,Sales)
22: SELECT 1,'Sudhir.Murthy',5,2000
23: UNION ALL
24: SELECT 2,'Tom.Hanks',6,3000
25: UNION ALL
26: SELECT 3,'Jim.Griffiths',5,1000
27: UNION ALL
28: SELECT 4,'Sue.Steely',7,1000
29: UNION ALL
30: SELECT 5,'Sudhir.Murthy',8,2000
31: UNION ALL
32: SELECT 6,'James.Martin',5,1000
33: UNION ALL
34: SELECT 7,'Paul.Smith',5,1000
35: UNION ALL
36: SELECT 8,'Sudhir.Murthy',6,1000
37: UNION ALL
38: SELECT 9,'Tom.Hanks',7,2000
39: UNION ALL
40: SELECT 10,'Sue.Steely',4,2000
41: UNION ALL
42: SELECT 11,'Jim.Griffiths',6,2000
43: UNION ALL
44: SELECT 12,'James.Martin',7,5000
45: UNION ALL
46: SELECT 13,'Paul.Smith',4,4000
47: UNION ALL
48: SELECT 14,'Sue.Steely',3,1000
49: UNION ALL
50: SELECT 15,'Tom.Hanks',5,1000
51: UNION ALL
52: SELECT 16,'Jim.Griffiths',6,1000
53: UNION ALL
54: SELECT 17,'Sudhir.Murthy',3,2000
55: UNION ALL
56: SELECT 18,'Tom.Hanks',2,5000
57: UNION ALL
58: SELECT 19,'Sue.Steely',2,2000
59: UNION ALL
60: SELECT 20,'Sudhir.Murthy',7,1000
61: 62: --Fetch from @TblOrderDetails
63: SELECT * FROM @TblOrderDetails
Your @TblOrderDetails table data should now look like this :-
Id UniqueOrderKey Employee ProductId Sales
----------------------------------------------------------------------------------------
1 BD8BAC49-B6A1-E211-BE97-68A86D1C3B19 Sudhir.Murthy 5 2000.000
2 BE8BAC49-B6A1-E211-BE97-68A86D1C3B19 Tom.Hanks 6 3000.000
3 BF8BAC49-B6A1-E211-BE97-68A86D1C3B19 Jim.Griffiths 5 1000.000
4 C08BAC49-B6A1-E211-BE97-68A86D1C3B19 Sue.Steely 7 1000.000
5 C18BAC49-B6A1-E211-BE97-68A86D1C3B19 Sudhir.Murthy 8 2000.000
6 C28BAC49-B6A1-E211-BE97-68A86D1C3B19 James.Martin 5 1000.000
7 C38BAC49-B6A1-E211-BE97-68A86D1C3B19 Paul.Smith 5 1000.000
8 C48BAC49-B6A1-E211-BE97-68A86D1C3B19 Sudhir.Murthy 6 1000.000
9 C58BAC49-B6A1-E211-BE97-68A86D1C3B19 Tom.Hanks 7 2000.000
10 C68BAC49-B6A1-E211-BE97-68A86D1C3B19 Sue.Steely 4 2000.000
11 C78BAC49-B6A1-E211-BE97-68A86D1C3B19 Jim.Griffiths 6 2000.000
12 C88BAC49-B6A1-E211-BE97-68A86D1C3B19 James.Martin 7 5000.000
13 C98BAC49-B6A1-E211-BE97-68A86D1C3B19 Paul.Smith 4 4000.000
14 CA8BAC49-B6A1-E211-BE97-68A86D1C3B19 Sue.Steely 3 1000.000
15 CB8BAC49-B6A1-E211-BE97-68A86D1C3B19 Tom.Hanks 5 1000.000
16 CC8BAC49-B6A1-E211-BE97-68A86D1C3B19 Jim.Griffiths 6 1000.000
17 CD8BAC49-B6A1-E211-BE97-68A86D1C3B19 Sudhir.Murthy 3 2000.000
18 CE8BAC49-B6A1-E211-BE97-68A86D1C3B19 Tom.Hanks 2 5000.000
19 CF8BAC49-B6A1-E211-BE97-68A86D1C3B19 Sue.Steely 2 2000.000
20 D08BAC49-B6A1-E211-BE97-68A86D1C3B19 Sudhir.Murthy 7 1000.000
Now, From the above schema of the table, you can find out how much sales has an employee made, how many products has he sold etc., For instance, the below query
1: SELECT Employee,
2: COUNT(ProductId) AS TotalProducts,
3: SUM(Sales) AS TotalSales
4: FROM @TblOrderDetails
5: GROUP BY Employee
6: ORDER BY Employee
fetches you the result :-
Employee TotalProducts TotalSales
--------------------------------------------
James.Martin 2 6000.000
Jim.Griffiths 3 4000.000
Paul.Smith 2 5000.000
Sudhir.Murthy 5 8000.000
Sue.Steely 4 6000.000
Tom.Hanks 4 11000.000
But, you would want to generate a cross tabular report, for instance, which give you a result of employees vs products, which looks like this ?
