The Row Comparison Functions - LEAD and LAG : LEAD LAG « Analytical Functions « Oracle PL/SQL Tutorial






See a previous row value on the same row as the current value.

SQL>
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('20060
725','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('19860
221','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('19900
315','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('19990
421','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('19980
808','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('19960
104','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('19980
212','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('20020
415','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 ROW_NUMBER() OVER(ORDER BY start_date) rn,
  2    city, start_date, salary,
  3    LAG(salary) OVER(ORDER BY start_date) Previous,
  4    LEAD(salary) OVER(ORDER BY start_date) Next
  5  FROM employee
  6  ORDER BY start_date;

        RN CITY       START_DAT     SALARY   PREVIOUS       NEXT
---------- ---------- --------- ---------- ---------- ----------
         1 Vancouver  21-MAR-76    6661.78               6544.78
         2 Vancouver  12-DEC-78    6544.78    6661.78    2344.78
         3 Vancouver  24-OCT-82    2344.78    6544.78    2334.78
         4 Vancouver  15-JAN-84    2334.78    2344.78    4322.78
         5 New York   30-JUL-87    4322.78    2334.78    7897.78
         6 New York   31-DEC-90    7897.78    4322.78    1234.56
         7 Toronto    25-JUL-96    1234.56    7897.78    1232.78
         8 Vancouver  17-SEP-96    1232.78    1234.56

8 rows selected.

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

Table dropped.

SQL>
SQL>
SQL>
SQL>
SQL>








16.17.LEAD LAG
16.17.1.Using the LAG() and LEAD() Functions
16.17.2.The Row Comparison Functions - LEAD and LAG
16.17.3.The row comparison function partitioned as with other aggregates
16.17.4.LAG and LEAD Options
16.17.5.Lag salary over, lead salary over