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.
- 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');
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