datediff1¶
This function is used to calculate the difference between date1 and date2 and return the difference in a specified datepart.
Similar function: datediff. The datediff function is used to calculate the difference between date1 and date2 but does not return the difference in a specified datepart.
Syntax¶
datediff1(string date1, string date2, string datepart)
Parameters¶
Parameter | Mandatory | Type | Description |
|---|---|---|---|
date1 | Yes | DATE or STRING | Minuend of the date difference between date1 and date2. The following formats are supported:
|
date2 | Yes | DATE or STRING | Subtrahend of the date difference between date1 and date2. The following formats are supported:
|
datepart | Yes | STRING | Unit of the time to be returned This parameter supports the following extended date formats: year, month or mon, day, and hour.
|
Return Values¶
The return value is of the BIGINT type.
Note
If the values of date1 and date2 are not of the DATE or STRING type, the error message "data type mismatch" is displayed.
If the values of date1 and date2 are of the DATE or STRING type but are not in one of the supported formats, NULL is returned.
If the value of date1 is smaller than that of date2, the return value is a negative number.
If the value of date1 or date2 is NULL, NULL is returned.
If the value of datepart is NULL, NULL is returned.
Example Code¶
The value 14400 is returned.
select datediff1('2023-06-30 00:00:00', '2023-06-20 00:00:00', 'mi');
The value 10 is returned.
select datediff1(date '2023-06-21', date '2023-06-11', 'dd');
The value NULL is returned.
select datediff1(date '2023-05-21', date '2023-05-10', null);
The value NULL is returned.
select datediff1(date '2023-05-21', null, 'dd');