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Subqueries, Table Expressions, and Ranking Functions
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With N Kollar , the subquery returns no values and is therefore NULL. The outer query obviously doesn t nd any orders for which empid = NULL and therefore returns an empty set. Note that the query doesn t break (fail) it s a valid query. With N D% , the subquery returns two values (1, 9), and because the outer query expects a scalar, it breaks at run time and generates the following error :
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Msg 512, Level 16, State 1, Line 1 Subquery returned more than 1 value. This is not permitted when the subquery follows =, !=, <, <= , >, >= or when the subquery is used as an expression.
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Logically, a self-contained subquery can be evaluated just once for the whole outer query. Physically, the optimizer can consider many different ways to achieve the same thing, so you shouldn t think in such strict terms. Now that we ve covered the essentials, let s move on to more sophisticated problems involving self-contained subqueries. I ll start with a problem belonging to a group of problems called relational division. Relational division problems have many nuances and many practical applications. Logically, it s like dividing one set by another, producing a result set. For example, from the InsideTSQL2008 database, return all customers for whom every employee from the USA has handled at least one order. In this case, you re dividing the set of all orders by the set of all employees from the USA, and you expect the set of matching customers back. Filtering here is not that simple because for each customer you need to inspect multiple rows to gure out whether you have a match. Here I ll show a technique using GROUP BY and DISTINCT COUNT to solve relational division problems. I ll show you other techniques later in the book. If you knew ahead of time the list of all employee IDs for USA employees, you could write the following query to solve the problem:
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SELECT custid FROM Sales.Orders WHERE empid IN(1, 2, 3, 4, 8) GROUP BY custid HAVING COUNT(DISTINCT empid) = 5;
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This query generates the following output:
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custid ----------5 9 20 24 34 35 37 38
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39 41 46 47 48 51 55 63 65 71 80 83 84 87 89
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This query nds all orders with one of the ve U.S. employee IDs, groups those orders by custid, and returns customer IDs that have (all) ve distinct empid values in their group of orders. To make the solution more dynamic and accommodate lists of employee IDs that are unknown ahead of time and also large lists even when known, you can use subqueries instead of literals:
SELECT custid FROM Sales.Orders WHERE empid IN (SELECT empid FROM HR.Employees WHERE country = N'USA') GROUP BY custid HAVING COUNT(DISTINCT empid) = (SELECT COUNT(*) FROM HR.Employees WHERE country = N'USA');
Another problem involving self-contained subqueries is returning all orders placed on the last actual order date of the month. Note that the last actual order date of the month might be different than the last date of the month for example, if a company doesn t place orders on weekends. So the last actual order date of the month has to be queried from the data. Here s the solution query:
SELECT orderid, custid, empid, orderdate FROM Sales.Orders WHERE orderdate IN (SELECT MAX(orderdate) FROM Sales.Orders GROUP BY YEAR(orderdate), MONTH(orderdate));
This query produces the following output:
orderid ----------10269 10294 custid ----------89 65 empid ----------5 4 orderdate ----------------------2006-07-31 00:00:00.000 2006-08-30 00:00:00.000
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