Unix shell脚本中的日期算术 [英] Date arithmetic in Unix shell scripts
本文介绍了Unix shell脚本中的日期算术的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我需要在Unix shell脚本中进行日期算术，用于控制第三方程序的执行。
我正在使用一个函数来增加一天，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&&& amp ; 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];那么
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。
这是最好的和更兼容的方式吗？
解决方案
p>我已经写了一个bash脚本，用于将用英语表达的日期转换为常规
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，我将每个调用
的输出放在与冒号（:)分开的调用相同的行上。运行 ComputeDate 时，下面显示的引号是不：
$ ComputeDate'昨天'：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'2010年10月29日'：10 / 29/2010
$ ComputeDate'this monday'：'this monday'has passed。你的意思是下一个星期？
$ ComputeDate'今后一周'：03/27/2010
$ ComputeDate'this satu'：03/20/2010
$ ComputeDate'next星期一：03/22/2010
$ ComputeDate'next thur'：03/25/2010
$ ComputeDate'mon in 2 weeks'：03/28/2010
$ ComputeDate'最后一天的日子：03/31/2010
$ ComputeDate'feb'的最后一天：2/28/2010
$ ComputeDate'feb 2000'的最后一天：2/29 / 2000
$ ComputeDate'从昨天起1周'：03/26/2010
$ ComputeDate'今天1周'：03/27/2010
$ ComputeDate'从明天起1周'： 03/28/2010
$ ComputeDate'从昨天起的2周'：4/2/2010
$ ComputeDate'今天2周'：4/3/2010
$ ComputeDate'2周从明天起：4/4/2010
$ ComputeDate'3月的最后一天后1周：4/7/2010
$ ComputeDate'下周四之后1周'：4/1/2010
$ ComputeDate'3月的最后一个星期之后'：4 / 14/2010
$ ComputeDate'3月的最后一天1周后的2周内：4/15/2010
$ ComputeDate'3月的最后一天后1天：4/1/2010
$ ComputeDate'今天之后1天后1天后1天之后'：03/24/2010
我已经将这个脚本作为这个问题的答案，因为它说明了如何通过一组bash函数对
进行日期算术，而这些函数可能对其他人来说可能是有用的
。它正确地处理闰年和闰年：
＃！ / bin / bash
＃ConvertDate - 将可读取的日期转换为MM / DD / YY日期
＃
＃Date :: =月/日/年
＃ |月/日
＃| DayOfWeek
＃| [this | next] DayOfWeek
＃| DayofWeek [in | in] [Number | next] weeks [s]
＃|日期中的[日|周] [s]日期
＃|本月最后一天
＃|月的最后一天
＃
＃Month :: = January |二月三月四月|五月... |十二月
＃January :: = jan |一月| 1
＃February :: = feb |一月| 2
＃...
＃十二月:: = dec |十二月| 12
＃Day :: = 1 | 2 | ... | 31
＃DayOfWeek :: = today |星期日|星期一|星期二| ... |星期六
＃星期日:: =太阳*
＃...
＃星期六:: = sat *
＃
＃Number :: = Day | a
＃
＃作者：Larry Morell
如果[$＃= 0];然后
printdirections $ 0
退出
fi
＃请求变量的值
GetVar（）{
Var = $ 1
echo -n$ Var = [$ {！Var}]：
local X
读取X
如果[！ -z $ X];那么
eval $ Var =$ X
fi
}
IsLeapYear（）{
local Year = $ 1
if [$ [20 $ Year％4] -eq 0];那么
echo yes
else
echo no
fi
}
＃AddToDate - 计算同一年内的另一个日期
DayNames =（mon tue wed thu fri sat sun）＃对应于'date'输出
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];那么
echo无法将$ dow转换为一个数字的天数
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）
＃返回月份（1 -12）从日期或月份名称
Month2Int（）{
ErrorFlag =
case $ 1
-e）
ErrorFlag = -e; shift
;;
esac
M = $ 1
月份= $ {M %% / *}＃删除/ ...
