Hello, I not sure if this is the right place to ask this... I am using
mysql.
What I need is a SQL statement that can find what years are in the
database that is greater than last year.
For example...
I have a table full of dates ranging from 2001 to 2005..
I am straining my brain to figure out a SQL statement that can give me
a list of years greater than last year..
In this case I need a list that has..
2003
2004
2005
If anyone could help me.. or give me a clue.. I would appreciate it. 2 26605
I finally did figure out how to do this.. I post my solution just in
case somebody else finds it useful or has a suggestion for improvment.
SELECT YEAR(startdate) AS year
FROM course_dates
WHERE YEAR(startdate) > (YEAR(NOW())-1)
GROUP BY YEAR(startdate)
Aggro <sp**********@y ahoo.com> wrote in message news:<AC******* *****@read3.ine t.fi>... Jason Tudisco wrote:
Hello, I not sure if this is the right place to ask this... I am using mysql.
What I need is a SQL statement that can find what years are in the database that is greater than last year.
For example...
I have a table full of dates ranging from 2001 to 2005..
I am straining my brain to figure out a SQL statement that can give me a list of years greater than last year..
In this case I need a list that has..
2003 2004 2005
If anyone could help me.. or give me a clue.. I would appreciate it.
Use mysql function year(date) to extract year from date, and compare years. For example
-------------------------------------------------- mysql> create table datetest(date_ date); Query OK, 0 rows affected (0.00 sec)
mysql> insert into datetest values( '2002-01-01' ); Query OK, 1 row affected (0.01 sec)
mysql> insert into datetest values( '2003-01-01' ); Query OK, 1 row affected (0.01 sec)
mysql> insert into datetest values( '2004-01-01' ); Query OK, 1 row affected (0.00 sec)
mysql> insert into datetest values( '2005-01-01' ); Query OK, 1 row affected (0.00 sec)
mysql> select * from datetest where year(date_) > (year(now())-1); +------------+ | date_ | +------------+ | 2003-01-01 | | 2004-01-01 | | 2005-01-01 | +------------+ 3 rows in set (0.03 sec) --------------------------------------------------
Read more about date and time functions from:
http://www.mysql.com/doc/en/Date_and...functions.html
I finally did figure out how to do this.. I post my solution just in
case somebody else finds it useful or has a suggestion for improvment.
SELECT YEAR(startdate) AS year
FROM course_dates
WHERE YEAR(startdate) > (YEAR(NOW())-1)
GROUP BY YEAR(startdate)
Aggro <sp**********@y ahoo.com> wrote in message news:<AC******* *****@read3.ine t.fi>... Jason Tudisco wrote:
Hello, I not sure if this is the right place to ask this... I am using mysql.
What I need is a SQL statement that can find what years are in the database that is greater than last year.
For example...
I have a table full of dates ranging from 2001 to 2005..
I am straining my brain to figure out a SQL statement that can give me a list of years greater than last year..
In this case I need a list that has..
2003 2004 2005
If anyone could help me.. or give me a clue.. I would appreciate it.
Use mysql function year(date) to extract year from date, and compare years. For example
-------------------------------------------------- mysql> create table datetest(date_ date); Query OK, 0 rows affected (0.00 sec)
mysql> insert into datetest values( '2002-01-01' ); Query OK, 1 row affected (0.01 sec)
mysql> insert into datetest values( '2003-01-01' ); Query OK, 1 row affected (0.01 sec)
mysql> insert into datetest values( '2004-01-01' ); Query OK, 1 row affected (0.00 sec)
mysql> insert into datetest values( '2005-01-01' ); Query OK, 1 row affected (0.00 sec)
mysql> select * from datetest where year(date_) > (year(now())-1); +------------+ | date_ | +------------+ | 2003-01-01 | | 2004-01-01 | | 2005-01-01 | +------------+ 3 rows in set (0.03 sec) --------------------------------------------------
Read more about date and time functions from:
http://www.mysql.com/doc/en/Date_and...functions.html This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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