linux 计算时间差

为了计算留存，需要知道多个从时间差，来获取不同的时间点。 以下代码在输入值与当前值，在同一月份时，不会有问题。但是如果是夸月份回溯数据，那么就会出现较大的问题。

#Date variables for
if [ -n "$1" ]; then
    TODAY=`date -d "" +"%Y%m%d"`
    base=$(($TODAY-$1))
elif [ -z "${DAYAGO1}" ]; then
    base=0
fi
arr=($base $(($base+1)) $(($base+2)) $(($base+8)) $(($base+31)))
dag=(0 1 2 8 31)

for i in {0..4}; do
    echo $i
    echo ${dag[$i]}
    echo ${arr[$i]}
    export "DAYAGO${dag[$i]}"=`date -d "${arr[$i]} days ago" +"%Y%m%d"`
done

为了修正bug，可以参考

http://stackoverflow.com/questions/3385003/shell-script-to-get-difference-in-two-dates

There's a solution that almost works: use the %s date format of GNU date, which prints the number of seconds since 1970-01-01 00:00. These can be subtracted to find the time difference between two dates.

echo $(( ($(date -d 2010-06-01 +%s) - $(date -d 2010-05-15 +%s)) / 86400))

But the following displays 0 in some locations:

echo $((($(date -d 2010-03-29 +%s) - $(date -d 2010-03-28 +%s)) / 86400))

Because of daylight savings time, there are only 23 hours between those times. You need to add at least one hour (and at most 23) to be safe.

echo $((($(date -d 2010-03-29 +%s) - $(date -d 2010-03-28 +%s) + 43200) / 86400))

Or you can tell date to work in a timezone without DST.

echo $((($(date -u -d 2010-03-29 +%s) - $(date -u -d 2010-03-28 +%s)) / 86400))

(POSIX says to call the reference timezone is UTC, but it also says not to count leap seconds, so the number of seconds in a day is always exactly 86400 in a GMT+xx timezone.)

最终结果就接近完美了，为：

#Date variables for
if [ -n "$1" ]; then
    TODAY=`date -d "" +"%Y%m%d"`
    BACKDAY=$1
    base=$((($(date -d $TODAY +%s) - $(date -d $BACKDAY +%s) + 43200) / 86400))
elif [ -z "${DAYAGO1}" ]; then
    base=1
fi
arr=($(($base)) $(($base+1)) $(($base+7)) $(($base+30)))
dag=(1 2 8 31)

for i in {0..3}; do
    echo ${dag[$i]}
    echo ${arr[$i]}
    export "DAYAGO${dag[$i]}"=`date -d "${arr[$i]} days ago" +"%Y%m%d"`
done

 shell 中的时间计算转为秒做相减运算

NOW=`date +"%Y-%m-%d %H:%M:%S"`
Back=$(ls -lt * "$dir"| line | awk '{print $6,$7,$8}')    #获取文件的修改时间
dn=`date -d  "$NOW" +%s`    #把当前时间转化为毫秒时间
db=`date -d  "$Back" +%s`
ddif=`expr $dn- $db`  #计算2个时间的差

另外一种，换个思路，从前往后推算也是可以的：

day='20120201'
this_day=$day
today='20120311'
i=-1
while [ $t_day != $today ]
do
        i=$(($i + 1))
        t_day=`date +%Y%0m%0d --date="$day $i day"`
        echo $t_day
done