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SQL - DATEDIFF() Function
The SQL DATEDIFF() function is an in-built function in SQL that is used to return the difference (as a signed integer value) between two dates or times.
This function accepts three parameters − the type of interval to measure (such as year, quarter, month, hour, minute, etc.), the start date or time that specifies the beginning of the period to measure, and the end date or time that specifies the end of the period to measure.
This function returns the difference in a 32-bit int data type, that can store values up to 2,147,483,647.
Syntax
Following is the syntax of the SQL DATEDIFF() function −
DATEDIFF(datepart, date1, date2)
Parameters
This function accepts three parameters. The same is described below −
datepart − This specifies the type of (date or time) segment to return. The following are the possible values −
- year, yyyy, yy = Year
- quarter, qq, q = Quarter
- month, mm, m = Month
- dayofyear, dy, y = Day of the year
- day, dd, d = Day
- week, ww, wk = Week
- weekday, dw, w = Weekday
- hour, hh = Hour
- minute, mi, n = Minute
- second, ss, s = Second
- millisecond, ms = Millisecond
- date1, date2 − These specify the two dates that we want to compare.
Example
We can calculate the difference between two date values in years using the following query −
SQL> SELECT DATEDIFF(YEAR, '2023/03/15', '2033/02/15') AS DATE_DIFF_IN_YEARS;
Output
When we execute the above query, the output is obtained as follows −
+--------------------+ | DATE_DIFF_IN_YEARS | +--------------------+ | 10 | +--------------------+
Example
Here, we are calculating the difference between two date values in quarters using the following query −
SQL> SELECT DATEDIFF(QUARTER, '2023/03/15', '2024/02/15') AS DATE_DIFF_IN_QUARTER;
Output
On executing the above query, the output is displayed as follows −
+----------------------+ | DATE_DIFF_IN_QUARTER | +----------------------+ | 4 | +----------------------+
Example
In the following example, we are trying to calculate the difference between two specified times in hours −
SQL> SELECT DATEDIFF(HOUR, '2023/03/15 07:30:00:000', '2023/03/16 06:00:00:000') AS TIME_DIFF_IN_HOUR;
Output
The output for the above query is produced as given below −
+-------------------+ | TIME_DIFF_IN_HOUR | +-------------------+ | 23 | +-------------------+
Example
Here, we are calculating the difference between two specified times in minutes using the following query −
SQL> SELECT DATEDIFF(MINUTE, '2023/03/01 07:30', '2023/04/01 07:30') AS TIME_DIFF_IN_MINUTE;
Output
If we execute the above query, the result is produced as follows −
+---------------------+ | TIME_DIFF_IN_MINUTE | +---------------------+ | 44640 | +---------------------+
Example
Assume we have created a table with the name EMPLOYEE in the SQL database using the CREATE statement as shown in the query below −
SQL> CREATE TABLE EMPLOYEE(ID INT NOT NULL, NAME VARCHAR (20) NOT NULL, DATE_OF_BIRTH VARCHAR (20));
Now, let us insert some records in the EMPLOYEE table using INSERT statements as shown in the query below −
SQL> INSERT INTO EMPLOYEE(ID, NAME, DATE_OF_BIRTH) VALUES(1, 'Dhruv', '2000-12-05'); INSERT INTO EMPLOYEE(ID, NAME, DATE_OF_BIRTH) VALUES(2, 'Arjun', '2000-03-01'); INSERT INTO EMPLOYEE(ID, NAME, DATE_OF_BIRTH) VALUES(3, 'Dev', '2001-03-15'); INSERT INTO EMPLOYEE(ID, NAME, DATE_OF_BIRTH) VALUES(4, 'Riya', '2003-12-05'); INSERT INTO EMPLOYEE(ID, NAME, DATE_OF_BIRTH) VALUES(5, 'Aarohi', '2000-05-02'); INSERT INTO EMPLOYEE(ID, NAME, DATE_OF_BIRTH) VALUES(6, 'Lisa', '1999-11-25'); INSERT INTO EMPLOYEE(ID, NAME, DATE_OF_BIRTH) VALUES(7, 'Roy', '2001-05-30');
We can verify whether the table is created or not using the following query −
SQL> SELECT * FROM EMPLOYEE;
The table EMPLOYEE is successfully created in the SQL database.
+-----+--------+---------------+ | ID | NAME | DATE_OF_BIRTH | +-----+--------+---------------+ | 1 | Dhruv | 2000-12-05 | | 2 | Arjun | 2000-03-01 | | 3 | Dev | 2001-03-15 | | 4 | Riya | 2003-12-05 | | 5 | Aarohi | 2000-05-02 | | 6 | Lisa | 1999-11-25 | | 7 | Roy | 2001-05-30 | +-----+--------+---------------+
Here, we are calculating the age of every employee in years using the following query −
Note − The CURRENT_TIMESTAMP retrieves the date and time at the moment.
