怎么样通过Shell使用Epoch进行日期时间转换和计算的几个函数？
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=$1+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a;
d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153);
dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10);
printf ("%4d-%02d-%02d\n",yy,mm,dd)
}'
}
days2date 14839
#日期转分钟
function date2minutes {
echo "$*" | 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=$1; nn=i%60; i=int(i/60); hh=i%24; dd=int(i/24); i=int(i/24);
a=i+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a;
d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153);
dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10);
printf ("%4d-%02d-%02d %02d:%02d\n",yy,mm,dd,hh,nn)
}'
}
minutes2date 21369299
#日期转秒数
function date2seconds {
echo "$*" | 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=$1; ss=i%60; i=int(i/60); nn=i%60; i=int(i/60); hh=i%24; dd=int(i/24); i=int(i/24);
a=i+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a;
d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153);
dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10);
printf ("%4d-%02d-%02d %02d:%02d:%02d\n",yy,mm,dd,hh,nn,ss)
}'
}
seconds2date 1282157959
#日期转毫秒
function date2milliseconds {
echo "$*" | 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%s\n",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.%03d\n",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
#!/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%02d\n",yy,mm,dd)
}'
}
year=`date +%Y`; month=`date +%m`; day=`date +%d`
days=`date2days $year $month $day`
N=5
let days-=$N
days2date $days
计算某天的N天之后的日期
#!/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%02d\n",yy,mm,dd)
}'
}
year=2010; month=01; day=20
days=`date2days $year $month $day`
let days+=5
days2date $days
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%02d\n",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=$1+2472632; b=int((4*a+3)/146097); c=int((-b*146097)/4)+a;
d=int((4*c+3)/1461); e=int((-1461*d)/4)+c; m=int((5*e+2)/153);
dd=-int((153*m+2)/5)+e+1; mm=int(-m/10)*12+m+3; yy=b*100+d-4800+int(m/10);
printf ("%4d%02d%02d\n",yy,mm,dd)
}'
}
num1=20105050
num2=20101001
arg1=`echo "$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:%02d\n",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%02d\n",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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