我需要在Unix shell脚本中进行日期算术,我用它来控制第三方程序的执行。
我正在使用一个函数来增加一天而另一个减少:
IncrementaDia(){
echo $1 | awk '
BEGIN {
diasDelMes[1] = 31
diasDelMes[2] = 28
diasDelMes[3] = 31
diasDelMes[4] = 30
diasDelMes[5] = 31
diasDelMes[6] = 30
diasDelMes[7] = 31
diasDelMes[8] = 31
diasDelMes[9] = 30
diasDelMes[10] = 31
diasDelMes[11] = 30
diasDelMes[12] = 31
}
{
anio=substr($1,1,4)
mes=substr($1,5,2)
dia=substr($1,7,2)
if((anio % 4 == 0 && anio % 100 != 0) || anio % 400 == 0)
{
diasDelMes[2] = 29;
}
if( dia == diasDelMes[int(mes)] ) {
if( int(mes) == 12 ) {
anio = anio + 1
mes = 1
dia = 1
} else {
mes = mes + 1
dia = 1
}
} else {
dia = dia + 1
}
}
END {
printf("%04d%02d%02d", anio, mes, dia)
}
'
}
if [ $# -eq 1 ]; then
tomorrow=$1
else
today=$(date +"%Y%m%d")
tomorrow=$(IncrementaDia $hoy)
fi
但现在我需要做更复杂的算术。
这是最好和更兼容的方法吗?
答案 0 :(得分:69)
答案 1 :(得分:17)
这是在shell脚本中进行日期计算的简便方法。
meetingDate='12/31/2011' # MM/DD/YYYY Format
reminderDate=`date --date=$meetingDate'-1 day' +'%m/%d/%Y'`
echo $reminderDate
以下是使用date
实用程序可以实现的日期计算的更多变体。
http://www.cyberciti.biz/tips/linux-unix-get-yesterdays-tomorrows-date.html
http://www.cyberciti.biz/faq/linux-unix-formatting-dates-for-display/
这对RHEL来说很有用。
答案 2 :(得分:15)
我编写了一个bash脚本,用于将用英语表示的日期转换为常规日期 mm / dd / yyyy日期。它被称为 ComputeDate 。
以下是其使用的一些示例。为简洁起见,我已经放置了每个调用的输出 在与调用相同的行上,由冒号(:)分隔。运行 ComputeDate 时,下面显示的引号不:
$ ComputeDate 'yesterday': 03/19/2010
$ ComputeDate 'yes': 03/19/2010
$ ComputeDate 'today': 03/20/2010
$ ComputeDate 'tod': 03/20/2010
$ ComputeDate 'now': 03/20/2010
$ ComputeDate 'tomorrow': 03/21/2010
$ ComputeDate 'tom': 03/21/2010
$ ComputeDate '10/29/32': 10/29/2032
$ ComputeDate 'October 29': 10/1/2029
$ ComputeDate 'October 29, 2010': 10/29/2010
$ ComputeDate 'this monday': 'this monday' has passed. Did you mean 'next monday?'