Employee Product2 Product3 Product4 Product5 Product6 Product7 Product8
------------------------------------------------------------------------------------------
In order to achieve the above desired result, we would have to transpose the rows in the select query to columns, in other worlds pivot the Employee and ProductId fields. This can be achieved using the CASE Construct and the PIVOT operator. Both of which are discussed extensively below.
1. Solution using CASE Constructs
Lets take a look at the old school CASE Construct in SQL Server. By using a CASE construct, We can rotate the rows into columns.
example 1.
1: /*Rotate Rows To Columns using CASE*/
2: 3: SELECT Employee,
4: (CASE WHEN ProductId=2 THEN ProductId ELSE NULL END) AS Product2,
5: (CASE WHEN ProductId=3 THEN ProductId ELSE NULL END) AS Product3,
6: (CASE WHEN ProductId=4 THEN ProductId ELSE NULL END) AS Product4,
7: (CASE WHEN ProductId=5 THEN ProductId ELSE NULL END) AS Product5,
8: (CASE WHEN ProductId=6 THEN ProductId ELSE NULL END) AS Product6,
9: (CASE WHEN ProductId=7 THEN ProductId ELSE NULL END) AS Product7,
10: (CASE WHEN ProductId=8 THEN ProductId ELSE NULL END) AS Product8
11: FROM @TblOrderDetailsexample 1 result.
1: Employee Product2 Product3 Product4 Product5 Product6 Product7 Product8 2: ------------------------------------------------------------------------------------ 3: Sudhir.Murthy NULL NULL NULL 5 NULL NULL NULL 4: Tom.Hanks NULL NULL NULL NULL 6 NULL NULL 5: Jim.Griffiths NULL NULL NULL 5 NULL NULL NULL 6: Sue.Steely NULL NULL NULL NULL NULL 7 NULL 7: Sudhir.Murthy NULL NULL NULL NULL NULL NULL 8 8: James.Martin NULL NULL NULL 5 NULL NULL NULL 9: Paul.Smith NULL NULL NULL 5 NULL NULL NULL 10: Sudhir.Murthy NULL NULL NULL NULL 6 NULL NULL 11: Tom.Hanks NULL NULL NULL NULL NULL 7 NULL 12: Sue.Steely NULL NULL 4 NULL NULL NULL NULL 13: Jim.Griffiths NULL NULL NULL NULL 6 NULL NULL 14: James.Martin NULL NULL NULL NULL NULL 7 NULL 15: Paul.Smith NULL NULL 4 NULL NULL NULL NULL 16: Sue.Steely NULL 3 NULL NULL NULL NULL NULL 17: Tom.Hanks NULL NULL NULL 5 NULL NULL NULL 18: Jim.Griffiths NULL NULL NULL NULL 6 NULL NULL 19: Sudhir.Murthy NULL 3 NULL NULL NULL NULL NULL 20: Tom.Hanks 2 NULL NULL NULL NULL NULL NULL 21: Sue.Steely 2 NULL NULL NULL NULL NULL NULL 22: Sudhir.Murthy NULL NULL NULL NULL NULL 7 NULLHowever, Note that employees repeat in the ‘Employee’ column. This is probably not the expected result. what we expect is to have an aggregate of all ProductIds over Employee.
In order to do that we take a look at the below example.
example 2.
