What is the correct aggregation expression to find the highest of all values in the invoice total column?

AGGREGATE FUNCTIONS

Aggregate functions are math or statistical functions that apply to sets or subsets of data rather than to individual instances (rows).

Examples

• Order Subtotals:  The (discounted) total $ amount for all items in each order
  

• Sales By Category:  The (discounted) total $ amount for each product, by category, for 1997
  

• Category Sales for 1997  (based on next query):  Total sales, by category, for 1997
  

• Product Sales for 1997:  Total product sales, by quarter, for 1997
  

The standard aggregate functions are  MIN, MAX, AVG, SUM, and COUNT.

They are almost always used with GROUP BY to create the subsets

             with WHERE before GROUP BY to set criteria before calculating

             and HAVING after GROUP BY to set criteria for the calculated results

 
The GROUP BY clause must include all columns in the SELECT clause except the calculated aggregate columns.  Grouping occurs from left to right, as written in the GROUP BY clause, and groups "break" (to a new group) as soon as a different column value appears in the left-to-right order.    For example, in the last query above, note the break between Chai and Chang, and the individual breaks for each quarter.  Scrolling down in the query, there is a break (a new group or subset) for each product category.

• What would you expect to see if you used an aggregate function in a query but did not include a GROUP BY clause?

Note on aggregate functions and nulls:  The aggregate functions do not include rows that have null values in the columns involved in the calculations; that is, nulls are not handled as if they were zero.  The only function that can return a value for all rows, regardless of the presence of nulls is the COUNT function:

            E.g.      COUNT(column_name) returns a count of all rows where the value of column_name is not null, but

                        COUNT(*) returns the count of all rows, even those with nulls in column_name

AGGREGATE FUNCTIONS

Exercise   - modify an existing query

• Run the Category Sales for 1997 query, change to SQL view, and save as Sales Over 100K

To see only sales totals greater than $100000, add the following criteria after the GROUP BY clause:

            HAVING Sum(ProductSales) > 100000

Run the query and compare the results with the original query - totals should be the same but there should be only two rows.

• Save the original query again as Product Sales First Quarter.  To see only totals for the first quarter, add the following criteria before the GROUP BY clause (ShippedQuarter is a column in the Product Sales for 1997 query):

            WHERE ShippedQuarter like "*1"

Run the query and compare the results with the original Category Sales for 1997 - all totals should be substantially lower.

• Change to SQL view and save as First Quarter Sales Over 20K.  To see only totals greater than $20000, add the following criteria after the GROUP BY clause:

            HAVING Sum(ProductSales) > 20000

Note:  No need to include the table or query name \226 the ProductSales column only appears in one of the tables or queries.

Run the query and compare the results with the previous query - only 4 values are selected.

Exercises using other aggregate functions

(Note:  None of these exercises require more than one table.)

•  How many different products are represented in the database?

• How many products are there in each category?

• Find the average freight charged by each shipper.

            Modify the original freight query:  Average freight by shipper for freight above $75

            Modify the original freight query:  Average freight by shipper showing only averages above $75

• Find the most popular product (as measured by how many ordered).

• Find the most expensive product.

• Practical example (contrived in this database): 

            Find duplicate (or more) product listings in Order Details.

            Find duplicate (or more) product-price pairings (same product and price) in Order Details

            (Compare with Find Duplicates wizard in each case)

COMPLETE SYNTAX FOR SELECT STATEMENT

  SELECT column list, function(), function(), ...

            FROM table1

                INNER JOIN table2

                ...

                ON table1.col1 = table2.col2

                ...

            WHERE criteria for row selection

            [AND criteriafor row selection]
            [OR criteria for row selection]

  GROUP BY column list

            HAVING criteria for function results

          
            ORDER BY column list

SUBQUERIES

Subqueries are used to structure queries.  In  many cases, a subquery can be used instead of a JOIN (and vice versa).

For database systems fully compliant with the SQL 92 standard, a subquery can also be used to provide one or more values in the SELECT clause.

In most database systems, subqueries are typically part of the WHERE clause, as follows:

• WHERE column IN (subquery

• WHERE column (subquery)

• WHERE EXISTS (subquery)

The subqueries themselves are complete SELECT statements, enclosed in parentheses.