$ ComputeDate 'a week after today': 03/27/2010
$ ComputeDate 'this satu': 03/20/2010
$ ComputeDate 'next monday': 03/22/2010
$ ComputeDate 'next thur': 03/25/2010
$ ComputeDate 'mon in 2 weeks': 03/28/2010
$ ComputeDate 'the last day of the month': 03/31/2010
$ ComputeDate 'the last day of feb': 2/28/2010
$ ComputeDate 'the last day of feb 2000': 2/29/2000
$ ComputeDate '1 week from yesterday': 03/26/2010
$ ComputeDate '1 week from today': 03/27/2010
$ ComputeDate '1 week from tomorrow': 03/28/2010
$ ComputeDate '2 weeks from yesterday': 4/2/2010
$ ComputeDate '2 weeks from today': 4/3/2010
$ ComputeDate '2 weeks from tomorrow': 4/4/2010
$ ComputeDate '1 week after the last day of march': 4/7/2010
$ ComputeDate '1 week after next Thursday': 4/1/2010
$ ComputeDate '2 weeks after the last day of march': 4/14/2010
$ ComputeDate '2 weeks after 1 day after the last day of march': 4/15/2010
$ ComputeDate '1 day after the last day of march': 4/1/2010
$ ComputeDate '1 day after 1 day after 1 day after 1 day after today': 03/24/2010
我已将此脚本作为此问题的答案包含在内,因为它说明了如何操作 通过一组bash函数进行日期算术,这些函数可能有用 为他人。它正确处理闰年和闰年:
#! /bin/bash
# ConvertDate -- convert a human-readable date to a MM/DD/YY date
#
# Date ::= Month/Day/Year
# | Month/Day
# | DayOfWeek
# | [this|next] DayOfWeek
# | DayofWeek [of|in] [Number|next] weeks[s]
# | Number [day|week][s] from Date
# | the last day of the month
# | the last day of Month
#
# Month ::= January | February | March | April | May | ... | December
# January ::= jan | january | 1
# February ::= feb | january | 2
# ...
# December ::= dec | december | 12
# Day ::= 1 | 2 | ... | 31
# DayOfWeek ::= today | Sunday | Monday | Tuesday | ... | Saturday
# Sunday ::= sun*
# ...
# Saturday ::= sat*
#
# Number ::= Day | a
#
# Author: Larry Morell
if [ $# = 0 ]; then
printdirections $0
exit
fi
# Request the value of a variable
GetVar () {
Var=$1
echo -n "$Var= [${!Var}]: "
local X
read X
if [ ! -z $X ]; then
eval $Var="$X"
fi
}
IsLeapYear () {
local Year=$1
if [ $[20$Year % 4] -eq 0 ]; then
echo yes
else
echo no
fi
}
# AddToDate -- compute another date within the same year
DayNames=(mon tue wed thu fri sat sun ) # To correspond with 'date' output
Day2Int () {
ErrorFlag=
case $1 in
-e )
ErrorFlag=-e; shift
;;
esac
local dow=$1
n=0
while [ $n -lt 7 -a $dow != "${DayNames[n]}" ]; do
let n++
done
if [ -z "$ErrorFlag" -a $n -eq 7 ]; then
echo Cannot convert $dow to a numeric day of wee
exit
fi
echo $[n+1]
}
Months=(31 28 31 30 31 30 31 31 30 31 30 31)
MonthNames=(jan feb mar apr may jun jul aug sep oct nov dec)
# Returns the month (1-12) from a date, or a month name
Month2Int () {
ErrorFlag=
case $1 in
-e )
ErrorFlag=-e; shift
;;
esac
M=$1
Month=${M%%/*} # Remove /...
case $Month in
[a-z]* )
Month=${Month:0:3}
M=0
while [ $M -lt 12 -a ${MonthNames[M]} != $Month ]; do
let M++
done
let M++
esac
if [ -z "$ErrorFlag" -a $M -gt 12 ]; then
echo "'$Month' Is not a valid month."