案例$ month
[az] *）
月份= $ {月：0：3}
M = 0
同时[$ M -lt 12 -a $ {MonthNames [M]}！= $ Month];做
让M ++
完成
让M ++
esac
如果[-z$ ErrorFlag-a $ M -gt 12];那么
echo'$ Month'不是一个有效的月份。
exit
fi
echo $ M
}
＃从合法日期检索月，日，年
GetMonth（）{
echo $ {1 %% / *}
}
GetDay（）{
echo $ 1 | col / 2
}
GetYear（）{
echo $ {1 ## * /}
}
AddToDate（）{
本地日期= $ 1
本地日= $ 2
本地月=`GetMonth $ Date`
本地日=`echo $ Date | col / 2`＃日期日期
local Year =`echo $ Date | col / 3`＃年份
local LeapYear =`IsLeapYear $ Year`
if [$ LeapYear =yes];然后
让月份[1] ++
fi
Day = $ [Day + days]
while [$ Day -gt $ {Months [$ Month-1]}] ; do
Day = $ [Day - $ {Months [$ Month-1]}]
let Month ++
done
echo$ Month / $ Day / $ Year
}
＃将日期转换为正常形式
NormalizeDate（）{
Date =`echo$ *| sed'sX * X / Xg'`
local Day =`date +％d`
local Month =`date +％m`
local年=`date +％Y`
#echo规范化日期= $ Date> / dev / tty
case $日期
* / * / *）
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`
;;
[az] *）＃更好的是一个月或一周的一天
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保持不变
;;
mon | tue | wed | thu | Fri | sat | sun）
＃计算下一个这样的一天
本地DayOfWeek =`date +％u`
D =`Day2Int $ Exp`
if [$ DayOfWeek -le $ D];那么
Date =`AddToDate $ Month / $ Day / $ Year $ [D-DayOfWeek]`
else
Date =`AddToDate $ Month / $ Day / $ Year $ [7 + D -DayOfWeek]`
fi
＃重置月/日/年
月=`echo $ Date | col / 1`
Day =`echo $ Date | col / 2`
Year =`echo $ Date | col / 3`
;;
*）echo$ Exp不是有效的一个月或者一天
exit
;;
esac
;;
*）echo$ Date不是有效的日期
exit
;;
esac
case $ Day in
[0-9] *）;; ＃天必须是数字
*）echo$ Date不是有效的日期
exit
;;
esac
[0-9] [0-9] [0-9] [0-9]）;; ＃年必须是4位数
[0-9] [0-9]）
年= 20 $年
;;
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（）{
本地日期= $ 1; shift
local Month =`GetMonth $ Date`
local Day =`echo $ Date | col / 2`
local Year =`echo $ Date | col / 3`
本地ExtensionExp =$ *
case $
* w * d *中的ExtensionExp＃如5周3天甚至5w2d
ExtensionExp =`echo $ ExtensionExp | sed's / [a-z] / / g'`
weeks =`echo $ ExtensionExp | col 1`
days =`echo $ ExtensionExp | col 2`
days = $ [7 * week + days]
Due =`AddToDate $ Month / $ Day / $ Year $ days`
;;
* d）＃喜欢5天或5d
ExtensionExp =`echo $ ExtensionExp | sed's / [a-z] / / g'`
days = $ ExtensionExp
Due =`AddToDate $ Month / $ Day / $ Year $ days`
;;
*）
到期= $ ExtensionExp
;;
esac
echo $ Due
}
＃Pop - 从数组中移除第一个元素并向左移动
Pop（）{
Var = $ 1
evalunset $ Var [0]
eval$ Var =（\ $ {$ Var [*]}）
}
ComputeDate（）{
local Date =`NormalizeDate $ 1`; shift
local Expression =`echo $ * | sed的/ ^ * a / 1 /; s /，/ /'| tr AZ az`
local Exp =（`echo $ Expression`）
