Introduction to SQL Grouping and Totaling

SQL excels in data aggregation, offering robust tools like grouping and totaling. This guide focuses on mastering SQL data aggregation through these methods.

SQL Aggregate Functions

In SQL, aggregation is the process of operating or calculating a set of values. The intent is to return a single summary value. SQL includes several very powerful Aggregate Functions such as AVG(), COUNT(), SUM(), MAX(), and MIN(). These functions, in turn, are most often found in SQL statements that implement a GROUP BY clause. However, these functions do not need to be associated with that clause.

This guide uses a CourseTaken table to demonstrate aggregate functions. From the command line, create the CourseTaken table.

CREATE TABLE CourseTaken (
    SSNumber CHAR(9) NOT NULL,
    CourseId CHAR(6) NOT NULL,
    NumericGrade INT NOT NULL,
    YearTaken INT NOT NULL
);

The CourseTaken table contains the following column data:

Use a SQL Aggregate Function to Calculate a Single Summary Value

The sections below provide different examples that use aggregate functions to return a single summary value. All examples use the CourseTaken table that was created in the Aggregate Functions section of the guide.

Example 1:

In this example, the aggregate function returns a numerical average grade for all students taking the course CSC101 in the year 2022.

SELECT AVG(NumericGrade) AS 'Avg Grade'
FROM CourseTaken
WHERE CourseId = 'CSC101
AND YearTaken = 2022;

SQL returns the following average grade:

Avg Grade
---------
92

Example 2:

The aggregate function below returns a count of the number of students that took course CSC101 before the year 2022.

SELECT COUNT(SSNumber) AS 'Student Count'
FROM CourseTaken
WHERE CourseId = 'CSC101'
AND YearTaken < 2022;

The following count is returned:

Student Count
---------
1

Example 3:

In this example, an aggregate function is used to obtain the maximum numeric grade recorded in any year by a Student taking CSC101.

SELECT MAX(NumericGrade) AS 'Max Grade'
FROM CourseTaken
WHERE CourseId = 'CSC101'

The returned maximum grade is the following:

Max Grade
---------
100

Aggregate Data Using Group Functions

The following examples demonstrate the use of the GROUP BY clause using the data from the CourseTaken table.

Example 1:

The example below determines the average grade for each student for all courses that they have taken to date. To execute this, use the SQL Group By clause to group by Student (in this case, the SSNumber column).

SELECT SSNumber, AVG(NumericGrade) AS 'Avg Grade'
FROM CourseTaken
GROUP BY SSNumber

The output returns the average grade for each student.

+-----------+----------+
| SSNumber  | Avg Grade|
+-----------+----------+
| 111111111 | 96.5     |
| 222222222 | 87.5     |
| 333333333 | 88       |
+-----------+----------+

Example 2:

The aggregate function below finds the average grade received across every CourseId in the CourseTaken table. To do this, group by CourseId within YearTaken with the following SQL code:

SELECT CourseId AS 'Course', YearTaken AS 'Year',
AVG(NumericGrade) AS 'Avg Grade'
FROM CourseTaken
GROUP BY CourseId, YearTaken
ORDER BY CourseId, YearTaken

You should see the following output:

+--------+------+-----------+
| Course | Year | Avg Grade |
+--------+------+-----------+
| CSC101 | 2021 | 98        |
| POL101 | 2021 | 92        |
| CSC101 | 2022 | 92        |
| EEE101 | 2022 | 75        |
| ENG101 | 2022 | 95        |
+--------+------+-----------+

Example 3:

From the SQL query in the previous example, you can restrict the number of rows that you operate by adding a WHERE clause to the query. For example, to generate the average grade received by students only for CourseId CSC101, group by CourseId within YearTaken. The following SQL code can accomplish this:

SELECT CourseId AS 'Course', YearTaken AS 'Year',
AVG(NumericGrade) AS 'Avg Grade'
FROM CourseTaken
WHERE CourseId = 'CSC101'
GROUP BY CourseId, YearTaken
ORDER BY CourseId, YearTaken

In the above SQL code, you are adding a condition (via the WHERE clause) before the actual group aggregation is performed (via the GROUP BY clause).

