Using MySQL where clause and ordering by avg() to find the average of duplicate individual elements

For this, use having clause instead of where. Let us first create a table −

mysql> create table DemoTable1338
   -> (
   -> Name varchar(10),
   -> Score int
   -> );
Query OK, 0 rows affected (1.54 sec)

Insert some records in the table using insert command. Here, we have inserted duplicate names with scores −

mysql> insert into DemoTable1338 values('Chris',8);
Query OK, 1 row affected (0.80 sec)
mysql> insert into DemoTable1338 values('Bob',4);
Query OK, 1 row affected (0.20 sec)
mysql> insert into DemoTable1338 values('Bob',9);
Query OK, 1 row affected (0.27 sec)
mysql> insert into DemoTable1338 values('Chris',6);
Query OK, 1 row affected (0.27 sec)
mysql> insert into DemoTable1338 values('David',5);
Query OK, 1 row affected (0.23 sec)
mysql> insert into DemoTable1338 values('David',7);
Query OK, 1 row affected (0.40 sec)

Display all records from the table using select statement −

mysql> select * from DemoTable1338;

This will produce the following output −

+-------+-------+
| Name  | Score |
+-------+-------+
| Chris |     8 |
| Bob   |     4 |
| Bob   |     9 |
| Chris |     6 |
| David |     5 |
| David |     7 |
+-------+-------+
6 rows in set (0.00 sec)

Following is the query to find the average of duplicate individual elements −

mysql> select Name,avg(Score) from DemoTable1338
   -> group by Name
   -> having avg(Score) < 9.5
   -> order by avg(Score);

This will produce the following output −

+-------+------------+
| Name  | avg(Score) |
+-------+------------+
| David |     6.0000 |
| Bob   |     6.5000 |
| Chris |     7.0000 |
+-------+------------+
3 rows in set (0.00 sec)