Deprecated: This feature is no longer recommended. Though some browsers might still support it, it may have already been removed from the relevant web standards, may be in the process of being dropped, or may only be kept for compatibility purposes. Avoid using it, and update existing code if possible; see the compatibility table at the bottom of this page to guide your decision. Be aware that this feature may cease to work at any time.
setYear() method sets the year for a specified date according to local time.
However, the way the legacy
setYear() method sets year values is different from how the preferred
setFullYear() method sets year values — and in some cases, also different from how
new Date() and
Date.parse() set year values. Specifically, given two-digit numbers, such as
setYear()interprets any two-digit number as an offset to
date.setYear(22)results in the year value being set to
date.setYear(61)results in the year value being set to
1961. (In contrast, while
new Date(61, 1)also results in the year value being set to
new Date("2/1/22")results in the year value being set to
2022; and similarly for
setFullYear()does no special interpretation but instead uses the literal two-digit value as-is to set the year; so
date.setFullYear(61)results in the year value being set to
date.setFullYear(22)results in the year value being set to
Because of those differences in behavior, you should no longer use the legacy
setYear() method, but should instead use the preferred
The number of milliseconds between 1 January 1970 00:00:00 UTC and the updated date.
yearValue is a number between 0 and 99 (inclusive), then the year for
dateObj is set to
1900 + yearValue. Otherwise, the year for
dateObj is set to
The first two lines set the year to 1996. The third sets the year to 2000.
const theBigDay = new Date(); theBigDay.setYear(96); theBigDay.setYear(1996); theBigDay.setYear(2000);
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