Su, Xiaomeng 76a5b1ee83 dli_sqlreference_20240227

Reviewed-by: Pruthi, Vineet <vineet.pruthi@t-systems.com>
Co-authored-by: Su, Xiaomeng <suxiaomeng1@huawei.com>
Co-committed-by: Su, Xiaomeng <suxiaomeng1@huawei.com>

2024-03-27 22:02:33 +00:00

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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

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