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