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Inside Microsoft SQL Server 2008: T-SQL Querying
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LISTING 9-7 Query Solution 1 to the Matching Current and Previous Occurrences problem
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SELECT C.empid, C.orderid AS curorderid, P.orderid AS prvorderid, C.orderdate AS curorderdate, P.orderdate AS prvorderdate, C.requireddate AS curreqdate, P.requireddate AS prvreqdate FROM Sales.Orders AS C LEFT OUTER JOIN Sales.Orders AS P ON P.orderid = (SELECT TOP (1) orderid FROM Sales.Orders AS O WHERE O.empid = C.empid AND (O.orderdate < C.orderdate OR (O.orderdate = C.orderdate AND O.orderid < C.orderid)) ORDER BY orderdate DESC, orderid DESC) ORDER BY C.empid, C.orderdate, C.orderid;
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The subquery s lter is a bit tricky because precedence is determined by two attributes: orderdate (ordering column) and orderid (tiebreaker). Had the request been for precedence based on a single column say, orderid alone the lter would have been much simpler O.orderid < C.orderid. Because two attributes are involved, previous rows are identi ed with a logical expression that says inner_sort_col < outer_sort_col or (inner_sort_col = outer_sort_col and inner_tiebreaker < outer_tiebreaker). This query generates the execution plan shown in Figure 9-6, with an I/O cost of 4,844 logical reads.
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FIGURE 9-6 Execution plan for the query in Listing 9-7
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The plan rst scans the covering index I created earlier on the key list (empid, orderdate, orderid), with the covered columns (custid, requireddate) speci ed as included columns. This scan s purpose is to return the current rows. For each current row, a Nested Loops operator initiates an Index Seek operation in the same index, driven by the subquery to fetch the key (orderid) of the previous row. For each returned previous orderid, another Nested Loops operator retrieves the requested list of attributes of the previous row. You realize that one of the two seek operations is super uous and that there s potential for a revised query that would issue only one seek per current order.
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TOP and APPLY
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You can try various query revisions that might improve performance. Listing 9-8 has an example of a query revision that generates the plan shown in Figure 9-7.
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LISTING 9-8 Query Solution 2 to the Matching Current and Previous Occurrences problem
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SELECT C.empid, C.orderid AS curorderid, P.orderid AS prvorderid, C.orderdate AS curorderdate, P.orderdate AS prvorderdate, C.requireddate AS curreqdate, P.requireddate AS prvreqdate FROM (SELECT empid, orderid, orderdate, requireddate, (SELECT TOP (1) orderid FROM Sales.Orders AS O2 WHERE O2.empid = O1.empid AND (O2.orderdate < O1.orderdate OR O2.orderdate = O1.orderdate AND O2.orderid < O1.orderid) ORDER BY orderdate DESC, orderid DESC) AS prvorderid FROM Sales.Orders AS O1) AS C LEFT OUTER JOIN Sales.Orders AS P ON C.prvorderid = P.orderid ORDER BY C.empid, C.orderdate, C.orderid;
FIGURE 9-7 Execution plan for the query in Listing 9-8
This plan incurs an I/O cost of 3,223 logical reads, but it also involves a sort operation. The solution creates a derived table called C that contains current orders, with an additional column (prvorderid) holding the orderid of the previous order as obtained by a correlated subquery. The outer query then joins C with another instance of Orders, aliased as P, which supplies the full list of attributes from the previous order. The lower I/O cost is mainly the result of the Merge join algorithm that the plan uses. In the graphical query plan, the upper input to the Merge Join operator is the result of an ordered scan of the clustered index on orderid, representing the previous orders, and this is the nonpreserved side of the outer join. The lower input is the result of scanning the covering index and fetching each previous orderid with a seek operation followed by a Top 1.
Inside Microsoft SQL Server 2008: T-SQL Querying
A merge join turned out to be cost effective here because the rows of the Orders table were presorted on the clustered index key column orderid and it was not too much work to sort the other input in preparation for the merge. In larger production systems, circumstances will most likely be different. With a much larger number of rows and a different clustered index on a column that frequently appears in range queries, perhaps you shouldn t expect to see the same query plan. This is where the APPLY operator comes in handy. It often leads to simple and ef cient plans that perform well even with large volumes of data. Using the APPLY operator in this case leads to a plan that scans the data once to get the current orders and performs a single index seek for each current order to fetch from the covering index all the attributes of the previous order at once. Listing 9-9 has the solution query, which generates the plan shown in Figure 9-8, with an I/O cost of 3,202 logical reads and no sorting involved.
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