linux计算时间差

发布时间：2014-09-05 14:37:23作者：知识屋

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

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，可以参考英文文章：

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

arr=($(($base)) $(($base+1)) $(($base+7)) $(($base+30)))

dag=(1 2 8 31)

for i in {0..3}; do

echo ${dag[$i]}