Examples

• IN

                     &nb sp; 

Find the price of all products in a particular category, for example condiments.

Type this query in the SQL window and check against the result shown below:

            SELECT ProductName, UnitPrice

            FROM Products

            WHERE CategoryID In

                         (SELECT CategoryID

                        FROM Categories

                        WHERE CategoryName = "Condiments");

This JOIN should give the same result:

            SELECT ProductName, UnitPrice

            FROM Products

                         INNER JOIN Categories

                         ON Products.CategoryID = Categories.CategoryID

            WHERE CategoryName = "Condiments";

• With comparison operator

Run the query Products Above Average Price; check in SQL view (shown here):

SELECT Products.ProductName, Products.UnitPrice
FROM Products
WHERE (((Products.UnitPrice) >
   (SELECT AVG([UnitPrice]) From Products)))
ORDER BY Products.UnitPrice DESC;

• Exercises

Modify the query to show products below average price.  Results (formatting removed):

Modify again to show products within plus or minus 10% of average price (requires some calculations)

• EXISTS

                     &nb sp; 

Create a list of suppliers from whom we buy dairy products.  Type the following query in the SQL window  - use the INNER JOIN ... ON syntax in the subquery, if you prefer.

            SELECT CompanyName

            FROM Suppliers AS s

            WHERE EXISTS

                  &nbs p;     (SELECT *

                   ;       FROM Products p, Categories c

                  &nbs p;     WHERE p.SupplierID = s.SupplierID

                  &nbs p;     AND p.CategoryID = c.CategoryID

                  &nbs p;     AND CategoryName LIKE "*Dairy*");

Result:

SUBQUERIES

Exercises

Create a list of customers located in the same city as a supplier, showing company name, city, and country (try with a subquery and check with a join).    (14 rows)

Find all freight charges that are greater than average and list along with the shipper's name (find the average first, as a check).  (242 rows)

How many shipments has each shipper made at charges greater than average?  (3 rows, 2 or 3 columns)

Which  suppliers do not sell dairy products?  (25 rows)

SUBQUERIES:  SYNTAX

            SELECT column list

            FROM table(s)

            WHERE

 [ IN   |

                     &nb sp;           <comparison>  |

                     &nb sp;           EXISTS]

                        (SELECT column list

                        FROM table(s)

                         ... )

            GROUP BY ...

            HAVING ...

            ORDER BY

SELF-JOIN

A self-join is a query in which a table is joined (compared) to itself.  Self-joins are used to compare values in a column with other values in the same column in the same table.  One practical use for self-joins:  obtaining running counts and running totals in an SQL query.

To write the query, select from the same table listed twice with different aliases, set up the comparison, and eliminate cases where a particular value would be equal to itself.

Example

Which customers are located in the same state (column name is Region)?  Type this statement in the SQL window:

            SELECT DISTINCT c1.ContactName, c1.Address, c1.City, c1.Region

            FROM Customers AS c1, Customers AS c2

            WHERE c1.Region = c2.Region

            AND c1.ContactName <> c2.ContactName

            ORDER BY c1.Region, c1.ContactName;

The result should look like this:

Exercise

Which customers are located in the same city?  (32 rows)

STATISTICS IN SQL

Note: Can be done in the database, but best done in a spreadsheet or using statistics software!

EXAMPLES

• Mean:  AVG (aggregate function)

•  Mode:  Count of most frequently occurring value  (combination of aggregate functions and subquery)

E.g.: Price frequency (upper result) and mode (lower result)

SQL for the two queries:

SELECT Products.UnitPrice, Count(*) AS Frequency
FROM Products
GROUP BY Products.UnitPrice
ORDER BY Count(*) DESC , UnitPrice;

SELECT UnitPrice, Count(*) AS Frequency
FROM Products
GROUP BY UnitPrice
HAVING Count(*) >= ALL
   (SELECT Count(*)
   FROM Products
   GROUP BY UnitPrice);

Note:  We need the subquery because aggregate functions cannot be nested.

