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Updated on 2023-10-25 GMT+08:00
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date_sub

This function is used to calculate the number of days in which start_date is subtracted by days.

To obtain the date with a specified change range based on the current date, use this function together with the current_date or getdate function.

Note that the logic of this function is opposite to that of the date_add function.

Syntax

date_sub(string startdate, int days)

Parameters

Table 1 Parameters

Parameter

Mandatory

Type

Description

start_date

Yes

DATE

or

STRING

Start date

The following formats are supported:

  • yyyy-mm-dd
  • yyyy-mm-dd hh:mi:ss
  • yyyy-mm-dd hh:mi:ss.ff3

days

Yes

BIGINT

Number of days to be reduced

  • If the value is greater than 0, the number of days is increased.
  • If the value of days is less than 0, the number of days is subtracted.
  • If the value is 0, the date does not change.
  • If the value is NULL, NULL is returned.

Return Values

The return value is of the DATE type.

  • If the value of start_date is not of the DATE or STRING type, the error message "data type mismatch" is displayed.
  • If the value of start_date is of the DATE or STRING type but is not in one of the supported formats, NULL is returned.
  • If the value of date is NULL, NULL is returned.
  • If the value of format is NULL, NULL is returned.

Example Code

The value 2023-08-12 is returned after two days are subtracted.

select date_sub('2023-08-14 17:00:00', 2);

The value 2023-08-15 is returned after one day is added.

select date_sub(date'2023-08-14', -1);

If the current time is 2023-08-14 17:00:00, 2022-08-13 is returned.

select date_sub(getdate(),1);

The value NULL is returned.

select date_sub('2023-08-14 17:00:00', null);
Parent topic: Date Functions