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To complete the lessons in this chapter, you must have:
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SQL Server 2008 installed The AdventureWorks database installed
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ne of the more common requirements that I come across, especially with data warehouses, is to find where data is missing. You might be expecting sales
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figures from a store on a daily basis, defects from a manufacturing line every hour, or inventory levels from a series of sequentially numbered bins. When trying to solve missing gaps problems, many developers turn to some kind of cursor operation. Either a database cursor is used, or the data is pulled back into an application that sorts the data and then moves across every record, keeping track of the previous and subsequent values to find all the missing values. One particular problem that I see is finding gaps in a sequence. I had to solve this type of problem several years ago for a customer who was trying to place advertisements automatically within a time sequence where you had to repeatedly find gaps that matched a set of criteria. The existing solution used a lot of trial and error with nested cursors. Using subqueries and unequal joins solved the problem with a 90 percent reduction in time and resources. A colleague of mine sent me an e-mail with a solution to the problem of figuring out a correlation between billboard postings and wildfires that used multiple WHILE loops and table variables. While the code worked and produced the required result, it could have been collapsed into a set of noncorrelated subqueries with unequal joins that would perform much better against a large result set. (Clint, I hope you found my solution useful.) Even as I was writing this chapter, another customer needed to validate data sets in a data warehouse to find data that was missing from the loads. While the one-row-at-a-time solutions get the job done, finding gaps within a large set of data can require a significant amount of time. Instead of using a cursor approach or multiple WHILE loops, you can solve the problem very efficiently either by using subqueries with unequal joins or by combining ranking functions with CTEs.
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Using Additional Query Techniques
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Lesson 1: building Recursive Queries with ctes
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CTEs provide a capability that is very similar to a derived table (derived tables are explained in Lesson 2 in this chapter). However, CTEs also allow you to iterate across a result set to solve one of the more difficult challenges within Transact-SQL (T-SQL), efficiently executing a recursive query. In this lesson, you learn how to build CTEs to solve recursive query problems.
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After this lesson, you will be able to:
Apply CTEs to recursive query problems.
Estimated lesson time: 20 minutes
Common Table Expressions
A CTE is defined with two parts:
A WITH clause containing a SELECT statement that generates a valid table An outer SELECT statement that references the table expression
The following CTE returns the number of employees that have a given title:
WITH EmpTitle AS (SELECT JobTitle, count(*) numtitles FROM HumanResources.Employee GROUP BY JobTitle) SELECT b.BusinessEntityID, b.JobTitle, a.numtitles FROM EmpTitle a INNER JOIN HumanResources.Employee b ON a.JobTitle = b.JobTitle; GO
A recursive CTE expands the definition of the table expression and consists of two parts:
An anchor query, which is the source of the recursion, along with a UNION ALL statement and a second query, which recurses across the anchor query. An outer query, which references the routine and specifies the number of recursion levels
For example, the following query allows you to specify an employee and then return each level of management above the employee in the organization across a maximum of 25 organization levels:
DECLARE @EmployeeToGetOrgFor INT = 126;
WITH EMP_cte(BusinessEntityID, OrganizationNode, FirstName, LastName, JobTitle, RecursionLevel) AS (SELECT e.BusinessEntityID, e.OrganizationNode, p.FirstName, p.LastName, e.JobTitle, 0 FROM HumanResources.Employee e INNER JOIN Person.Person as p ON p.BusinessEntityID = e.BusinessEntityID WHERE e.BusinessEntityID = @EmployeeToGetOrgFor UNION ALL
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