Bash - 给定日期的小时数

时间:2017-01-18 18:28:33

标签: linux bash date time dst

在Linux上使用Bash shell,并给出日期时间,如何确定该特定日期的小时数?

日期时间属于某些具有夏令时的时区,例如: MET。

2 个答案:

答案 0 :(得分:1)

为了完全考虑所有场景,您需要考虑以下几点:

  • 并非每个当地日都有午夜,如果您在其中一天传递日期,date命令将失败,除非您还传递时间和UTC的偏移量。这主要发生在春季前进过渡日。例如:

    $ TZ=America/Sao_Paulo date -d '2016-10-16'
    date: invalid date '2016-10-16'
    
  • 并非每个DST过渡都是1小时。 America/Lord_Howe切换30分钟。 Bash只执行整数除法,因此如果需要小数,则必须使用one of these techniques

这是一个解释这些问题的函数:

seconds_in_day() {
  # Copy input date to local variable
  date=$1

  # 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.

答案 1 :(得分:0)

10月30日是英国最后一次夏季时间的变化。那天我可以通过这种方式从shell中获得25个小时:

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

我不完全确定这会在每个bash shell中都有效,可能需要稍微调整一下。