MySQL - CURDATE() Function

The DATE, DATETIME and TIMESTAMP datatypes in MySQL are used to store the date, date and time, time stamp values respectively. Where a time stamp is a numerical value representing the number of milliseconds from '1970-01-01 00:00:01' UTC (epoch) to the specified time. MySQL provides a set of functions to manipulate these values.
The MYSQL CURDATE() is used to get the current days date. The resultant value is a string or a numerical value based on the context and, the date returned will be in the 'YYYY-MM-DD' or YYYYMMDD format.
Following is the syntax of the above function –
CURDATE();

Following example demonstrates the usage of the CURDATE() function –
mysql> SELECT CURDATE();
+------------+
|  CURDATE() |
+------------+
| 2021-07-11 |
+------------+
1 row in set (0.00 sec)
Following is an example of this function in numerical context –
mysql> SELECT CURDATE() +0;
+--------------+
| CURDATE() +0 |
+--------------+
|     20210711 |
+--------------+
1 row in set (0.00 sec)
You can add days to the current date as shown below –
mysql> SELECT CURDATE()+12;
+--------------+
| CURDATE()+12 |
+--------------+
|     20210723 |
+--------------+
1 row in set (0.00 sec)
We can also subtract the desired number of days from the current date using this function –
mysql> SELECT CURDATE()-22213;
+-----------------+
| CURDATE()-22213 |
+-----------------+
|        20188498 |
+-----------------+
1 row in set (0.00 sec)
Let us create a table with name MyPlayers in MySQL database using CREATE statement as shown below –
mysql> CREATE TABLE MyPlayers(
	ID INT,
	First_Name VARCHAR(255),
	Last_Name VARCHAR(255),
	Date_Of_Birth date,
	Place_Of_Birth VARCHAR(255),
	Country VARCHAR(255),
	PRIMARY KEY (ID)
);
Now, we will insert 7 records in MyPlayers table using INSERT statements −
mysql> insert into MyPlayers values(1, 'Shikhar', 'Dhawan', DATE('1981-12-05'), 'Delhi', 'India');
mysql> insert into MyPlayers values(2, 'Jonathan', 'Trott', DATE('1981-04-22'), 'CapeTown', 'SouthAfrica');
mysql> insert into MyPlayers values(3, 'Kumara', 'Sangakkara', DATE('1977-10-27'), 'Matale', 'Srilanka');
mysql> insert into MyPlayers values(4, 'Virat', 'Kohli', DATE('1988-11-05'), 'Delhi', 'India');
mysql> insert into MyPlayers values(5, 'Rohit', 'Sharma', DATE('1987-04-30'), 'Nagpur', 'India');
mysql> insert into MyPlayers values(6, 'Ravindra', 'Jadeja', DATE('1988-12-06'), 'Nagpur', 'India');
mysql> insert into MyPlayers values(7, 'James', 'Anderson', DATE('1982-06-30'), 'Burnley', 'England');
Following query calculates the age of the players in days —
mysql> SELECT First_Name, Last_Name, Date_Of_Birth, Country, DATEDIFF(CURDATE(),Date_Of_Birth) as Age_In_Days FROM MyPlayers;
+------------+------------+---------------+-------------+-------------+
| First_Name |  Last_Name | Date_Of_Birth |     Country | Age_In_Days |
+------------+------------+---------------+-------------+-------------+
|    Shikhar |     Dhawan |    1981-12-05 |       India |       14463 |
|   Jonathan |      Trott |    1981-04-22 | SouthAfrica |       14690 |
|     Kumara | Sangakkara |    1977-10-27 |    Srilanka |       15963 |
|      Virat |      Kohli |    1988-11-05 |       India |       11936 |
|      Rohit |     Sharma |    1987-04-30 |       India |       12491 |
|   Ravindra |     Jadeja |    1988-12-06 |       India |       11905 |
|      James |   Anderson |    1982-06-30 |     England |       14256 |
+------------+------------+---------------+-------------+-------------+
7 rows in set (0.11 sec)
Let us create another table with name Sales in MySQL database using CREATE statement as follows
mysql> CREATE TABLE sales(
	ID INT,
	ProductName VARCHAR(255),
	CustomerName VARCHAR(255),
	DispatchDate date,
	DispatchTime time,
	Price INT,
	Location VARCHAR(255)
);
Query OK, 0 rows affected (2.22 sec)
Now, we will insert 5 records in Sales table using INSERT statements −
insert into sales values (1, 'Key-Board', 'Raja', DATE('2019-09-01'), TIME('11:00:00'), 7000, 'Hyderabad');
insert into sales values (2, 'Earphones', 'Roja', DATE('2019-05-01'), TIME('11:00:00'), 2000, 'Vishakhapatnam');
insert into sales values (3, 'Mouse', 'Puja', DATE('2019-03-01'), TIME('10:59:59'), 3000, 'Vijayawada');
insert into sales values (4, 'Mobile', 'Vanaja', DATE('2019-03-01'), TIME('10:10:52'), 9000, 'Chennai');
insert into sales values (5, 'Headset', 'Jalaja', DATE('2019-04-06'), TIME('11:08:59'), 6000, 'Goa');
Following is another example of this function —
mysql> SELECT ProductName, CustomerName, DispatchDate, Price, DATEDIFF(CURDATE(), DispatchDate) as difference_in_days FROM sales;
+-------------+--------------+--------------+-------+--------------------+
| ProductName | CustomerName | DispatchDate | Price | difference_in_days |
+-------------+--------------+--------------+-------+--------------------+
|   Key-Board |         Raja |   2019-09-01 |  7000 |                679 |
|   Earphones |         Roja |   2019-05-01 |  2000 |                802 |
|       Mouse |         Puja |   2019-03-01 |  3000 |                863 |
|      Mobile |       Vanaja |   2019-03-01 |  9000 |                863 |
|     Headset |       Jalaja |   2019-04-06 |  6000 |                827 |
+-------------+--------------+--------------+-------+--------------------+
5 rows in set (0.00 sec)
Suppose we have created a table named Subscribers with 5 records in it using the following queries –
mysql> CREATE TABLE Subscribers(
	SubscriberName VARCHAR(255),
	PackageName VARCHAR(255),
	SubscriptionDate date
);
insert into Subscribers values('Raja', 'Premium', Date('2020-10-21'));
insert into Subscribers values('Roja', 'Basic', Date('2020-11-26'));
insert into Subscribers values('Puja', 'Moderate', Date('2021-03-07'));
insert into Subscribers values('Vanaja', 'Basic', Date('2021-02-21'));
insert into Subscribers values('Jalaja', 'Premium', Date('2021-01-30'));
Following query calculates and displays the remaining number of days for the subscription to complete —
mysql> SELECT SubscriberName, PackageName, SubscriptionDate, DATEDIFF(CURDATE(), SubscriptionDate) as Remaining_Days FROM Subscribers;
+----------------+-------------+------------------+----------------+
| SubscriberName | PackageName | SubscriptionDate | Remaining_Days |
+----------------+-------------+------------------+----------------+
|           Raja |     Premium |       2020-10-21 |            263 |
|           Roja |       Basic |       2020-11-26 |            227 |
|           Puja |    Moderate |       2021-03-07 |            126 |
|         Vanaja |       Basic |       2021-02-21 |            140 |
|         Jalaja |     Premium |       2021-01-30 |            162 |
+----------------+-------------+------------------+----------------+
5 rows in set (0.11 sec)

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