linux - 为什么 `date -d @1270787111` 中需要前缀 @ ？

Question

我想将纪元时间转换为人类可读的格式。我在网上搜索，找到了一个

示例:date -d @1270787111。

我重读了日期的手册。

-d, --date=STRING display time described by STRING, not 'now'

和

DATE STRING The --date=STRING is a mostly free...

我没有找到关于这种格式的任何信息。我错过了什么吗？

Answer 1

您所指的同一“DATE STRING”部分以以下句子结尾:

The date string format is more complex than is easily documented here but is fully described in the info documentation.

换句话说，您必须引用“信息”页面。您可以通过在终端中输入 info date 来完成此操作，或者在线查看这些文档。输入格式描述得很好here ，例如。

那么为什么date -d @1270787111 中需要前缀@？手册页示例中提到的带有 @ 的特定格式是 described in §28.8 :

28.8 Seconds since the Epoch

If you precede a number with @', it represents an internal time stamp
as a count of seconds. The number can contain an internal decimal
point (either.' or `,'); any excess precision not supported by the internal representation is truncated toward minus infinity. Such a number cannot be combined with any other date item, as it specifies a complete time stamp.

Internally, computer times are represented as a count of seconds since an epoch--a well-defined point of time. On GNU and POSIX systems, the epoch is 1970-01-01 00:00:00 UTC, so @0' represents this
time,@1' represents 1970-01-01 00:00:01 UTC, and so forth. GNU and most other POSIX-compliant systems support such times as an extension to POSIX, using negative counts, so that `@-1' represents 1969-12-31 23:59:59 UTC.

Traditional Unix systems count seconds with 32-bit two's-complement integers and can represent times from 1901-12-13 20:45:52 through 2038-01-19 03:14:07 UTC. More modern systems use 64-bit counts of seconds with nanosecond subcounts, and can represent all the times in the known lifetime of the universe to a resolution of 1 nanosecond.

On most hosts, these counts ignore the presence of leap seconds. For example, on most hosts @915148799' represents 1998-12-31 23:59:59
UTC,@915148800' represents 1999-01-01 00:00:00 UTC, and there is no way to represent the intervening leap second 1998-12-31 23:59:60 UTC.