Bash

Question

在 Linux 上使用 Bash shell 并给定一个日期时间，我如何确定那一天有多少个小时？

日期时间属于夏令时的某个时区，例如遇见

Answer 1

10 月 30 日是英国最后一次更改夏令时。我可以通过这种方式从 shell 获得那天的 25 小时：

t1=$(TZ='Europe/London' date --date='20161030' +%s)
t2=$(TZ='Europe/London' date --date='20161031' +%s)
echo $((($t2 - $t1) / 3600))

我不确定这是否适用于每个 bash shell，可能需要稍微调整一下。

Answer 2

为了完全考虑所有场景，您需要考虑以下几点：

并非每个本地日都有午夜，如果您传递这些日子之一的日期，date 命令将失败，除非您还传递时间和与 UTC 的偏移量.这主要发生在 spring-forward 过渡日。例如：
```
$ TZ=America/Sao_Paulo date -d '2016-10-16'
date: invalid date '2016-10-16'
```
并非每个 DST 转换都是 1 小时。 America/Lord_Howe 切换 30 分钟。 Bash只进行整数除法，所以如果你想要小数，你必须使用one of these techniques。

这里有一个函数可以解决这些问题：

seconds_in_day() {
  # Copy input date to local variable
  date=

  # Start with the offset at noon on the given date.
  # Noon will almost always exist (except Samoa on 2011-12-30)
  offset1=$(date -d "$date 12:00" +%z)

  # Next get the offset for midnight.  If it doesn't exist, the time will jump back to 23:00 and we'll get a different offset.
  offset1=$(date -d "$date 00:00 $offset1" +%z)

  # Next get the offset for the next day at midnight.  Again, if it doesn't exist, it will jump back an hour.
  offset2=$(date -d "$date 00:00 $offset1 + 1 day" +%z)

  # Get the unix timestamps for both the current date and the next one, at midnight with their respective offsets.
  unixtime1=$(date -d "$date 00:00 $offset1" +%s)
  unixtime2=$(date -d "$date 00:00 $offset2 + 1 day" +%s)

  # Calculate the difference in seconds and hours.  Use awk for decimal math.
  seconds=$((unixtime2 - unixtime1))
  hours=$(awk -v seconds=$seconds 'BEGIN { print seconds / 3600 }')

  # Print the output
  echo "$date had $seconds secs in $TZ, or $hours hours."
}

示例：

$ TZ=America/Los_Angeles seconds_in_day 2016-03-12
2016-03-12 had 86400 secs in America/Los_Angeles, or 24 hours.
$ TZ=America/Los_Angeles seconds_in_day 2016-03-13
2016-03-13 had 82800 secs in America/Los_Angeles, or 23 hours.
$ TZ=America/Los_Angeles seconds_in_day 2016-03-14
2016-03-14 had 86400 secs in America/Los_Angeles, or 24 hours.

$ TZ=America/Los_Angeles seconds_in_day 2016-11-05
2016-11-05 had 86400 secs in America/Los_Angeles, or 24 hours.
$ TZ=America/Los_Angeles seconds_in_day 2016-11-06
2016-11-06 had 90000 secs in America/Los_Angeles, or 25 hours.
$ TZ=America/Los_Angeles seconds_in_day 2016-11-07
2016-11-07 had 86400 secs in America/Los_Angeles, or 24 hours.

$ TZ=America/Sao_Paulo seconds_in_day 2016-02-19
2016-02-19 had 86400 secs in America/Sao_Paulo, or 24 hours.
$ TZ=America/Sao_Paulo seconds_in_day 2016-02-20
2016-02-20 had 90000 secs in America/Sao_Paulo, or 25 hours.
$ TZ=America/Sao_Paulo seconds_in_day 2016-02-21
2016-02-21 had 86400 secs in America/Sao_Paulo, or 24 hours.

$ TZ=America/Sao_Paulo seconds_in_day 2016-10-15
2016-10-15 had 86400 secs in America/Sao_Paulo, or 24 hours.
$ TZ=America/Sao_Paulo seconds_in_day 2016-10-16
2016-10-16 had 82800 secs in America/Sao_Paulo, or 23 hours.
$ TZ=America/Sao_Paulo seconds_in_day 2016-10-17
2016-10-17 had 86400 secs in America/Sao_Paulo, or 24 hours.

$ TZ=Australia/Lord_Howe seconds_in_day 2016-04-02
2016-04-02 had 86400 secs in Australia/Lord_Howe, or 24 hours.
$ TZ=Australia/Lord_Howe seconds_in_day 2016-04-03
2016-04-03 had 88200 secs in Australia/Lord_Howe, or 24.5 hours.
$ TZ=Australia/Lord_Howe seconds_in_day 2016-04-04
2016-04-04 had 86400 secs in Australia/Lord_Howe, or 24 hours.

$ TZ=Australia/Lord_Howe seconds_in_day 2016-10-01
2016-10-01 had 86400 secs in Australia/Lord_Howe, or 24 hours.
$ TZ=Australia/Lord_Howe seconds_in_day 2016-10-02
2016-10-02 had 84600 secs in Australia/Lord_Howe, or 23.5 hours.
$ TZ=Australia/Lord_Howe seconds_in_day 2016-10-03
2016-10-03 had 86400 secs in Australia/Lord_Howe, or 24 hours.

Bash - 给定日期的小时数

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