Reference · Dates

date.of

date.of(int, int, int) -> str

What it does

The date with a year, a month and a day, as ISO text "YYYY-MM-DD". The month and the day are 1-based, and the day must exist in that month of that year. It is the way from numbers to a date; date.parts goes back.

Examples

print(date.of(2026, 10, 27), date.of(2024, 2, 29))let { year, month } = date.parts("2026-10-27")print(date.of(year, month + 1, 1))

2026-10-27 2024-02-292026-11-01

Edge cases

• Years are 1 to 9999, padded to four digits: date.of(1, 1, 1) is "0001-01-01".
• There is no rolling over: day 32 or month 13 is an error, not the next month or year; use date.add_days and date.add_months for that.
• A float with a whole value is taken as an int.

print(date.of(1, 1, 1), date.of(9999, 12, 31), date.of(2026.0, 1, 1))print(date.of(2026, 12, 31) | date.add_days(1))

0001-01-01 9999-12-31 2026-01-012027-01-01

Errors

A day that does not exist in the month, a month outside 1 to 12 or a year outside 1 to 9999 is a runtime error, not an error value, as is an argument that is not an int.

print(date.of(2023, 2, 29))

line 1: `date.of`: there is no date 2023-2-29

Chaining

The year comes first in a pipeline, which is rarely what is wanted; it is more often the end of one:

let [y, m, d] = "2026-10-27" | split("-") | map(int)print(date.of(y, m, d) | date.format("D MMMM YYYY"))print(range(1, 4) | map((m) => date.of(2026, m, 1) | date.end_of_month))

27 October 2026["2026-01-31", "2026-02-28", "2026-03-31"]

See also

date.parts, date.parse, date.valid, date.days_in_month, date.add_days

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