1: /*Rotate Rows To Columns using COUNT ( CASE..)*/
2: 3: SELECT Employee,
4: COUNT(CASE WHEN ProductId=2 THEN ProductId ELSE 0 END) AS Product2,
5: COUNT(CASE WHEN ProductId=3 THEN ProductId ELSE 0 END) AS Product3,
6: COUNT(CASE WHEN ProductId=4 THEN ProductId ELSE 0 END) AS Product4,
7: COUNT(CASE WHEN ProductId=5 THEN ProductId ELSE 0 END) AS Product5,
8: COUNT(CASE WHEN ProductId=6 THEN ProductId ELSE 0 END) AS Product6,
9: COUNT(CASE WHEN ProductId=7 THEN ProductId ELSE 0 END) AS Product7,
10: COUNT(CASE WHEN ProductId=8 THEN ProductId ELSE 0 END) AS Product8
11: FROM @TblOrderDetails 12: GROUP BY Employee 13: example 2 result.
Employee Product2 Product3 Product4 Product5 Product6 Product7 Product8
-----------------------------------------------------------------------------------
James.Martin 2 2 2 2 2 2 2
Jim.Griffiths 3 3 3 3 3 3 3
Paul.Smith 2 2 2 2 2 2 2
Sudhir.Murthy 5 5 5 5 5 5 5
Sue.Steely 4 4 4 4 4 4 4
Tom.Hanks 4 4 4 4 4 4 4
Here as expected, The COUNT() aggregate function aggregates the ProductIds over Employees giving you an overview of how many products each employee sold. Also, in daily usage and of course dependent on problems, a most common scenario is to use the aggregate function SUM(). I will leave that for you to fiddle around.
2. Solution using PIVOT
PIVOT rotates a table-valued expression by turning the unique values form one column in the expression into multiple columns in the output, and performs aggregations where they are required on any remaining column values that are wanted in the final output.
UNPIVOT performs the opposite operation to PIVOT by rotating the columns of a table valued expression into row values.
NOTE :- IMPORTANT
When PIVOT and UNPIVOT are used against databases that are upgraded to SQL Server 2005 or later, the compatibility level of the database must be set to 90 or higher.
Syntax for using PIVOT
The following is annotated syntax for PIVOT.
SELECT <;non-pivoted column>,
[first pivoted column] AS <;column name>,
[second pivoted column] AS <;column name>,
...
[last pivoted column] AS <;column name>
FROM
(<;SELECT query that produces the data>)
AS <;alias for the source query>
PIVOT
(
<;aggregation function>(<column being aggregated>)
FOR
[<;column that contains the values that will become column headers>]
IN ( [first pivoted column], [second pivoted column],
... [last pivoted column])
) AS <;alias for the pivot table>
<optional ORDER BY clause>;
example 3.
Here, I’m going to demonstrate the PIVOT operation to pivot table valued expression rows to columns. The SQL Server PIVOT keyword / construct is only available from SQL Server 2005 onwards.
In it’s simple form, The PIVOT construct looks as below. It has a Aggregate function, which is COUNT() here in our case. We are then using this to Count (ProductId) From the Query and pivotting the distinct row values of the ProductId which are [2],[3],[4],[5],[6],[7],[8] to columns in the Select Query.
/*********************************************************************
Query to yeild result set of employees vs all products that they sold.
Here we are pivotting the ProductIds.
**********************************************************************/
SELECT Employee,
[2] AS Product2,
[3] AS Product3,
[4] AS Product4,
[5] AS Product5,
[6] AS Product6,
[7] AS Product7,
[8] AS Product8
FROM @TblOrderDetails
PIVOT (
COUNT(ProductId)
FOR ProductId IN ([2],[3],[4],[5],[6],[7],[8])
)Pvt
/*Notice that after pivotting the ProductId column,
It will not be available in the SELECT statement.*/
example 3 result.