exit
fi
echo $M
}
# Retrieve month,day,year from a legal date
GetMonth() {
echo ${1%%/*}
}
GetDay() {
echo $1 | col / 2
}
GetYear() {
echo ${1##*/}
}
AddToDate() {
local Date=$1
local days=$2
local Month=`GetMonth $Date`
local Day=`echo $Date | col / 2` # Day of Date
local Year=`echo $Date | col / 3` # Year of Date
local LeapYear=`IsLeapYear $Year`
if [ $LeapYear = "yes" ]; then
let Months[1]++
fi
Day=$[Day+days]
while [ $Day -gt ${Months[$Month-1]} ]; do
Day=$[Day - ${Months[$Month-1]}]
let Month++
done
echo "$Month/$Day/$Year"
}
# Convert a date to normal form
NormalizeDate () {
Date=`echo "$*" | sed 'sX *X/Xg'`
local Day=`date +%d`
local Month=`date +%m`
local Year=`date +%Y`
#echo Normalizing Date=$Date > /dev/tty
case $Date in
*/*/* )
Month=`echo $Date | col / 1 `
Month=`Month2Int $Month`
Day=`echo $Date | col / 2`
Year=`echo $Date | col / 3`
;;
*/* )
Month=`echo $Date | col / 1 `
Month=`Month2Int $Month`
Day=1
Year=`echo $Date | col / 2 `
;;
[a-z]* ) # Better be a month or day of week
Exp=${Date:0:3}
case $Exp in
jan|feb|mar|apr|may|june|jul|aug|sep|oct|nov|dec )
Month=$Exp
Month=`Month2Int $Month`
Day=1
#Year stays the same
;;
mon|tue|wed|thu|fri|sat|sun )
# Compute the next such day
local DayOfWeek=`date +%u`
D=`Day2Int $Exp`
if [ $DayOfWeek -le $D ]; then
Date=`AddToDate $Month/$Day/$Year $[D-DayOfWeek]`
else
Date=`AddToDate $Month/$Day/$Year $[7+D-DayOfWeek]`
fi
# Reset Month/Day/Year
Month=`echo $Date | col / 1 `
Day=`echo $Date | col / 2`
Year=`echo $Date | col / 3`
;;
* ) echo "$Exp is not a valid month or day"
exit
;;
esac
;;
* ) echo "$Date is not a valid date"
exit
;;
esac
case $Day in
[0-9]* );; # Day must be numeric
* ) echo "$Date is not a valid date"
exit
;;
esac
[0-9][0-9][0-9][0-9] );; # Year must be 4 digits
[0-9][0-9] )
Year=20$Year
;;
esac
Date=$Month/$Day/$Year
echo $Date
}
# NormalizeDate jan
# NormalizeDate january
# NormalizeDate jan 2009
# NormalizeDate jan 22 1983
# NormalizeDate 1/22
# NormalizeDate 1 22
# NormalizeDate sat
# NormalizeDate sun
# NormalizeDate mon
ComputeExtension () {
local Date=$1; shift
local Month=`GetMonth $Date`
local Day=`echo $Date | col / 2`
local Year=`echo $Date | col / 3`
local ExtensionExp="$*"
case $ExtensionExp in
*w*d* ) # like 5 weeks 3 days or even 5w2d
ExtensionExp=`echo $ExtensionExp | sed 's/[a-z]/ /g'`
weeks=`echo $ExtensionExp | col 1`
days=`echo $ExtensionExp | col 2`
days=$[7*weeks+days]
Due=`AddToDate $Month/$Day/$Year $days`
;;
*d ) # Like 5 days or 5d
ExtensionExp=`echo $ExtensionExp | sed 's/[a-z]/ /g'`
days=$ExtensionExp
Due=`AddToDate $Month/$Day/$Year $days`
;;
* )
Due=$ExtensionExp
;;
esac
echo $Due
}
# Pop -- remove the first element from an array and shift left
Pop () {
Var=$1
eval "unset $Var[0]"
eval "$Var=(\${$Var[*]})"
}
ComputeDate () {
