Using the ORDER BY Clause to Sort Groups : Group By « Query Select « Oracle PL/SQL Tutorial

For example, the following query sorts the output using ORDER BY COUNT (city):

SQL> -- create demo table
SQL> create table Employee(
  2    ID                 VARCHAR2(4 BYTE)         NOT NULL,
  3    First_Name         VARCHAR2(10 BYTE),
  4    Last_Name          VARCHAR2(10 BYTE),
  5    Start_Date         DATE,
  6    End_Date           DATE,
  7    Salary             Number(8,2),
  8    City               VARCHAR2(10 BYTE),
  9    Description        VARCHAR2(15 BYTE)
 10  )
 11  /

Table created.

SQL>
SQL> -- prepare data
SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2               values ('01','Jason',    'Martin',  to_date('19960725','YYYYMMDD'), to_date('20060725','YYYYMMDD'), 1234.56, 'Toronto',  'Programmer')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('02','Alison',   'Mathews', to_date('19760321','YYYYMMDD'), to_date('19860221','YYYYMMDD'), 6661.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('03','James',    'Smith',   to_date('19781212','YYYYMMDD'), to_date('19900315','YYYYMMDD'), 6544.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('04','Celia',    'Rice',    to_date('19821024','YYYYMMDD'), to_date('19990421','YYYYMMDD'), 2344.78, 'Vancouver','Manager')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary,  City,       Description)
  2                values('05','Robert',   'Black',   to_date('19840115','YYYYMMDD'), to_date('19980808','YYYYMMDD'), 2334.78, 'Vancouver','Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('06','Linda',    'Green',   to_date('19870730','YYYYMMDD'), to_date('19960104','YYYYMMDD'), 4322.78,'New York',  'Tester')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('07','David',    'Larry',   to_date('19901231','YYYYMMDD'), to_date('19980212','YYYYMMDD'), 7897.78,'New York',  'Manager')
  3  /

1 row created.

SQL> insert into Employee(ID,  First_Name, Last_Name, Start_Date,                     End_Date,                       Salary, City,        Description)
  2                values('08','James',    'Cat',     to_date('19960917','YYYYMMDD'), to_date('20020415','YYYYMMDD'), 1232.78,'Vancouver', 'Tester')
  3  /

1 row created.

SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
  2  /

ID   FIRST_NAME LAST_NAME  START_DAT END_DATE      SALARY CITY       DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01   Jason      Martin     25-JUL-96 25-JUL-06    1234.56 Toronto    Programmer
02   Alison     Mathews    21-MAR-76 21-FEB-86    6661.78 Vancouver  Tester
03   James      Smith      12-DEC-78 15-MAR-90    6544.78 Vancouver  Tester
04   Celia      Rice       24-OCT-82 21-APR-99    2344.78 Vancouver  Manager
05   Robert     Black      15-JAN-84 08-AUG-98    2334.78 Vancouver  Tester
06   Linda      Green      30-JUL-87 04-JAN-96    4322.78 New York   Tester
07   David      Larry      31-DEC-90 12-FEB-98    7897.78 New York   Manager
08   James      Cat        17-SEP-96 15-APR-02    1232.78 Vancouver  Tester

8 rows selected.

SQL>
SQL>
SQL> SELECT city, COUNT(first_name)
  2  FROM employee
  3  GROUP BY city
  4  ORDER BY COUNT(first_name);

CITY       COUNT(FIRST_NAME)
---------- -----------------
Toronto                    1
New York                   2
Vancouver                  5

SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
  2  /

Table dropped.

SQL>
SQL>

2.5.Group By
	2.5.1.	Using Multiple Columns in a Group
	2.5.2.	GROUP by with NULL value
	2.5.3.	Use the aggregate functions with the GROUP BY clause
	2.5.4.	By default, GROUP BY sorts the rows in ascending order
	2.5.5.	Using the ORDER BY Clause to Sort Groups
	2.5.6.	You don't have to include the columns used in the GROUP BY clause in your SELECT clause
	2.5.7.	GROUP BY used on a column without the column name appearing in the result set
	2.5.8.	Any row-level variable (i.e., a column name) in the result set must be mentioned in the GROUP BY clause for the query to make sense.
	2.5.9.	Grouping at Multiple Levels
	2.5.10.	ORA-00979: not a GROUP BY expression
	2.5.11.	Getting Values and Subtotals in One Go
	2.5.12.	Getting Values and Subtotals in One Go with COALESCE
	2.5.13.	Getting Values and Subtotals in One Go with UNION
	2.5.14.	Group all ids with averager value more than
	2.5.15.	Group and count employeem and display only if its count is more than 4
	2.5.16.	Count employee, group by department id and job title
	2.5.17.	Count and group by department id
	2.5.18.	Count distinct with group by
	2.5.19.	Born after '1960-01-01', group by department number with count(*) >= 4;
	2.5.20.	Group by case