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[英]Trying to add 1 day to a timestamp in bash script but it's only adding 19 hours
[英]Bash - number of hours in given day
在 Linux 上使用 Bash shell,並給定日期時間,我如何確定那一天有多少小時?
日期時間與某些具有夏令時的時區有關,例如 MET。
為了完全考慮所有場景,您需要考慮以下幾點:
並非每個本地日都有午夜,如果您在這些日子之一傳遞日期,則date
命令將失敗,除非您還傳遞了時間和與 UTC 的偏移量。 這主要發生在春季向前的過渡日。 例如:
$ TZ=America/Sao_Paulo date -d '2016-10-16' date: invalid date '2016-10-16'
並非每個 DST 過渡都是 1 小時。 America/Lord_Howe
切換 30 分鍾。 Bash 只執行整數除法,所以如果你想要小數,你必須使用這些技術之一。
這是一個解釋這些的函數:
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."
}
例子:
$ 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.
10 月 30 日是英國最后一個夏令時。 我可以通過這種方式從 shell 獲得那天的 25 小時:
t1=$(TZ='Europe/London' date --date='20161030' +%s)
t2=$(TZ='Europe/London' date --date='20161031' +%s)
echo $((($t2 - $t1) / 3600))
我不完全確定這會在每個 bash shell 中工作,並且可能需要稍微調整一下。
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