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How can we use functions to calculate dates in MySQL?

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Release: 2023-08-30 21:41:02
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How can we use functions to calculate dates in MySQL?

In MySQL, we can use the following functions to calculate dates-

  • CURDATE() function- Basically It returns the computer's current date.
  • YEAR() Function - It returns the year for the specified date.
  • MONTH () Function - Returns the month of the specified date.
  • DAY() function - Returns the day of the specified date.
  • RIGHT() function - It returns the number of characters specified in the function since the given date. The part of the expression that compares the return value of the RIGHT() function evaluates to 1 or 0.

To understand it, consider the data from the table named "Collegedetail" as shown below -

mysql> Select * from Collegedetail;
+------+---------+------------+
| ID   | Country | Estb       |
+------+---------+------------+
| 111  | INDIA   | 2010-05-01 |
| 130  | INDIA   | 1995-10-25 |
| 139  | USA     | 1994-09-25 |
| 1539 | UK      | 2001-07-23 |
| 1545 | Russia  | 2010-07-30 |
+------+---------+------------+
5 rows in set (0.00 sec)
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In the query below, we are using all the different date functions Calculated DATE -

mysql> Select ID, Estb, CURDATE(), YEAR(Estb), MONTH(Estb), DAY(Estb), (RIGHT(CURDATE(),5) < RIGHT(estb,5))As &#39;Return&#39; FROM Collegedetail;
+------+------------+------------+------------+-------------+-----------+--------+
| ID   | Estb       | CURDATE()  | YEAR(Estb) | MONTH(Estb) | DAY(Estb) | Return |
+------+------------+------------+------------+-------------+-----------+--------+
| 111  | 2010-05-01 | 2017-11-30 | 2010       | 5           | 1         | 0      |
| 130  | 1995-10-25 | 2017-11-30 | 1995       | 10          | 25        | 0      |
| 139  | 1994-09-25 | 2017-11-30 | 1994       | 9           | 25        | 0      |
| 1539 | 2001-07-23 | 2017-11-30 | 2001       | 7           | 23        | 0      |
| 1545 | 2010-07-30 | 2017-11-30 | 2010       | 7           | 30        | 0      |
+------+------------+------------+------------+-------------+-----------+--------+
5 rows in set (0.00 sec)

mysql> Select ID, estb, CURDATE(),((YEAR(CURDATE())-YEAR(estb))-(RIGHT(CURDATE(),5)<RIGHT(estb,5))) AS &#39;YEARS_OLD&#39; from collegedetail;
+------+------------+------------+-----------+
| ID   | estb       | CURDATE()  | YEARS_OLD |
+------+------------+------------+-----------+
| 111  | 2010-05-01 | 2017-11-30 | 7         |
| 130  | 1995-10-25 | 2017-11-30 | 22        |
| 139  | 1994-09-25 | 2017-11-30 | 23        |
| 1539 | 2001-07-23 | 2017-11-30 | 16        |
| 1545 | 2010-07-30 | 2017-11-30 | 7         |
+------+------------+------------+-----------+
5 rows in set (0.00 sec)
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source:tutorialspoint.com
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