date substraction

Shell Programming and Scripting
1

hello
i have obtained the current date ..
current_date=date "+%m/%d%y"

and i have another date ,stored in my log file which i have already retrieved. i want to store the subtraction in a varible called diff.

diff=log_date - currentdate
ex: log_date=01/28/11
current_date= 02/25/11
then diff=28

2

Hi, Try the below bash script ,
Date_Difference_Days.sh

#!/bin/sh
cur_date=`date "+%m/%d/%y"`
log_date='01/28/11'
echo $log_date $cur_date | awk '
BEGIN {
        d[1] = 31
        d[2] = 28
        d[3] = 31
        d[4] = 30
        d[5] = 31
        d[6] = 30
        d[7] = 31
        d[8] = 31
        d[9] = 30
        d[10] = 31
        d[11] = 30
        d[12] = 31
}
{
        yr=substr($1,7,2);mon=substr($1,1,2);day=substr($1,4,2)
        yr1=substr($2,7,2);mon1=substr($2,1,2);day1=substr($2,4,2)

        if((yr % 4 == 0 && yr % 100 != 0) || yr % 400 == 0) { d["02"] = 29;}

        if ((yr1 - yr)==0 && (mon1-mon)==0) {print day1-day;exit}

        if ((yr1 - yr)==0 && (mon1-mon)!=0)
        {
        for (i=mon+1;i<int(mon1);i++){dayadd+=d}
        print day1+(d[int(mon)]-day)+dayadd
        exit
        }

        if ((yr1 - yr)!=0)
        {
        mon_new=mon
        for(j=yr;j<yr1;j++)
        {
                for (i=mon_new+1;i<=12;i++){dayadd+=d}
                mon_new=0
        }
        for (i=1;i<int(mon1);i++){dayadd+=d}
        print "Days Diff :- " day1+(d[int(mon)]-day)+dayadd
        exit
        }
}'
3

one more version , but this does have some limitations , like year<=2099 etc.

 
#!/bin/sh
get_JD () {
scale=0
echo " $1-32075+1461*($3+4800+($2-14)/12)/4+367*($2-2-($2-14)/12*12)/12-3*(($3+4900+($2-14)/12)/100)/4 " | bc
}
first=`get_JD 1 1 2011`
second=`get_JD 1 3 2011`
diff=`expr $second - $first`
echo $diff

I googled it on my intereset. The logic applied here seems to be an algo used in US Navy. I don't have much idea how this is working!!!

4

Using gnu date

from="20100401" 
to="20100503" 
start=$(date --date="$from" +"%s") 
end=$(date --date="$to" +"%s") 
days=$(( (end - start) / 86400 )) 
echo $days 

from="04/01/11" 
to="05/03/11" 
start=$(date --date="$from" +"%s") 
end=$(date --date="$to" +"%s") 
days=$(( (end - start) / 86400 )) 
echo $days

Using ksh93 builtin printf

((day=24*60*60))  # 86400 s
fromdate="04/01/11"
todate="05/03/11"
# datestr => epoc 
epoc1=$(printf "%(%#)T" "$fromdate") 
epoc2=$(printf "%(%#)T" "$todate") 
# length of day (seconds) = 86400 
echo $(( (epoc2-epoc1) /day)) 

fromdate="2011-04-01"
todate="2011-05-03"
# datestr => epoc 
epoc1=$(printf "%(%#)T" "$fromdate") 
epoc2=$(printf "%(%#)T" "$todate") 
echo $(( (epoc2-epoc1) /day))

Or using calculating like panyam.

More about calculating
Is there a formula for calculating the Julian day number?

You can do panyam solution without any external commands, using builtin in every posix compatible shells (bash, ksh93, dash, ...)

get_JD()
{
        d="$1"
        m="$2"
        y="$3"
        echo $(( d-32075+1461*(y+4800+(m-14)/12)/4+367*(m-2-(m-14)/12*12)/12-3*((y+4900+(y-14)/12)/100)/4 ))
}

first=$( get_JD 1 4 2011 )       # day month year
second=$( get_JD 4 5 2011 )
diff=$((second - first))
echo $diff