Bash 将字符串转换为时间戳
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Bash convert string to timestamp
提问by user3979986
I have a string in format 20141225093000which represents Dec 25, 2014 09:30:00and I want to convert the original format to a unix timestamp format so i can do time operations on it.How would I do this in bash?
我有一个20141225093000表示格式的字符串，Dec 25, 2014 09:30:00我想将原始格式转换为 unix 时间戳格式，以便我可以对其进行时间操作。我将如何在 bash 中执行此操作？
I can easily parse out the values with exprbut I was hoping to be able to identify a format like YYYYmmddHHMMSS and then convert it based on that.
我可以轻松地解析出这些值，expr但我希望能够识别像 YYYYmmddHHMMSS 这样的格式，然后根据它进行转换。
回答by Charles Duffy
With GNU date, you can convert YYYY-MM-DDTHH:MM:SSto epoch time (seconds since 1-1-1970) easily, like so:
使用 GNU 日期，您可以轻松转换YYYY-MM-DDTHH:MM:SS为纪元时间（自 1-1-1970 以来的秒数），如下所示：
date -d '2014-12-25T09:30:00' +%s
To do this starting without any delimiters:
要在没有任何分隔符的情况下执行此操作：
in=20141225093000
rfc_form="${in:0:4}-${in:4:2}-${in:6:2}T${in:8:2}:${in:10:2}:${in:12:2}"
epoch_time=$(date -d "$rfc_form" +%s)
回答by Tiago Lopo
You need to transform the string before calling date:
您需要在调用之前转换字符串date：
#!/bin/bash
s="20141225093000"
s=$(perl -pe 's/(\d{4})(\d{2})(\d{2})(\d{2})(\d{2})(\d{2})/-- ::/g' <<< "$s")
date -d "$s" +%s
Yet another way:
还有一种方式：
perl -MDate::Parse -MPOSIX -le '$s="20141225093000"; $s =~ s/(\d{4})(\d{2})(\d{2})(\d{2})(\d{2})(\d{2})/-- ::/g ; print str2time($s);'
1419499800
回答by glenn Hymanman
GNU awk:
GNU awk：
gawk -v t=20141225093000 'BEGIN {gsub(/../, "& ", t); sub(/ /,"",t); print mktime(t)}'
If GNU date is not available, then it's likely GNU awk may not be. Perl probably has the highest chance of being available. This snippet uses strptimeso you don't have to parse the time string at all:
如果 GNU 日期不可用，则很可能 GNU awk 不可用。Perl 可能有最高的可用机会。使用此代码段strptime，您根本不必解析时间字符串：
perl -MTime::Piece -E 'say Time::Piece->strptime(shift, "%Y%m%d%H%M%S")->epoch' 20141225093000
回答by pjh
This Bash function does the conversion with builtins:
这个 Bash 函数使用内置函数进行转换：
# Convert UTC datetime string (YYYY-MM-DD hh:mm:ss) to Unix epoch seconds
function ymdhms_to_epoch
{
local -r ymdhms=${1//[!0-9]} # Remove non-digits
if (( ${#ymdhms} != 14 )) ; then
echo "error - '$ymdhms' is not a valid datetime" >&2
return 1
fi
# Extract datetime components, possibly with leading zeros
local -r year=${ymdhms:0:4}
local -r month_z=${ymdhms:4:2}
local -r day_z=${ymdhms:6:2}
local -r hour_z=${ymdhms:8:2}
local -r minute_z=${ymdhms:10:2}
local -r second_z=${ymdhms:12:2}
# Remove leading zeros from datetime components to prevent them
# being treated as octal values
local -r month=${month_z#0}
local -r day=${day_z#0}
local -r hour=${hour_z#0}
local -r minute=${minute_z#0}
local -r second=${second_z#0}
# Calculate Julian Day Number (jdn)
# (See <http://en.wikipedia.org/wiki/Julian_day>, Calculation)
local -r -i a='(14-month)/12'
local -r -i y=year+4800-a
local -r -i m=month+12*a-3
local -r -i jdn='day+(153*m+2)/5+365*y+(y/4)-(y/100)+(y/400)-32045'
# Calculate days since the Unix epoch (1 Jan. 1970)
local -r -i epoch_days=jdn-2440588
local -r -i epoch_seconds='((epoch_days*24+hour)*60+minute)*60+second'
echo $epoch_seconds
return 0
}
Example usage:
用法示例：
$ ymdhms_to_epoch '1970-01-01 00:00:00'
0
$ ymdhms_to_epoch '2014-10-18 00:10:06'
1413591006
$ ymdhms_to_epoch '2014-12-25 09:30:00'
1419499800
