Shell使用Epoch进行日期时间转换和计算的几个小函数

摘要：核心代码当你遇到一个date命令不给力的系统时，可以试试这几个小函数。#日期转天数functiondate2days{echo"$*"|aw...

核心代码

当你遇到一个date命令不给力的系统时，可以试试这几个小函数。 #日期转天数 function date2days { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; print j }' } date2days `echo "2010-08-18 18:59:19" | sed 's/-/ /g;s/:/ /g'` #天数转日期 function days2date { echo "$1" | awk '{ a=$1+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a; d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153); dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10); printf ("%4d-%02d-%02dn",yy,mm,dd) }' } days2date 14839 #日期转分钟 function date2minutes { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; j=j*1440+$4*60+$5 print j }' } date2minutes `echo "2010-08-18 18:59:19" | sed 's/-/ /g;s/:/ /g'` #分钟转日期 function minutes2date { echo "$1" | awk '{ i=$1; nn=i%60; i=int(i/60); hh=i%24; dd=int(i/24); i=int(i/24); a=i+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a; d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153); dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10); printf ("%4d-%02d-%02d %02d:%02dn",yy,mm,dd,hh,nn) }' } minutes2date 21369299 #日期转秒数 function date2seconds { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; j=j*86400+$4*3600+$5*60+$6 print j }' } date2seconds `echo "2010-08-18 18:59:19" | sed 's/-/ /g;s/:/ /g'` #秒数转日期 function seconds2date { echo "$1" | awk '{ i=$1; ss=i%60; i=int(i/60); nn=i%60; i=int(i/60); hh=i%24; dd=int(i/24); i=int(i/24); a=i+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a; d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153); dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10); printf ("%4d-%02d-%02d %02d:%02d:%02dn",yy,mm,dd,hh,nn,ss) }' } seconds2date 1282157959 #日期转毫秒 function date2milliseconds { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; j=j*86400+$4*3600+$5*60+$6 printf ("%d%sn",j,$7) }' } date2milliseconds `echo "2010-08-18 18:59:19.073" | sed 's/-/ /g;s/:/ /g;s/./ /g'` #毫秒转日期 function milliseconds2date { echo "$1" | awk '{ i=$1; ms=i%1000; i=int(i/1000); ss=i%60; i=int(i/60); nn=i%60; i=int(i/60); hh=i%24; dd=int(i/24); i=int(i/24); a=i+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a; d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153); dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10); printf ("%4d-%02d-%02d %02d:%02d:%02d.%03dn",yy,mm,dd,hh,nn,ss,ms) }' } milliseconds2date 1282157959073

应用实例：

计算今天的N天之后的日期

计算某天的N天之后的日期

计算上一个星期的全部日期

#!/bin/bash function date2days { echo "$1 $2 $3" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; print j }' } function days2date { echo "$1" | awk '{ a=$1+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a; d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153); dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10); printf ("%4d%02d%02dn",yy,mm,dd) }' } function date2week { echo "$1 $2 $3" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; dow=(int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472629)%7; print dow }' } year=`date +%Y`; month=`date +%m`; day=`date +%d` days=`date2days $year $month $day` week=`date2week $year $month $day` let dateEnd=$days-$week-1 let dateBegin=$dateEnd-6 for ((i=$dateBegin;i<=$dateEnd;i++)); do days2date $i done

日期时间转换成毫秒

function date2milliseconds { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; j=j*86400+$4*3600+$5*60+$6 print j$7 }' } date2milliseconds `echo "2010-08-18 18:59:19.073" | /usr/xpg4/bin/awk -F'[:.-]+' '$1=$1'`

日期时间转换成秒

function date2seconds { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; j=j*86400+$4*3600+$5*60+$6 print j }' } date2seconds `echo "2010-07-21 00:00:00" | sed 's/-/ /g;s/:/ /g'`

判断一个数字是否为合法日期

function date2days { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; print j }' } function days2date { echo "$1" | awk '{ a=$1+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a; d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153); dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10); printf ("%4d%02d%02dn",yy,mm,dd) }' } num1=20105050 num2=20101001 arg1=`echo "$num1" | sed -r 's/(....)(..)(..)/1 2 3/g'` arg2=`echo "$num2" | sed -r 's/(....)(..)(..)/1 2 3/g'` days1=`date2days $arg1` date1=`days2date $days1` days2=`date2days $arg2` date2=`days2date $days2` [ "$num1" -eq "$date1" ] && echo "$num1 is valid date" || echo "$num1 is invalid date" [ "$num2" -eq "$date2" ] && echo "$num2 is valid date" || echo "$num2 is invalid date"

计算10分钟之前的时间

function date2minutes { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; j=j*1440+$4*60+$5 print j }' } function minutes2date { echo "$1" | awk '{ i=$1; nn=i%60; i=int(i/60); hh=i%24; dd=int(i/24); i=int(i/24); a=i+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a; d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153); dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10); printf ("%4d-%02d-%02d %02d:%02dn",yy,mm,dd,hh,nn) }' } now=`date "+%Y-%m-%d %H:%M" | sed 's/-/ /g;s/:/ /g'` minutes=`date2minutes $now` let minutes-=10 minutes2date $minutes

计算指定日期和当前系统日期之家相差多少天

#!/bin/bash function date2days { echo "$*" | awk '{ z=int((14-$2)/12); y=$1+4800-z; m=$2+12*z-3; j=int((153*m+2)/5)+$3+y*365+int(y/4)-int(y/100)+int(y/400)-2472633; print j }' } echo "Enter your date:" read input InpuDays=$(date2days ${input:0:4} ${input:4:2} ${input:6:2}) SysDays=$(date2days `date +"%Y %m %d"`) let result=$InpuDays-$SysDays echo $result

#./test.sh Enter your date: 20110605 25

上个星期周一的日期