Oracle PL/SQL Tutorial/Date Timestamp Functions/LAST DAY
Содержание
Calculating the Number of Days of Summer in June
<source lang="sql">
SQL> SELECT LAST_DAY("20-JUN-99") "Last_Day",
2 LAST_DAY("20-JUN-99") - TO_DATE("20-JUN-99") "Days_Summer"
3 from DUAL;
Last_Day Days_Summer
-----------
30-JUN-99 10 SQL></source>
Combine LAST_DAY and ADD_MONTHS together
<source lang="sql">
SQL> SQL> SQL> SQL> -- create demo table SQL> create table Employee(
2 ID VARCHAR2(4 BYTE) NOT NULL primary key, 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 LAST_DAY(ADD_MONTHS(start_date, 3))+1 RESTOCK_DATE
2 FROM employee;
RESTOCK_D
01-NOV-96 01-JUL-76 01-APR-79 01-FEB-83 01-MAY-84 01-NOV-87 01-APR-91 01-JAN-97 8 rows selected. SQL> SQL> SQL> -- clean the table SQL> drop table Employee
2 /
Table dropped.</source>
Finding the Last Day of the Month Starting Summer
<source lang="sql">
SQL> SELECT TO_CHAR(LAST_DAY("30-JUN-99"),"MM/DD/YYYY HH:MM:SS AM") "Last_Day"
2 from DUAL;
Last_Day
06/30/1999 12:06:00 AM SQL></source>
LAST_DAY(column value)+1
<source lang="sql">
SQL> SQL> SQL> -- create demo table SQL> create table Employee(
2 ID VARCHAR2(4 BYTE) NOT NULL primary key, 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 LAST_DAY(start_date)+1 INSURANCE_START_DATE FROM employee;
INSURANCE
01-AUG-96 01-APR-76 01-JAN-79 01-NOV-82 01-FEB-84 01-AUG-87 01-JAN-91 01-OCT-96 8 rows selected. SQL> SQL> -- clean the table SQL> drop table Employee
2 /
Table dropped.</source>
SELECT LAST_DAY("01-JAN-2005") FROM dual;
<source lang="sql">
SQL> SELECT LAST_DAY("01-JAN-2005") FROM dual; LAST_DAY(
31-JAN-05</source>
The LAST_DAY function returns the last day of any month
<source lang="sql">
SQL> SQL> SELECT TO_CHAR(LAST_DAY("23SEP2006"))FROM dual; TO_CHAR(L
30-SEP-06 SQL></source>
