Shell使用Epoch进行日期时间转换和计算的几个小函数

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

#!/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 $day`
let date=$days-$week-7
days2date $date
								 
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