SUBDATE

Subtract 10 days from a given date and return the resulting date.

SELECT SUBDATE(“2017-06-15”, INTERVAL 10 DAY);

Definition and Usage

The SUBDATE() function subtracts a specified time/date interval from a given date and returns the resulting date.

SUBDATE(date, INTERVAL value unit)

OR:

SUBDATE(date, days)

Parameter	Description
date	Required: The initial/original date
days	Required: The number of days to subtract from the original date
value	Required: The value of the time/date interval to subtract. Both positive and negative values are accepted.
unit	Required: The type of interval, which can be one of the following values: MICROSECOND SECOND MINUTE HOUR DAY WEEK MONTH QUARTER YEAR SECOND_MICROSECOND MINUTE_MICROSECOND MINUTE_SECOND HOUR_MICROSECOND HOUR_SECOND HOUR_MINUTE DAY_MICROSECOND DAY_SECOND DAY_MINUTE DAY_HOUR YEAR_MONTH

Works in:

From MySQL version 4.0

Subtract 15 minutes from a given date and return the resulting date.

SELECT SUBDATE(“2017-06-15 09:34:21”, INTERVAL 15 MINUTE);

Subtract 3 hours from a given date and return the resulting date.

SELECT SUBDATE(“2017-06-15 09:34:21”, INTERVAL 3 HOUR);

Add 2 months to a given date and return the resulting date.

SELECT SUBDATE(“2017-06-15”, INTERVAL –2 MONTH);