duration calculation

Shell Programming and Scripting
1

I have a file which has 3 coloumns emp_name, Joining_date, Designation.

abc	12/1/2001	SSE
def	2/25/2007	SE
ghi	3/18/2009	SA
abc	8/1/2008	SSE
def	2/13/2007	SE
ghi	3/24/2005	SA

I need to find out the emp who has been in the company for longest period(Till date).

Can I have any suggestion, how to calculate the duration till date?
Thanks

2

convert the dates from the 'MM/DD/YYYY' to 'YYYYMMDD' format and get the lowest number. The rest should be easy.

3

if you are using Linux, below is my generic script for calculating # of days different between 2 given dates. using a -today flag as many calcs have today's date as one of the inputs. It should accept about any UNIX date format, at least all I've tried.

#!/usr/bin/bash

if [ $# -ne "2" ]; then
        echo "incorrect format:   $0 date date"
        echo "example: ./list_times 01/16/2009 04/01/2009"
        echo "example: ./list_times 2009-01-16 04/01/2009"
        exit 1
fi

if [ $2 = "-today" ]; then
        TODAY=`date +%m/%d/%Y`
        /usr/bin/echo -n  $1 " "

                 echo "("$(date -d $1 +%s)-$(date -d $TODAY +%s)")/86400" | bc
 else
                 echo "("$(date -d $1 +%s)-$(date -d $2 +%s)")/86400" | bc
fi

# for your example, I simply created a file called emp, and run it through a for loop, but should give you an idea how to incorporate into a script.

for EMP in `awk '{ print $2 }' emp`
  do 
      ./list_times $EMP -today
done

output:
8/1/2008 -620
3/24/2005 -1846
3/18/2009 -391
2/25/2007 -1142
2/13/2007 -1154
12/1/2001 -3054

4

Assuming the data is in empl.txt

>cat empl.txt | tr "/" " " | awk '{printf "%04d%02d%02d,%-40s,%-40s\n", $4,$2,$3,$1,$5}' | sort
20011201,abc                                     ,SSE
20050324,ghi                                     ,SA
20070213,def                                     ,SE
20070225,def                                     ,SE
20080801,abc                                     ,SSE
20090318,ghi                                     ,SA