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== Second Tuesday of the Month==



<source lang="sql">
SQL>
SQL> CREATE TABLE monthdates(monthdate date NOT NULL PRIMARY KEY);
Table created.
SQL> INSERT INTO monthdates(monthdate) VALUES (DATE "2007-04-01");
1 row created.
SQL> INSERT INTO monthdates(monthdate) VALUES (DATE "2007-05-01");
1 row created.
SQL> INSERT INTO monthdates(monthdate) VALUES (DATE "2007-06-01");
1 row created.
SQL>
SQL> SELECT monthdate AS first_day_of_month
2 , monthdate
3 + MOD( ( 10 - TO_CHAR(monthdate,"d") ), 7 ) + 7
4 AS second_tuesday_of_month
5 FROM monthdates
6
SQL> DROP TABLE monthdates;
Table dropped.
SQL></source>

== select date "1996-01-29" + interval "1" month as col_1==



<source lang="sql">
SQL>
SQL>
SQL> select date "1996-01-29" + interval "1" month as col_1
2 from dual;
COL_1
---------
29-FEB-96</source>

== select date "1997-08-11" - interval "3" month as col_3==



<source lang="sql">
SQL>
SQL>
SQL> select date "1997-08-11" - interval "3" month as col_3
2 from dual;
COL_3
---------
11-MAY-97
SQL></source>

== Subtract one date from another, yielding the number of days between the two dates.==



<source lang="sql">
The following example subtracts July 31, 2003, from August 2, 2003:
SQL> SELECT TO_DATE("02-AUG-2003") - TO_DATE("31-JUL-2003") FROM dual;
TO_DATE("02-AUG-2003")-TO_DATE("31-JUL-2003")
---------------------------------------------
2
SQL></source>

== Subtracts two days from August 2, 2003==



<source lang="sql">
SQL>
SQL>
SQL> SELECT TO_DATE("02-AUG-2003") - 2 FROM dual;
TO_DATE("
---------
31-JUL-03
SQL></source>

== Using Date Arithmetic==

You can use the addition and subtraction operators with date type value.
You can add a number, representing a number of days, to a date.
The following example adds 2 days to July 31, 2003, and displays the resulting date:



<source lang="sql">
SQL> SELECT TO_DATE("31-JUL-2003") + 2 FROM dual;
TO_DATE("
---------
02-AUG-03
SQL></source>

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в 13:45, 26 мая 2010

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