• Median:   Value such that the number of values above and below it is the same ("middle" value, not necessarily same as average or mean).  (Moved after running count and running sums because it requires a running count and is more complex than either.)

• Ranking; running count: Compare table to itself (self-join) and generate count based on the comparison.

           E.g.:  Assign sequential numbers to customers based on their customer ID.

Note:  CustomerNum is not stored in the database - it is generated whenever needed by running the query.
SQL for the query:

SELECT count(p1.CustomerID) AS CustomerNum, p1.CompanyName,  p1.CustomerID
FROM Customers  p1
   INNER JOIN Customers p2
   ON p1.CustomerID >= p2.CustomerID
GROUP BY p1.CompanyName, p1.CustomerID
ORDER BY 1;

•  Running sum:  Compare table to itself (self-join) and generate totals based on the comparison.
  

E.g.:  Calculate the running total $ amount by date for orders placed by a particular customer (ID RATTC). To simplify the SQL, first calculate total for this customer without generating a running total and then the running total based on the first query (results below, on the left):

The result on the right includes a running count based on date as well as the running total.

SQL for the three queries:

Query at top left (RATTC_ForRunSum)

SELECT a.OrderID, a.OrderDate, Ssum(a.ExtendedPrice) AS OrderTotal
FROM Invoices AS a
WHERE a.CustomerID = "RATTC"
GROUP BY a.OrderID, a.OrderDate;

Query at lower left with running total only (RATTC_RunningTotal):

SELECT a.OrderID, a.OrderDate, Sum(b.OrderTotal) AS RunningTotal
FROM RATTC_ForRunSum AS a, RATTC_ForRunSum AS b
WHERE b.OrderDate <= a.OrderDate
GROUP BY a.OrderID, a.OrderDate;

Query at lower right with running total and count (RATTC_RunningTotalAndCount):

SELECT Count(b.OrderDate) AS DateCount, a.OrderID, a.OrderDate, Sum(b.OrderTotal) AS RunningTotal
FROM RATTC_ForRunSum AS a, RATTC_ForRunSum AS b
WHERE b.OrderDate <= a.OrderDate
GROUP BY a.OrderID, a.OrderDate;

• Median:   Value such that the number of values above and below it is the same ("middle" value, not necessarily same as average or mean).  The solution offered here is an approximation - may not work in all cases.  Also, other database systems have better functions for finding the "middle" of a sorted list.

E.g.:  Find the median unit price for all products.  Build the query in three steps (results below the third SQL statement):

• Number the product list based on UnitPrice:

SELECT Count(a.ProductID) AS PriceNum, a.ProductName, a.UnitPrice
FROM Products AS a
   INNER JOIN Products AS b
   ON a.UnitPrice>=b.UnitPrice
GROUP BY a.ProductID, a.ProductName, a.UnitPrice
ORDER BY a.UnitPrice;

• Find the middle one or two numbers (added clauses highlighted):

SELECT Count(a.ProductID) AS PriceNum, a.ProductName, a.UnitPrice
FROM Products AS a
   INNER JOIN Products AS b
   ON a.UnitPrice >= b.UnitPrice
GROUP BY a.ProductID, a.ProductName, a.UnitPrice
HAVING Count(a.ProductID)    
   Between (SELECT (Count(ProductID)/2)-0.5                  
                 &nbs p;  FROM Products)   
   And        (SELECT (Count(a.ProductID)/2)+0.5            
                 &nbs p;  FROM Products)
ORDER BY a.UnitPrice;

• Average the results (added clauses highlighted):

SELECT Avg(UnitPrice) AS MedianPrice
FROM
   (SELECT Count(a.ProductID) AS PriceNum, a.ProductName, a.UnitPrice
   FROM Products AS a
      INNER JOIN Products AS b
      ON a.UnitPrice>=b.UnitPrice
   GROUP BY a.ProductID, a.ProductName, a.UnitPrice
   HAVING Count(a.ProductID)        
      Between
         (SELECT (Count(ProductID)/2)-0.5                   & nbsp;                   
          FROM Products)       
      And       
         (SELECT (Count(a.ProductID)/2)+0.5                    ;              
         FROM Products));

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