SQL GROUP BYStatements
The GROUP BY statement can be used with some aggregate functions.
GROUP BY Statement
The GROUP BY statement is used in conjunction with aggregate functions to group the result set by one or more columns.
SQL GROUP BY Syntax
SELECT column_name, aggregate_function(column_name)
FROM table_name
WHERE column_name operator value
GROUP BY column_name;
FROM table_name
WHERE column_name operator value
GROUP BY column_name;
Demonstration Database
In this tutorial, we will use the EXAMPLE sample database.
Below is data selected from the "Websites" table:
+----+--------------+---------------------------+-------+---------+ | id | name | url | alexa | country | +----+--------------+---------------------------+-------+---------+ | 1 | Google | https://www.google.cm/ | 1 | USA | | 2 | 淘宝 | https://www.taobao.com/ | 13 | CN | | 3 | Example | http://www.example.com/ | 4689 | CN | | 4 | 微博 | http://weibo.com/ | 20 | CN | | 5 | Facebook | https://www.facebook.com/ | 3 | USA | | 7 | stackoverflow | http://stackoverflow.com/ | 0 | IND | +----+---------------+---------------------------+-------+---------+
Below is the data from the "access_log" website access log table:
mysql> SELECT * FROM access_log; +-----+---------+-------+------------+ | aid | site_id | count | date | +-----+---------+-------+------------+ | 1 | 1 | 45 | 2016-05-10 | | 2 | 3 | 100 | 2016-05-13 | | 3 | 1 | 230 | 2016-05-14 | | 4 | 2 | 10 | 2016-05-14 | | 5 | 5 | 205 | 2016-05-14 | | 6 | 4 | 13 | 2016-05-15 | | 7 | 3 | 220 | 2016-05-15 | | 8 | 5 | 545 | 2016-05-16 | | 9 | 3 | 201 | 2016-05-17 | +-----+---------+-------+------------+ 9 rows in set (0.00 sec)
Simple Application of GROUP BY
Count the visits of each site_id in access_log:
Examples
SELECT site_id, SUM(access_log.count) AS
nums
FROM access_log GROUP BY site_id;
FROM access_log GROUP BY site_id;
Executing the above SQL produces the following output:
SQL GROUP BY Multi-Table Join
The following SQL statement counts the number of records for websites that have records:
Example
SELECT Websites.name,COUNT(access_log.aid) AS nums FROM
access_log
LEFT JOIN Websites
ON access_log.site_id=Websites.id
GROUP BY Websites.name;
LEFT JOIN Websites
ON access_log.site_id=Websites.id
GROUP BY Websites.name;
Executing the above SQL produces the following output: