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==An Example Query that Illustrates the Use of GROUPING_ID()==



<source lang="sql">

SQL> CREATE TABLE employee(
2 employee_id INTEGER,
3 division_id CHAR(3),
4 job_id CHAR(3),
5 first_name VARCHAR2(10) NOT NULL,
6 last_name VARCHAR2(10) NOT NULL,
7 salary NUMBER(6, 0)
8 );
Table created.
SQL>
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(1, "BUS","PRE","James","Smith","800000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(2, "SAL","MGR","Ron","Johnson","350000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(3, "SAL","WOR","Fred","Hobbs","140000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(4, "SUP","MGR","Susan","Jones","200000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(5, "SAL","WOR","Rob","Green","350000");
1 row created.
SQL>
SQL> select * from employee;
EMPLOYEE_ID DIV JOB FIRST_NAME LAST_NAME SALARY
----------- --- --- ---------- ---------- ----------
1 BUS PRE James Smith 800000
2 SAL MGR Ron Johnson 350000
3 SAL WOR Fred Hobbs 140000
4 SUP MGR Susan Jones 200000
5 SAL WOR Rob Green 350000
SQL>
SQL>
SQL>
SQL>
SQL> --An Example Query that Illustrates the Use of GROUPING_ID()
SQL>
SQL> --The following example passes division_id and job_id to GROUPING_ID().
Notice the output from the GROUPING_ID() function agrees with the expected returned
values documented in the previous section:
SQL>
SQL> SELECT
2 division_id, job_id,
3 GROUPING(division_id) AS DIV_GRP,
4 GROUPING(job_id) AS JOB_GRP,
5 GROUPING_ID(division_id, job_id) AS grp_id,
6 SUM(salary)
7 FROM employee
8 GROUP BY CUBE(division_id, job_id);
DIV JOB DIV_GRP JOB_GRP GRP_ID SUM(SALARY)
--- --- ---------- ---------- ---------- -----------
1 1 3 1840000
MGR 1 0 2 550000
PRE 1 0 2 800000
WOR 1 0 2 490000
BUS 0 1 1 800000
BUS PRE 0 0 0 800000
SAL 0 1 1 840000
SAL MGR 0 0 0 350000
SAL WOR 0 0 0 490000
SUP 0 1 1 200000
SUP MGR 0 0 0 200000
11 rows selected.
SQL>
SQL>
SQL>
SQL>
SQL>
SQL> drop table employee;
Table dropped.
SQL>

</source>

==Eliminate duplicate rows using a HAVING clause that only allows rows whose GROUP_ID() is 0==



<source lang="sql">

SQL> CREATE TABLE employee(
2 employee_id INTEGER,
3 division_id CHAR(3),
4 job_id CHAR(3),
5 first_name VARCHAR2(10) NOT NULL,
6 last_name VARCHAR2(10) NOT NULL,
7 salary NUMBER(6, 0)
8 );
Table created.
SQL>
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(1, "BUS","PRE","James","Smith","800000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(2, "SAL","MGR","Ron","Johnson","350000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(3, "SAL","WOR","Fred","Hobbs","140000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(4, "SUP","MGR","Susan","Jones","200000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(5, "SAL","WOR","Rob","Green","350000");
1 row created.
SQL>
SQL> select * from employee;
EMPLOYEE_ID DIV JOB FIRST_NAME LAST_NAME SALARY
----------- --- --- ---------- ---------- ----------
1 BUS PRE James Smith 800000
2 SAL MGR Ron Johnson 350000
3 SAL WOR Fred Hobbs 140000
4 SUP MGR Susan Jones 200000
5 SAL WOR Rob Green 350000
SQL>
SQL>
SQL> --Eliminate duplicate rows using a HAVING clause that only allows rows whose
GROUP_ID() is 0
SQL>
SQL> SELECT division_id, job_id, GROUP_ID(), SUM(salary)
2 FROM employee
3 GROUP BY division_id, ROLLUP(division_id, job_id)
4 HAVING GROUP_ID() = 0;
DIV JOB GROUP_ID() SUM(SALARY)
--- --- ---------- -----------
BUS PRE 0 800000
SAL MGR 0 350000
SAL WOR 0 490000
SUP MGR 0 200000
BUS 0 800000
SAL 0 840000
SUP 0 200000
7 rows selected.
SQL>
SQL>
SQL>
SQL> drop table employee;
Table dropped.
SQL>

</source>

==One useful application of GROUPING_ID() is to filter rows using a HAVING clause==