local Date=`NormalizeDate $1`; shift
local Expression=`echo $* | sed 's/^ *a /1 /;s/,/ /' | tr A-Z a-z `
local Exp=(`echo $Expression `)
local Token=$Exp # first one
local Ans=
#echo "Computing date for ${Exp[*]}" > /dev/tty
case $Token in
*/* ) # Regular date
M=`GetMonth $Token`
D=`GetDay $Token`
Y=`GetYear $Token`
if [ -z "$Y" ]; then
Y=$Year
elif [ ${#Y} -eq 2 ]; then
Y=20$Y
fi
Ans="$M/$D/$Y"
;;
yes* )
Ans=`AddToDate $Date -1`
;;
tod*|now )
Ans=$Date
;;
tom* )
Ans=`AddToDate $Date 1`
;;
the )
case $Expression in
*day*after* ) #the day after Date
Pop Exp; # Skip the
Pop Exp; # Skip day
Pop Exp; # Skip after
#echo Calling ComputeDate $Date ${Exp[*]} > /dev/tty
Date=`ComputeDate $Date ${Exp[*]}` #Recursive call
#echo "New date is " $Date > /dev/tty
Ans=`AddToDate $Date 1`
;;
*last*day*of*th*month|*end*of*th*month )
M=`date +%m`
Day=${Months[M-1]}
if [ $M -eq 2 -a `IsLeapYear $Year` = yes ]; then
let Day++
fi
Ans=$Month/$Day/$Year
;;
*last*day*of* )
D=${Expression##*of }
D=`NormalizeDate $D`
M=`GetMonth $D`
Y=`GetYear $D`
# echo M is $M > /dev/tty
Day=${Months[M-1]}
if [ $M -eq 2 -a `IsLeapYear $Y` = yes ]; then
let Day++
fi
Ans=$[M]/$Day/$Y
;;
* )
echo "Unknown expression: " $Expression
exit
;;
esac
;;
next* ) # next DayOfWeek
Pop Exp
dow=`Day2Int $DayOfWeek` # First 3 chars
tdow=`Day2Int ${Exp:0:3}` # First 3 chars
n=$[7-dow+tdow]
Ans=`AddToDate $Date $n`
;;
this* )
Pop Exp
dow=`Day2Int $DayOfWeek`
tdow=`Day2Int ${Exp:0:3}` # First 3 chars
if [ $dow -gt $tdow ]; then
echo "'this $Exp' has passed. Did you mean 'next $Exp?'"
exit
fi
n=$[tdow-dow]
Ans=`AddToDate $Date $n`
;;
[a-z]* ) # DayOfWeek ...
M=${Exp:0:3}
case $M in
jan|feb|mar|apr|may|june|jul|aug|sep|oct|nov|dec )
ND=`NormalizeDate ${Exp[*]}`
Ans=$ND
;;
mon|tue|wed|thu|fri|sat|sun )
dow=`Day2Int $DayOfWeek`
Ans=`NormalizeDate $Exp`
if [ ${#Exp[*]} -gt 1 ]; then # Just a DayOfWeek
#tdow=`GetDay $Exp` # First 3 chars
#if [ $dow -gt $tdow ]; then
#echo "'this $Exp' has passed. Did you mean 'next $Exp'?"
#exit
#fi
#n=$[tdow-dow]
#else # DayOfWeek in a future week
Pop Exp # toss monday
Pop Exp # toss in/off
if [ $Exp = next ]; then
Exp=2
fi
n=$[7*(Exp-1)] # number of weeks
n=$[n+7-dow+tdow]
Ans=`AddToDate $Date $n`
fi
;;
esac
;;
[0-9]* ) # Number weeks [from|after] Date
n=$Exp
Pop Exp;
case $Exp in
w* ) let n=7*n;;
esac
Pop Exp; Pop Exp
#echo Calling ComputeDate $Date ${Exp[*]} > /dev/tty
Date=`ComputeDate $Date ${Exp[*]}` #Recursive call
#echo "New date is " $Date > /dev/tty
Ans=`AddToDate $Date $n`
;;
esac
echo $Ans
}
Year=`date +%Y`
Month=`date +%m`
Day=`date +%d`
DayOfWeek=`date +%a |tr A-Z a-z`
Date="$Month/$Day/$Year"
ComputeDate $Date $*
此脚本广泛使用我编写的另一个脚本(称为 col ...对使用Linux提供的标准 col 的用户表示道歉)。这个版本 col 简化了从标准输入中提取列的过程。因此,
$ echo a b c d e | col 5 3 2
打印
e c b
这是 col 脚本:
#!/bin/sh
# col -- extract columns from a file
# Usage:
# col [-r] [c] col-1 col-2 ...
# where [c] if supplied defines the field separator
# where each col-i represents a column interpreted according to the presence of -r as follows:
# -r present : counting starts from the right end of the line
# -r absent : counting starts from the left side of the line
Separator=" "
Reverse=false
case "$1" in
-r ) Reverse=true; shift;
;;
[0-9]* )
;;
* )Separator="$1"; shift;
;;
esac
case "$1" in
-r ) Reverse=true; shift;
;;
[0-9]* )
;;
* )Separator="$1"; shift;
;;
esac
# Replace each col-i with $i
Cols=""
for f in $*
do
if [ $Reverse = true ]; then
Cols="$Cols \$(NF-$f+1),"
else
Cols="$Cols \$$f,"
fi
done
Cols=`echo "$Cols" | sed 's/,$//'`
#echo "Using column specifications of $Cols"
awk -F "$Separator" "{print $Cols}"
它还使用 printdirections 在不正确地调用脚本时打印出路线:
#!/bin/sh
#
# printdirections -- print header lines of a shell script
#
# Usage:
# printdirections path
# where
# path is a *full* path to the shell script in question
# beginning with '/'
#
# To use printdirections, you must include (as comments at the top
# of your shell script) documentation for running the shell script.
if [ $# -eq 0 -o "$*" = "-h" ]; then
printdirections $0
exit
fi
# Delete the command invocation at the top of the file, if any
# Delete from the place where printdirections occurs to the end of the file
# Remove the # comments
# There is a bizarre oddity here.
sed '/#!/d;/.*printdirections/,$d;/ *#/!d;s/# //;s/#//' $1 > /tmp/printdirections.$$
# Count the number of lines
numlines=`wc -l /tmp/printdirections.$$ | awk '{print $1}'`
# Remove the last line
numlines=`expr $numlines - 1`
head -n $numlines /tmp/printdirections.$$
rm /tmp/printdirections.$$
要分别使用文件 ComputeDate , col 和 printdirections 中的三个脚本。将文件放在PATH命名的目录中,通常为〜/ bin。然后使用以下命令使它们可执行:
$ chmod a+x ComputeDate col printdirections
有问题吗?发给我一些电子邮件:morell AT cs.atu.edu在主题中放置 ComputeDate 。
答案 3 :(得分:7)
对于BSD / OS X兼容性,您还可以使用带有-j
和-v
的日期实用程序来进行日期数学运算。请参阅FreeBSD manpage for date。您可以将以前的Linux答案与此答案结合起来,这可能会为您提供足够的兼容性。
在BSD上,作为Linux,运行date
将为您提供当前日期:
$ date
Wed 12 Nov 2014 13:36:00 AEDT
现在有了BSD的约会,您可以使用-v
进行数学运算,例如列出明天的日期(+1d
加上一天):
$ date -v +1d
Thu 13 Nov 2014 13:36:34 AEDT
您可以使用现有日期作为基础,并可选择使用strftime指定解析格式,并确保使用-j
,这样您就不会更改系统日期:
$ date -j -f "%a %b %d %H:%M:%S %Y %z" "Sat Aug 09 13:37:14 2014 +1100"
Sat 9 Aug 2014 12:37:14 AEST
您可以将此作为日期计算的基础:
$ date -v +1d -f "%a %b %d %H:%M:%S %Y %z" "Sat Aug 09 13:37:14 2014 +1100"
Sun 10 Aug 2014 12:37:14 AEST
请注意,-v
隐含-j
。
可以按顺序提供多项调整:
$ date -v +1m -v -1w
Fri 5 Dec 2014 13:40:07 AEDT
有关详细信息,请参阅联机帮助页。
答案 4 :(得分:6)
要在UNIX上对日期进行算术运算,请将日期作为自UNIX纪元以来的秒数,进行一些计算,然后转换回可打印的日期格式。 date命令应该能够为您提供自纪元以来的秒数,并从该数字转换回可打印日期。我的本地日期命令执行此操作,