The following output is returned:

+--------+------+-----------+
| Course | Year | Avg Grade |
+--------+------+-----------+
| CSC101 | 2021 | 98        |
| CSC101 | 2022 | 92        |
+--------+------+-----------+

Example 4:

From the SQL query in Example 2, you can apply a condition before the final result is returned. To accomplish this use the SQL Having clause. You can determine the average grade across every CourseId, where the aggregated average grade is greater than 90. You can again group by CourseId within YearTaken. The following SQL code can accomplish this:

SELECT CourseId AS ‘Course’, YearTaken AS ‘Year’,
AVG(NumericGrade) AS ‘Avg Grade’
FROM CourseTaken
GROUP BY CourseId, YearTaken
HAVING AVG(NumericGrade) > 90
ORDER BY CourseId, YearTaken

The output is the following:

+--------+------+-----------+
| Course | Year | Avg Grade |
+--------+------+-----------+
| CSC101 | 2021 | 98        |
| POL101 | 2021 | 92        |
| CSC101 | 2022 | 92        |
| ENG101 | 2022 | 95        |
+--------+------+-----------+

The row for CourseId EEE101 was not returned. This is because the Having clause filtered it out after the GROUP BY clause was executed (CourseId EEE101’s average grade is below 90).

Example 5:

Building upon the SQL code from Example 3 and Example 4, you can create aggregation queries that utilize both the Where and Having clause. For example, you can determine the courses that were taken in 2021, where the average grade for those courses taken was greater than 93. Here, the Where clause filters out results before the Group By data aggregation is performed, and the Having clause filters out results returned after the Group By data aggregation is performed. The following SQL code can accomplish this:

SELECT CourseId AS ‘Course’, YearTaken AS ‘Year’,
AVG(NumericGrade) AS ‘Avg Grade’
FROM CourseTaken
WHERE YearTaken = 2021
GROUP BY CourseId, YearTaken
HAVING AVG(NumericGrade) > 93
ORDER BY CourseId

The output returned is the following:

+--------+------+-----------+
| Course | Year | Avg Grade |
+--------+------+-----------+
| CSC101 | 2021 | 98        |
+--------+------+-----------+

Example 6:

You can count the number of rows associated with each Group By aggregation in a query. Building upon the previous example SQL code, you can generate the average grade received by Students for only CourseId CSC101, grouped by CourseId within YearTaken. The code should provide the number of students (count) associated with each group. The following SQL code can accomplish this:

SELECT CourseId AS ‘Course’, YearTaken AS ‘Year’,
AVG(NumericGrade) AS ‘Avg Grade’,
Count(SSNumber) AS ‘Count’
FROM CourseTaken
WHERE CourseId = ‘CSC101’
GROUP BY CourseId, YearTaken
ORDER BY CourseId, YearTaken

The Count(SSNumber) in the SELECT clause could have been specified as Count(*). The difference between the two syntaxes is that Count(*) includes rows that have NULL values in them as well. As per the CourseTaken table definition above, all columns in the CourseTaken table must contain non-null values (the NOT NULL attribute assures this). The Count(SSNumber) and Count(*) would be functionally equivalent in this example.

The following output is returned:

+--------+------+-----------+-------+
| Course | Year | Avg Grade | Count |
+--------+------+-----------+-------+
| CSC101 | 2021 | 98        | 1     |
| CSC101 | 2022 | 92        | 2     |
+--------+------+-----------+-------+

Conclusion

This guide explores the foundational aspects of SQL’s robust data aggregation capabilities for grouping and totaling. It emphasizes the use of the WHERE clause to restrict values before aggregation and the HAVING clause to filter grouped results after aggregation in SQL queries.

For further SQL knowledge, refer to our guides on SQL data types, SQL joins, and SQL user management security.