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Max of Date DD-MM-YYYY 17:30

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# 1  
Old 03-13-2012
Max of Date DD-MM-YYYY 17:30
Hi,

I have two sets of dates as below. Need to find the maximum of them.

14-Feb-2006 17:30,02-Feb-2006 14:46
14-Feb-2006 17:30,02-Feb-2006 14:46
14-Feb-2006 17:30,02-Feb-2006 14:46
23-May-2006 18:25,02-Feb-2006 14:46
13-Feb-2006 12:00,02-Feb-2006 14:46
23-May-2006 18:25,02-Feb-2006 14:46
13-Feb-2006 12:00,02-Feb-2006 14:46
13-Feb-2006 12:00,02-Feb-2006 14:46
05-Oct-2007 09:19,02-Feb-2006 14:46
13-Feb-2006 12:00,02-Feb-2006 14:46
05-Oct-2007 09:19,02-Feb-2006 14:46
13-Feb-2006 12:00,02-Feb-2006 14:46
13-Feb-2006 12:00,02-Feb-2006 14:46

The differnce may be even in minutes

Please Suggest
# 2  
Old 03-13-2012
With GNU date:

Code:
#! /bin/bash
x=0
while IFS=',' read t1 t2
do
    if [ $x -lt $((($(date -d "$t1" +%s) - $(date -d "$t2" +%s)) / 60)) ]
    then
        x=$((($(date -d "$t1" +%s) - $(date -d "$t2" +%s)) / 60))
        y="${t1},${t2}"
    fi
done < inputfile
echo "$y"
echo "$x"

# 3  
Old 03-13-2012
Using ksh93
Code:
#!/bin/ksh93

maxdiff=0
while IFS=',' read date1 date2
do
   diff=$(( $(printf "%(%s)T" "$date1") - $(printf "%(%s)T" "$date2") ))
   (( diff > maxdiff )) && maxdiff=$diff
done < infile

mins=$((maxdiff % (60 * 60) / 60))
hours=$((maxdiff / (60 * 60)))

printf "%02d:%02d\n" $hours $mins

# 4  
Old 03-15-2012
Hi
Thanks for you Reply balajesuri and fpmurphy.

But Both are not working .
I am getting error. I am usinh ksh in solaris.

for balajesuri version
Quote:
date: illegal option -- d
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
date: illegal option -- d
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
./maxdate.sh: line 6: ( - ) / 60: syntax error: operand expected (error token is ") / 60")
For fpmurphy Version,
Quote:
./maxdate.sh[7]: ()T - ()T : syntax error
# 5  
Old 03-15-2012
Please explain what you mean by 'maximum of them' -- the maximum absolute date, or the widest range?
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