Oracle PL/SQL Tutorial/Analytical Functions/GROUPING ID — различия между версиями

Текущая версия на 10:06, 26 мая 2010

An Application of GROUPING_ID()

GROUPING_ID() function filters rows using a HAVING clause to exclude rows that don"t contain a subtotal or total.
GROUPING_ID() function accepts one or more columns and returns the decimal equivalent of the GROUPING bit vector.
GROUPING bit vector is computed by combining the results of a call to the GROUPING() function for each column in order.
One 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.

Quote from:

Oracle Database 10g SQL (Osborne ORACLE Press Series) (Paperback)

# Paperback: 608 pages

# Publisher: McGraw-Hill Osborne Media; 1st edition (February 20, 2004)

# Language: English

# ISBN-10: 0072229810

# ISBN-13: 978-0072229813

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>
SQL> SELECT
  2   city, description,
  3   GROUPING_ID(city, description) AS grp_id,
  4   SUM(salary)
  5  FROM employee
  6  GROUP BY CUBE(city, description)
  7  HAVING GROUPING_ID(city, description) > 0;
CITY       DESCRIPTION         GRP_ID SUM(SALARY)
---------- --------------- ---------- -----------
                                    3    32574.02
           Tester                   2     21096.9
           Manager                  2    10242.56
           Programmer               2     1234.56
Toronto                             1     1234.56
New York                            1    12220.56
Vancouver                           1     19118.9
7 rows selected.
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
  2  /
Table dropped.
SQL>
SQL>

group_id with cube

SQL> create table employees(
  2    empno      NUMBER(4)
  3  , ename      VARCHAR2(8)
  4  , init       VARCHAR2(5)
  5  , job        VARCHAR2(8)
  6  , mgr        NUMBER(4)
  7  , bdate      DATE
  8  , msal       NUMBER(6,2)
  9  , comm       NUMBER(6,2)
 10  , deptno     NUMBER(2) ) ;
Table created.
SQL>
SQL>
SQL> insert into employees values(1,"Jason",  "N",  "TRAINER", 2,   date "1965-12-18",  800 , NULL,  10);
1 row created.
SQL> insert into employees values(2,"Jerry",  "J",  "SALESREP",3,   date "1966-11-19",  1600, 300,   10);
1 row created.
SQL> insert into employees values(3,"Jord",   "T" , "SALESREP",4,   date "1967-10-21",  1700, 500,   20);
1 row created.
SQL> insert into employees values(4,"Mary",   "J",  "MANAGER", 5,   date "1968-09-22",  1800, NULL,  20);
1 row created.
SQL> insert into employees values(5,"Joe",    "P",  "SALESREP",6,   date "1969-08-23",  1900, 1400,  30);
1 row created.
SQL> insert into employees values(6,"Black",  "R",  "MANAGER", 7,   date "1970-07-24",  2000, NULL,  30);
1 row created.
SQL> insert into employees values(7,"Red",    "A",  "MANAGER", 8,   date "1971-06-25",  2100, NULL,  40);
1 row created.
SQL> insert into employees values(8,"White",  "S",  "TRAINER", 9,   date "1972-05-26",  2200, NULL,  40);
1 row created.
SQL> insert into employees values(9,"Yellow", "C",  "DIRECTOR",10,  date "1973-04-27",  2300, NULL,  20);
1 row created.
SQL> insert into employees values(10,"Pink",  "J",  "SALESREP",null,date "1974-03-28",  NULL, 0,     30);
1 row created.
SQL>
SQL>
SQL> select   deptno, job
  2  ,        grouping_id(deptno, job) gid
  3  from     employees
  4  group by CUBE(deptno, job);
    DEPTNO JOB             GID
---------- -------- ----------
                             3
           MANAGER           2
           TRAINER           2
           DIRECTOR          2
           SALESREP          2
        10                   1
        10 TRAINER           0
        10 SALESREP          0
        20                   1
        20 MANAGER           0
        20 DIRECTOR          0
        20 SALESREP          0
        30                   1
        30 MANAGER           0
        30 SALESREP          0
        40                   1
        40 MANAGER           0
        40 TRAINER           0
18 rows selected.
SQL>
SQL> drop table employees;
Table dropped.

grouping_id function with case statement

SQL>
SQL> create table employees(
  2    empno      NUMBER(4)
  3  , ename      VARCHAR2(8)
  4  , init       VARCHAR2(5)
  5  , job        VARCHAR2(8)
  6  , mgr        NUMBER(4)
  7  , bdate      DATE
  8  , msal       NUMBER(6,2)
  9  , comm       NUMBER(6,2)
 10  , deptno     NUMBER(2) ) ;
Table created.
SQL>
SQL>
SQL> insert into employees values(1,"Jason",  "N",  "TRAINER", 2,   date "1965-12-18",  800 , NULL,  10);
1 row created.
SQL> insert into employees values(2,"Jerry",  "J",  "SALESREP",3,   date "1966-11-19",  1600, 300,   10);
1 row created.
SQL> insert into employees values(3,"Jord",   "T" , "SALESREP",4,   date "1967-10-21",  1700, 500,   20);
1 row created.
SQL> insert into employees values(4,"Mary",   "J",  "MANAGER", 5,   date "1968-09-22",  1800, NULL,  20);
1 row created.
SQL> insert into employees values(5,"Joe",    "P",  "SALESREP",6,   date "1969-08-23",  1900, 1400,  30);
1 row created.
SQL> insert into employees values(6,"Black",  "R",  "MANAGER", 7,   date "1970-07-24",  2000, NULL,  30);
1 row created.
SQL> insert into employees values(7,"Red",    "A",  "MANAGER", 8,   date "1971-06-25",  2100, NULL,  40);
1 row created.
SQL> insert into employees values(8,"White",  "S",  "TRAINER", 9,   date "1972-05-26",  2200, NULL,  40);
1 row created.
SQL> insert into employees values(9,"Yellow", "C",  "DIRECTOR",10,  date "1973-04-27",  2300, NULL,  20);
1 row created.
SQL> insert into employees values(10,"Pink",  "J",  "SALESREP",null,date "1974-03-28",  NULL, 0,     30);
1 row created.
SQL>
SQL>
SQL> select   deptno
  2  ,        case grouping_id(deptno,job)
  3                when 0 then job
  4                when 1 then "**dept **"
  5                when 3 then "**total**"
  6           end  job
  7  ,        count(empno) headcount
  8  from     employees
  9  group by ROLLUP(deptno, job);
    DEPTNO JOB        HEADCOUNT
---------- --------- ----------
        10 TRAINER            1
        10 SALESREP           1
        10 **dept **          2
        20 MANAGER            1
        20 DIRECTOR           1
        20 SALESREP           1
        20 **dept **          3
        30 MANAGER            1
        30 SALESREP           2
        30 **dept **          3
        40 MANAGER            1
        40 TRAINER            1
        40 **dept **          2
           **total**         10
14 rows selected.
SQL>
SQL> drop table employees;
Table dropped.
SQL>

Oracle PL/SQL Tutorial/Analytical Functions/GROUPING ID — различия между версиями

Текущая версия на 10:06, 26 мая 2010

An Application of GROUPING_ID()

group_id with cube

grouping_id function with case statement

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