diffYears365
diffYears365 returns the difference in years (defined as 365 days) between two date-time values, as a double.
Syntax
diffYears365(start, end)
Parameters
| Parameter | Type | Description |
|---|---|---|
| start | Instant | The start time of the range. |
| start | ZonedDateTime | The start time of the range. |
| end | Instant | The end time of the range. |
| end | ZonedDateTime | The end time of the range. |
Returns
Returns the difference in years between the supplied start and end values.
Examples
d1 = parseInstant("2022-01-01T00:00:00 ET")
d2 = parseInstant("2023-01-02T00:00:00 ET")
difference = diffYears365(d1, d2)
println difference
- Log