local Token = $ Exp＃first one
local Ans =
#echo$ {Exp [*]}> / dev / tty
case $令牌
* / *）＃正常日期
M =`GetMonth $ Token`
D =`GetDay $ Token`
Y =`GetYear $ Token`
if [-z$ Y];那么
Y = $ Year
elif [$ {＃Y} -eq 2];那么
Y = 20 $ Y
fi
Ans =$ M / $ D / $ Y
;;
yes *）
Ans =`AddToDate $ Date -1`
;;
tod * | now）
Ans = $ Date
;;
tom *）
Ans =`AddToDate $ Date 1`
;;
）
case $表达式
* day * after *）＃日期后的日期
Pop Exp; ＃跳过
Pop Exp; ＃跳过天
Pop Exp; ＃跳过
#echo调用ComputeDate $ Date $ {Exp [*]}> / dev / tty
Date =`ComputeDate $ Date $ {Exp [*]}`#Recursive call
#echo新的日期是$ Date> / dev / tty
Ans =`AddToDate $ Date 1`
;;
*最后*日* * *月* *结束* * *月）
M =`date +％m`
Day = $ {Months [M-1] }
if [$ M -eq 2 -a`IsLeapYear $ Year` = yes];那么
let Day ++
fi
Ans = $ Month / $ Day / $ Year
;;
* last * day * of *）
D = $ {Expression ## * of}
D =`NormalizeDate $ D`
M =`GetMonth $ D`
Y =`GetYear $ D`
＃echo M是$ M> / dev / tty
Day = $ {Months [M-1]}
如果[$ M -eq 2 -a`IsLeapYear $ Y` = yes];然后
let Day ++
fi
Ans = $ [M] / $ Day / $ Y
;;
*）
echo未知表达式：$ expression
exit
;;
esac
;;
next *）＃next DayOfWeek
Pop Exp
dow =`Day2Int $ DayOfWeek`＃前3个字符
tdow =`Day2Int $ {Exp：0：3}＃＃ 3个字符
n = $ [7-dow + tdow]
Ans =`AddToDate $ Date $ n`
;;
this *）
Pop Exp
dow =`Day2Int $ DayOfWeek`
tdow =`Day2Int $ {Exp：0：3}`＃前3个字符
if [$ dow -gt $ tdow];那么
echo'这个$ Exp'已经过去了，你的意思是'next $ Exp？'
exit
fi
n = $ [tdow-dow]
Ans =`AddToDate $ Date $ n`
;;
[az] *）＃DayOfWeek ...
M = $ {Exp：0：3}
case $ M
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];然后＃Just a DayOfWeek
＃tdow =`GetDay $ Exp`＃前3个字符
#if [$ dow -gt $ tdow];那么
#echo'这个$ Exp'已经过去了，你的意思是'下一个$ Exp'？
#exit
#fi
#n = $ [tdow-dow]
#else＃DayOfWeek在未来的一周
Pop Exp＃toss monday
Pop Exp＃toss in / off
if [$ Exp = next];那么
Exp = 2
fi
n = $ [7 *（Exp-1）]＃周数
n = $ [n + 7-dow + tdow]
Ans =`AddToDate $ Date $ n`
fi
;;
esac
;;
[0-9] *）＃Number星期[from | after]日期
n = $ Exp
Pop Exp;
case $ Exp in
w *）let n = 7 * n ;;
esac
Pop Exp; Pop Exp
#echo调用ComputeDate $ Date $ {Exp [*]}> / dev / tty
Date =`ComputeDate $ Date $ {Exp [*]}`#Recursive call
#echo新的日期是$ Date> / dev / tty
Ans =`AddToDate $ Date $ n`
;;
esac
echo $ Ans
}
年份=`日期+％Y`
月份=日期+％m`
日=`date +％d`
DayOfWeek =`date +％a | tr AZ az`
Date =$ Month / $ Day / $ Year
ComputeDate $ Date $ *
此脚本广泛使用我写的另一个脚本（称为 col col 的用户，许多人表示歉意）。此版本的
col 简化了从stdin提取列。因此，
$ echo a b c d e | col 5 3 2
打印
ecb
这里是 col 脚本：
＃！/ bin / sh
＃col - 从文件中提取列
＃用法：
＃col [-r] [c] col-1 col-2 ...