Employee Product2 Product3 Product4 Product5 Product6 Product7 Product8
--------------------------------------------------------------------------------------------
Sudhir.Murthy 0 0 0 1 0 0 0
Tom.Hanks 0 0 0 0 1 0 0
Jim.Griffiths 0 0 0 1 0 0 0
Sue.Steely 0 0 0 0 0 1 0
Sudhir.Murthy 0 0 0 0 0 0 1
James.Martin 0 0 0 1 0 0 0
Paul.Smith 0 0 0 1 0 0 0
Sudhir.Murthy 0 0 0 0 1 0 0
Tom.Hanks 0 0 0 0 0 1 0
Sue.Steely 0 0 1 0 0 0 0
Jim.Griffiths 0 0 0 0 1 0 0
James.Martin 0 0 0 0 0 1 0
Paul.Smith 0 0 1 0 0 0 0
Sue.Steely 0 1 0 0 0 0 0
Tom.Hanks 0 0 0 1 0 0 0
Jim.Griffiths 0 0 0 0 1 0 0
Sudhir.Murthy 0 1 0 0 0 0 0
Tom.Hanks 1 0 0 0 0 0 0
Sue.Steely 1 0 0 0 0 0 0
Sudhir.Murthy 0 0 0 0 0 1 0
As you can notice, this is not quite what we aimed for initially, The employees are repeated here. We would ideally want to have a consolidated view of all *distinct* employees and the number of products that they’ve sold.
example 4.
In this example, we are going to extend our query from example 3 to achieve what we call as ‘Grouping’ using PIVOT. As you can see from the query below, We have used the Aggregate function COUNT() inside the PIVOT construct and we have a select query to select Employee and ProductId’s. But notice the construct has changed in the From clause. By using the Employee & ProductId in the From Clause here, every employee servers as a value column against which the unique rows - [2],[3],[4],[5],[6],[7] & [8] are grouped to columns.
/***************************************************************
Query to yeild the consolidated result set of employees vs
products that they sold.
****************************************************************/
SELECT Employee,
[2] AS Product2,
[3] AS Product3,
[4] AS Product4,
[5] AS Product5,
[6] AS Product6,
[7] AS Product7,
[8] AS Product8
FROM (
SELECT Employee,ProductId FROM @TblOrderDetails
) AS Qry
PIVOT (
COUNT(ProductId)
FOR ProductId IN ([2],[3],[4],[5],[6],[7],[8])
) Pvt
/*Notice that after pivotting the ProductId column,
It will not be available in the SELECT statement.*/
example 4 result.
1: Employee Product2 Product3 Product4 Product5 Product6 Product7 Product8 2: ---------------------------------------------------------------------------------------- 3: James.Martin 0 0 0 1 0 1 0 4: Jim.Griffiths 0 0 0 1 2 0 0 5: Paul.Smith 0 0 1 1 0 0 0 6: Sudhir.Murthy 0 1 0 1 1 1 1 7: Sue.Steely 1 1 1 0 0 1 0 8: Tom.Hanks 1 0 0 1 1 1 0Note: When aggregate functions are used with PIVOT, the presence of any null values in the value column are not considered when computing an aggregation.
UNPIVOT example.
UNPIVOT is the reverse operation of PIVOT. i.e It converts the table valued expression to rows by rotating columns into rows. Suppose the table produced in the previous example is stored in the table ‘@TblEmployeeProducts’, and you want to rotate the column identifiers Product2, Product3, Product4, Product5, Product6, Product7 and Product8 into row values that correspond to a particular Employee.
1: /*Declare the table variable = @TblEmployeeProducts2: Please Note: The table is only designed this way to
3: keep the example simple and straightforward. 4: In an ideal world, Products will not be repeated this
5: way to violate Third and BC Normal Forms.