% date -n
1219371462
% date 1219371462
Thu Aug 21 22:17:42 EDT 2008
%
请参阅您当地的date(1)
手册页。
要增加一天,请添加86400秒。
答案 5 :(得分:4)
date --date='1 days ago' '+%a'
这不是一个非常兼容的解决方案。它只适用于Linux。至少,它在Aix和Solaris中没有用。
适用于RHEL:
date --date='1 days ago' '+%Y%m%d'
20080807
答案 6 :(得分:4)
为什么不使用像perl或python这样的语言来编写脚本呢?这更自然地支持复杂的日期处理?当然你可以在bash中完成所有操作,但我认为你也可以使用python在平台之间获得更多的一致性,只要你能确保安装了perl或python。
我应该补充说,将python和perl脚本连接到包含的shell脚本中非常容易。
答案 7 :(得分:4)
我已经碰到过几次了。我的想法是:
示例脚本(检查某些用户文件的年龄以查看是否可以删除该帐户):
#!/usr/local/bin/perl
$today = time();
$user = $ARGV[0];
$command="awk -F: '/$user/ {print \$6}' /etc/passwd";
chomp ($user_dir = `$command`);
if ( -f "$user_dir/.sh_history" ) {
@file_dates = stat("$user_dir/.sh_history");
$sh_file_date = $file_dates[8];
} else {
$sh_file_date = 0;
}
if ( -f "$user_dir/.bash_history" ) {
@file_dates = stat("$user_dir/.bash_history");
$bash_file_date = $file_dates[8];
} else {
$bash_file_date = 0;
}
if ( $sh_file_date > $bash_file_date ) {
$file_date = $sh_file_date;
} else {
$file_date = $bash_file_date;
}
$difference = $today - $file_date;
if ( $difference >= 3888000 ) {
print "User needs to be disabled, 45 days old or older!\n";
exit (1);
} else {
print "OK\n";
exit (0);
}
答案 8 :(得分:3)
进一步研究,我认为你可以简单地使用日期。 我在OpenBSD上尝试了以下内容:我采用了2008年2月29日的日期和随机时间(以080229301535的形式),并在当天的部分添加了+1,如下所示:
$ date -j 0802301535
Sat Mar 1 15:35:00 EST 2008
如您所见,日期正确格式化了时间......
HTH
答案 9 :(得分:3)
如果你想继续使用awk,那么mktime和strftime函数很有用:
BEGIN { dateinit }
{ newdate=daysadd(OldDate,DaysToAdd)}
# daynum: convert DD-MON-YYYY to day count
#-----------------------------------------
function daynum(date, d,m,y,i,n)
{
y=substr(date,8,4)
m=gmonths[toupper(substr(date,4,3))]
d=substr(date,1,2)
return mktime(y" "m" "d" 12 00 00")
}
#numday: convert day count to DD-MON-YYYY
#-------------------------------------------
function numday(n, y,m,d)
{
m=toupper(substr(strftime("%B",n),1,3))
return strftime("%d-"m"-%Y",n)
}
# daysadd: add (or subtract) days from date (DD-MON-YYYY), return new date (DD-MON-YYYY)
#------------------------------------------
function daysadd(date, days)
{
return numday(daynum(date)+(days*86400))
}
#init variables for date calcs
#-----------------------------------------
function dateinit( x,y,z)
{
# Stuff for date calcs
split("JAN:1,FEB:2,MAR:3,APR:4,MAY:5,JUN:6,JUL:7,AUG:8,SEP:9,OCT:10,NOV:11,DEC:12", z)
for (x in z)
{
split(z[x],y,":")
gmonths[y[1]]=y[2]
}
}
答案 10 :(得分:3)
Chris F.A. Johnson的书“Shell Script Recipes:A Problem Solution Approach”(ISBN:978-1-59059-471-1)有一个日期函数库,可能会有所帮助。源代码在http://apress.com/book/downloadfile/2146处可用(日期函数位于tar文件中的第08章/ data-funcs-sh中)。
答案 11 :(得分:0)
如果GNU版本的日期适合您,为什么不抓取源代码并在AIX和Solaris上编译它?