<source lang="sql">

SQL> CREATE TABLE employee(
2 employee_id INTEGER,
3 division_id CHAR(3),
4 job_id CHAR(3),
5 first_name VARCHAR2(10) NOT NULL,
6 last_name VARCHAR2(10) NOT NULL,
7 salary NUMBER(6, 0)
8 );
Table created.
SQL>
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(1, "BUS","PRE","James","Smith","800000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(2, "SAL","MGR","Ron","Johnson","350000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(3, "SAL","WOR","Fred","Hobbs","140000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(4, "SUP","MGR","Susan","Jones","200000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(5, "SAL","WOR","Rob","Green","350000");
1 row created.
SQL>
SQL> select * from employee;
EMPLOYEE_ID DIV JOB FIRST_NAME LAST_NAME SALARY
----------- --- --- ---------- ---------- ----------
1 BUS PRE James Smith 800000
2 SAL MGR Ron Johnson 350000
3 SAL WOR Fred Hobbs 140000
4 SUP MGR Susan Jones 200000
5 SAL WOR Rob Green 350000
SQL>
SQL>
SQL>
SQL>
SQL> --A Useful Application of GROUPING_ID()
SQL>
SQL> --One useful application of GROUPING_ID() is to filter rows using a HAVING clause.
Your HAVING clause can exclude rows that don�t contain a subtotal or total by simply checking
if GROUPING_ID() returns a value greater than 0. For example:
SQL>
SQL> SELECT
2 division_id, job_id,
3 GROUPING_ID(division_id, job_id) AS grp_id,
4 SUM(salary)
5 FROM employee
6 GROUP BY CUBE(division_id, job_id)
7 HAVING GROUPING_ID(division_id, job_id) > 0;
DIV JOB GRP_ID SUM(SALARY)
--- --- ---------- -----------
3 1840000
MGR 2 550000
PRE 2 800000
WOR 2 490000
BUS 1 800000
SAL 1 840000
SUP 1 200000
7 rows selected.
SQL>
SQL>
SQL>
SQL>
SQL> drop table employee;

</source>

==You can use the GROUP_ID() function to remove duplicate rows returned by a GROUP BY clause==



<source lang="sql">

SQL> CREATE TABLE employee(
2 employee_id INTEGER,
3 division_id CHAR(3),
4 job_id CHAR(3),
5 first_name VARCHAR2(10) NOT NULL,
6 last_name VARCHAR2(10) NOT NULL,
7 salary NUMBER(6, 0)
8 );
Table created.
SQL>
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(1, "BUS","PRE","James","Smith","800000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(2, "SAL","MGR","Ron","Johnson","350000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(3, "SAL","WOR","Fred","Hobbs","140000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(4, "SUP","MGR","Susan","Jones","200000");
1 row created.
SQL> insert into employee (EMPLOYEE_ID,division_id,JOB_ID,FIRST_NAME,LAST_NAME,SALARY)
2 values(5, "SAL","WOR","Rob","Green","350000");
1 row created.
SQL>
SQL> select * from employee;
EMPLOYEE_ID DIV JOB FIRST_NAME LAST_NAME SALARY
----------- --- --- ---------- ---------- ----------
1 BUS PRE James Smith 800000
2 SAL MGR Ron Johnson 350000
3 SAL WOR Fred Hobbs 140000
4 SUP MGR Susan Jones 200000
5 SAL WOR Rob Green 350000
SQL>
SQL> --Using the GROUP_ID() Function
SQL>
SQL> --You can use the GROUP_ID() function to remove duplicate rows returned by a GROUP BY clause.
GROUP_ID() doesn�t accept any parameters.
If n duplicates exist for a particular grouping, GROUP_ID returns numbers in the range 0 to n-1.
SQL>
SQL>
SQL> SELECT division_id, job_id, GROUP_ID(), SUM(salary)
2 FROM employee
3 GROUP BY division_id, ROLLUP(division_id, job_id);
DIV JOB GROUP_ID() SUM(SALARY)
--- --- ---------- -----------
BUS PRE 0 800000
SAL MGR 0 350000
SAL WOR 0 490000
SUP MGR 0 200000
BUS 0 800000
SAL 0 840000
SUP 0 200000
BUS 1 800000
SAL 1 840000
SUP 1 200000
10 rows selected.
SQL>
SQL>
SQL>
SQL> drop table employee;
Table dropped.
SQL>

</source>

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в 13:45, 26 мая 2010

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