6: */ 7: 8: DECLARE @TblEmployeeProducts TABLE( 9: Id INT NOT NULL PRIMARY KEY,10: UniqueOrderKey UNIQUEIDENTIFIER DEFAULT(NEWSEQUENTIALID()),
11: Employee NVARCHAR(50) NOT NULL, 12: Product2 INT NOT NULL, 13: Product3 INT NOT NULL, 14: Product4 INT NOT NULL, 15: Product5 INT NOT NULL, 16: Product6 INT NOT NULL, 17: Product7 INT NOT NULL, 18: Product8 INT NOT NULL 19: ) 20: 21: --Insert Test Data into @TblEmployeeProducts 22: 23: INSERT @TblEmployeeProducts 24: (Id,Employee,Product2,Product3,Product4,Product5,Product6,Product7,Product8) 25: SELECT 1,'James.Martin',0,0,0,1,0,1,0
26: UNION ALL27: SELECT 2,'Jim.Griffiths',0,0,0,1,2,0,0
28: UNION ALL29: SELECT 3,'Paul.Smith',0,0,1,1,0,0,0
30: UNION ALL31: SELECT 4,'Sudhir.Murthy',0,1,0,1,1,1,0
32: UNION ALL33: SELECT 5,'Sue.Steely',1,1,1,0,0,1,0
34: UNION ALL35: SELECT 6,'Tom.Hanks',1,0,0,1,1,1,0
36: 37: --------------------------------------- 38: 39: SELECT Employee,
40: Product2 as Prod2, 41: Product3 as Prod3, 42: Product4 as Prod4, 43: Product5 as Prod5, 44: Product6 as Prod6, 45: Product7 as Prod7, 46: Product8 as Prod8 47: FROM @TblEmployeeProducts 48: After you execute the select statement (above) , Your table should look like.,
Employee Prod2 Prod3 Prod4 Prod5 Prod6 Prod7 Prod8
----------------------------------------------------------------------------
James.Martin 0 0 0 1 0 1 0
Jim.Griffiths 0 0 0 1 2 0 0
Paul.Smith 0 0 1 1 0 0 0
Sudhir.Murthy 0 1 0 1 1 1 0
Sue.Steely 1 1 1 0 0 1 0
Tom.Hanks 1 0 0 1 1 1 0
Now, To rotate the columns of the above table values into row values, We can use UNPIVOT. This means that you must identify two additional columns. The column that will contain the column values that you are rotating (Product2, Product3,...) will be called ProductId, and the column that will hold the values that currently reside under the columns being rotated will be called TotalOrders. These columns correspond to the pivot_column and value_column, respectively, in the Transact-SQL definition.
Here is the query:-
example 5.
1: /*Query to unpivot the Employee and Products to
2: rows of Employee, Products and TotalOrders*/ 3: 4: SELECT Employee,ProductId,TotalOrders
5: FROM (6: SELECT Employee,
7: Product2, 8: Product3, 9: Product4, 10: Product5, 11: Product6, 12: Product7, 13: Product8 14: FROM @TblEmployeeProducts 15: )Qry 16: UNPIVOT (17: TotalOrders FOR ProductId IN(Product2,
18: Product3, 19: Product4, 20: Product5, 21: Product6, 22: Product7, 23: Product8) 24: )AS UnPvt 25: 26: example 5 result.
1: Employee ProductId TotalOrders 2: ----------------------------------------- 3: James.Martin Product2 0 4: James.Martin Product3 0 5: James.Martin Product4 0 6: James.Martin Product5 1 7: James.Martin Product6 0 8: James.Martin Product7 1 9: James.Martin Product8 0 10: Jim.Griffiths Product2 0 11: Jim.Griffiths Product3 0 12: Jim.Griffiths Product4 0 13: Jim.Griffiths Product5 1 14: Jim.Griffiths Product6 2 15: Jim.Griffiths Product7 0 16: Jim.Griffiths Product8 0 17: Paul.Smith Product2 0 18: Paul.Smith Product3 0 19: Paul.Smith Product4 1 20: Paul.Smith Product5 1 21: Paul.Smith Product6 0 22: Paul.Smith Product7 0 23: Paul.Smith Product8 0 24: Sudhir.Murthy Product2 0 25: Sudhir.Murthy Product3 1 26: Sudhir.Murthy Product4 0 27: Sudhir.Murthy Product5 1 28: Sudhir.Murthy Product6 1 29: Sudhir.Murthy Product7 1 30: Sudhir.Murthy Product8 0 31: Sue.Steely Product2 1 32: Sue.Steely Product3 1 33: Sue.Steely Product4 1 34: Sue.Steely Product5 0 35: Sue.Steely Product6 0 36: Sue.Steely Product7 1 37: Sue.Steely Product8 0 38: Tom.Hanks Product2 1 39: Tom.Hanks Product3 0 40: Tom.Hanks Product4 0 41: Tom.Hanks Product5 1 42: Tom.Hanks Product6 1 43: Tom.Hanks Product7 1 44: Tom.Hanks Product8 0Note on UNPIVOT
Notice that UNPIVOT is not the exact reverse of PIVOT. PIVOT performs an aggregation and, therefore, merges possible multiple rows into a single row in the output. UNPIVOT does not reproduce the original table-valued expression result because rows have been merged. Besides, null values in the input of UNPIVOT disappear in the output, whereas there may have been original null values in the input before the PIVOT operation.
References : 1. Microsoft, MSDN.
Source Code : Complete SQL Script Here
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