http://www.gnu.org/software/coreutils/
在任何情况下,如果您要编写自己的代码,源代码应该帮助您更正日期算术。
顺便说一句,评论说“这个解决方案很好,但你肯定可以注意到它不是那么好。似乎没有人想到在构建Unix时修改日期。”真的没有把我们带到任何地方。到目前为止,我发现每个建议都非常有用且符合目标。
答案 12 :(得分:0)
以下是我的两个便士 - 一个使用date
和grep
的脚本包装。
示例用法
> sh ./datecalc.sh "2012-08-04 19:43:00" + 1s
2012-08-04 19:43:00 + 0d0h0m1s
2012-08-04 19:43:01
> sh ./datecalc.sh "2012-08-04 19:43:00" - 1s1m1h1d
2012-08-04 19:43:00 - 1d1h1m1s
2012-08-03 18:41:59
> sh ./datecalc.sh "2012-08-04 19:43:00" - 1d2d1h2h1m2m1s2sblahblah
2012-08-04 19:43:00 - 1d1h1m1s
2012-08-03 18:41:59
> sh ./datecalc.sh "2012-08-04 19:43:00" x 1d
Bad operator :-(
> sh ./datecalc.sh "2012-08-04 19:43:00"
Missing arguments :-(
> sh ./datecalc.sh gibberish + 1h
date: invalid date `gibberish'
Invalid date :-(
脚本
#!/bin/sh
# Usage:
#
# datecalc "<date>" <operator> <period>
#
# <date> ::= see "man date", section "DATE STRING"
# <operator> ::= + | -
# <period> ::= INTEGER<unit> | INTEGER<unit><period>
# <unit> ::= s | m | h | d
if [ $# -lt 3 ]; then
echo "Missing arguments :-("
exit; fi
date=`eval "date -d \"$1\" +%s"`
if [ -z $date ]; then
echo "Invalid date :-("
exit; fi
if ! ([ $2 == "-" ] || [ $2 == "+" ]); then
echo "Bad operator :-("
exit; fi
op=$2
minute=$[60]
hour=$[$minute*$minute]
day=$[24*$hour]
s=`echo $3 | grep -oe '[0-9]*s' | grep -m 1 -oe '[0-9]*'`
m=`echo $3 | grep -oe '[0-9]*m' | grep -m 1 -oe '[0-9]*'`
h=`echo $3 | grep -oe '[0-9]*h' | grep -m 1 -oe '[0-9]*'`
d=`echo $3 | grep -oe '[0-9]*d' | grep -m 1 -oe '[0-9]*'`
if [ -z $s ]; then s=0; fi
if [ -z $m ]; then m=0; fi
if [ -z $h ]; then h=0; fi
if [ -z $d ]; then d=0; fi
ms=$[$m*$minute]
hs=$[$h*$hour]
ds=$[$d*$day]
sum=$[$s+$ms+$hs+$ds]
out=$[$date$op$sum]
formattedout=`eval "date -d @$out +\"%Y-%m-%d %H:%M:%S\""`
echo $1 $2 $d"d"$h"h"$m"m"$s"s"
echo $formattedout
答案 13 :(得分:-1)
这对我有用:
TZ=GMT+6;
export TZ
mes=`date --date='2 days ago' '+%m'`
dia=`date --date='2 days ago' '+%d'`
anio=`date --date='2 days ago' '+%Y'`
hora=`date --date='2 days ago' '+%H'`