＃其中[c]如果提供定义字段分隔符
＃其中每个col-i表示列根据-r的存在如下解释：
＃-r present：从行的右端开始计数
＃-r缺席：从行左边开始计数
Separator =
Reverse = false
-r中的$ 1）Reverse = true;转移;
;;
[0-9] *）
;;
*）Separator =$ 1;转移;
;;
esac
案例$ 1in
-r）Reverse = true;转移;
;;
[0-9] *）
;;
*）Separator =$ 1;转移;
;;
esac
＃用$ i
替换每个col-i cols =
for $ in $ *
do
if [ $ Reverse = true];那么
Cols =$ Cols \ $（NF- $ f + 1）
else
Cols =$ Cols \ $$ f
fi
done
Cols =`echo$ Cols| sed's /，$ //'`
#echo使用$ Cols的列规格
awk -F$ Separator{print $ Cols}
当脚本调用不正确时，还会使用 printdirections 打印出路线：
＃！/ bin / sh
＃
＃printdirections - 打印shell脚本的标题行
＃
＃用法：
＃printdirections路径
＃其中
＃路径是有问题的shell脚本的一个* full *路径
＃以'/'开头
＃
＃要使用printdirections，您必须包含（作为您的shell脚本的顶部
＃上的注释）运行shell脚本的文档。
if [$＃-eq 0 -o$ *=-h];然后
printdirections $ 0
exit
fi
＃删除文件顶部的命令调用，如果有
＃从打印方向发生到最后的位置删除的文件
＃删除＃comments
＃这里有一个奇怪的奇怪。
sed'/#!/d;/.*printdirections/,$d;/ *＃/！d; s /＃//; s /＃//'$ 1> /tmp/printdirections.$$
＃计算行数
numlines =`wc -l /tmp/printdirections.$$ | awk'{print $ 1}'`
＃删除最后一行
numlines =`expr $ numlines - 1`
head -n $ numlines /tmp/printdirections.$$
rm /tmp/printdirections.$$
请将这三个脚本分别放在文件 ComputeDate ， col 和 printdirections 中。将文件放在由PATH命名的目录中，通常为〜/ bin。然后使它们执行：
$ chmod a + x ComputeDate col printdirections/ pre>
问题？给我一些emaiL：morell AT cs.atu.edu在该主题中的 ComputeDate 。
I need to do date arithmetic in Unix shell scripts that I use to control the execution of third party programs.
I'm using a function to increment a day and another to decrement:
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) fibut now I need to do more complex arithmetic.
What it's the best and more compatible way to do this?
解决方案I have written a bash script for converting dates expressed in English into conventional mm/dd/yyyy dates. It is called ComputeDate.
Here are some examples of its use. For brevity I have placed the output of each invocation on the same line as the invocation, separarted by a colon (:). The quotes shown below are not necessary when running ComputeDate:
$ 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': 03/24/2010I have included this script as an answer to this problem because it illustrates how to do date arithmetic via a set of bash functions and these functions may prove useful for others. It handles leap years and leap centuries correctly:
#! /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 $*This script makes extensive use of another script I wrote (called col ... many apologies to those who use the standard col supplied with Linux). This version of col simplifies extracting columns from the stdin. Thus,
$ echo a b c d e | col 5 3 2prints
e c bHere it the col script:
#!/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}"It also uses printdirections for printing out directions when the script is invoked improperly:
#!/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.$$To use this place the three scripts in the files ComputeDate, col, and printdirections, respectively. Place the file in directory named by your PATH, typically, ~/bin. Then make them executable with:
$ chmod a+x ComputeDate col printdirectionsProblems? Send me some emaiL: morell AT cs.atu.edu Place ComputeDate in the subject.
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