Question

怎么写一个shell脚本实现求出两个日期的间隔天数

Answer 1

#!/bin/sh

day_count()
{
Date=$1
year=${Date:0:4}
month=${Date:4:2}
day=${Date:6:2}

month_temp=1
Datenum=$day
while [ $month_temp -lt $month ]
do

case $month_temp in
1|3|5|7|8|10|12) Datenum=`expr 31 + $Datenum` ;;
4|6|9|11) Datenum=`expr 30 + $Datenum` ;;
2) if [ `expr $year % 100` != 0 ] && [ `expr $year % 4` == 0 ] || [ `expr $year % 400` == 0 ];then
Datenum=`expr 29 + $Datenum`
else
Datenum=`expr 28 + $Datenum`
fi ;;
esac
month_temp=$(($month_temp + 1))
done

echo $Datenum
}

read -p "输入第一个日期(日期格式:YYYYMMDD如19991219)：" Date1
read -p "输入第二个日期(日期格式:YYYYMMDD如19991219)：" Date2

Datenum1=$(day_count $Date1)
Datenum2=$(day_count $Date2)

if [ $Date1 -ge $Date2 ];then
yearnum=`expr ${Date1:0:4} - ${Date2:0:4}`
yearnum=`expr $yearnum \* 365`
Datenum1=`expr $Datenum1 + $yearnum `
echo "the number is `expr $Datenum1 - $Datenum2`"
else
yearnum=`expr ${Date2:0:4} - ${Date1:0:4}`
yearnum=`expr $yearnum \* 365`
Datenum2=`expr $Datenum2 + $yearnum `
echo "the number is `expr $Datenum2 - $Datenum1`"
fi

Answer 2

#!/bin/bash
end_time="2016-03-17"
start_time="2016-03-15"
time1=$(($(date +%s -d "${end_time}") - $(date +%s -d "${start_time}")));
echo $((${time1}/(3600*24)+1))