SQL> SELECT ID, NAME, DATE_OF_BIRTH, DATEDIFF(YEAR, DATE_OF_BIRTH, CURRENT_TIMESTAMP) AS AGE_IN_YEARS FROM EMPLOYEE;
Output
When we execute the above query, the output is obtained as follows −
+-----+--------+---------------+--------------+ | ID | NAME | DATE_OF_BIRTH | AGE_IN_YEARS | +-----+--------+---------------+--------------+ | 1 | Dhruv | 2000-12-05 | 23 | | 2 | Arjun | 2000-03-01 | 23 | | 3 | Dev | 2001-03-15 | 22 | | 4 | Riya | 2003-12-05 | 20 | | 5 | Aarohi | 2000-05-02 | 23 | | 6 | Lisa | 1999-11-25 | 24 | | 7 | Roy | 2001-05-30 | 22 | +-----+--------+---------------+--------------+
Example
Let us create another table with the name OTT in the SQL database using the CREATE statement as shown in the query below −
SQL> CREATE TABLE OTT(ID INT NOT NULL, SUBSCRIBER_NAME VARCHAR (200) NOT NULL, MEMBERSHIP VARCHAR (200), SUBCRIPTION_DATE DATE NOT NULL);
Now, let us insert some records in the OTT table using INSERT statements as shown in the query below −
SQL> INSERT INTO OTT(ID, SUBSCRIBER_NAME, MEMBERSHIP, SUBCRIPTION_DATE) VALUES(1, 'Dhruv', 'Silver', '2022-12-05'); INSERT INTO OTT(ID, SUBSCRIBER_NAME, MEMBERSHIP, SUBCRIPTION_DATE) VALUES(2, 'Arjun','Platinum', '2021-03-01'); INSERT INTO OTT(ID, SUBSCRIBER_NAME, MEMBERSHIP, SUBCRIPTION_DATE) VALUES(3, 'Dev','Silver', '2021-03-15'); INSERT INTO OTT(ID, SUBSCRIBER_NAME, MEMBERSHIP, SUBCRIPTION_DATE) VALUES(4, 'Riya','Gold', '2022-12-05'); INSERT INTO OTT(ID, SUBSCRIBER_NAME, MEMBERSHIP, SUBCRIPTION_DATE) VALUES(5, 'Aarohi','Platinum', '2020-05-02'); INSERT INTO OTT(ID, SUBSCRIBER_NAME, MEMBERSHIP, SUBCRIPTION_DATE) VALUES(6, 'Lisa','Platinum', '2022-11-25'); INSERT INTO OTT(ID, SUBSCRIBER_NAME, MEMBERSHIP, SUBCRIPTION_DATE) VALUES(7, 'Roy','Gold', '2021-05-30');
We can verify whether the table OTT is created or not using the following query −
SQL> SELECT * FROM OTT;
The table OTT is successfully created in the SQL database.
+-----+-----------------+------------+------------------+ | ID | SUBSCRIBER_NAME | MEMBERSHIP | SUBSCRIPTION_DATE| +-----+-----------------+------------+------------------+ | 1 | Dhruv | Silver | 2022-12-05 | | 2 | Arjun | Platinum | 2021-03-01 | | 3 | Dev | Silver | 2021-03-15 | | 4 | Riya | Gold | 2022-12-05 | | 5 | Aarohi | Platinum | 2020-05-02 | | 6 | Lisa | Platinum | 2022-11-25 | | 7 | Roy | Gold | 2021-05-30 | +-----+-----------------+------------+------------------+
Here, we are trying to display the remaining number of days for the subscription plans to be complete using the following query −
SQL> SELECT SUBSCRIBER_NAME, SUBCRIPTION_DATE, DATEDIFF(DAY, SUBCRIPTION_DATE, CURRENT_TIMESTAMP) AS REMAINING_DAYS FROM OTT;
Output
If we execute the above query, the result is produced as follows −
+-----------------+-------------------+-----------------+ | SUBSCRIBER_NAME | SUBSCRIPTION_DATE | REMAINING_DAYS | +-----------------+-------------------+-----------------+ | Dhruv | 2022-12-05 | 72 | | Arjun | 2021-03-01 | 716 | | Dev | 2021-03-15 | 702 | | Riya | 2022-12-05 | 72 | | Aarohi | 2020-05-02 | 1019 | | Lisa | 2022-11-25 | 82 | | Roy | 2021-05-30 | 626 | +-----------------+-------------------+-----------------+
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