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Shell script (KSH) to list ONLY the ID of male employees whose last loging time was during the last

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Top Forums Shell Programming and Scripting Shell script (KSH) to list ONLY the ID of male employees whose last loging time was during the last
# 1  
Old 02-13-2012
Shell script (KSH) to list ONLY the ID of male employees whose last loging time was during the last
Can someone help me on my issue please Smilie


I have a table like this format:

Code:
# cat myfile.txt
 
 Employee  Gender    NAME      Last Login 
    ID                                       Time        
-------------------------------------------------
210125       M         ABC        02/03/2012 08:07 
451235       F         EFG         02/012/2012 16:53  
241567       M         GHI         02/01/2012 14:03  
314584       M         JKL         02/02/2012 12:43

*****************

I need shell script (KSH) to list ONLY the ID of male employees whose last loging time was during the last two days.


the output should be list of IDs whic meet the requirements

Moderator's Comments:
Mod Comment Please use code tags. Please use a descriptive title for a thread:
https://www.unix.com/unix-dummies-que...om-forums.html
Last edited by Sara_84; 02-13-2012 at 08:33 AM.. Reason: more descriptive title
# 2  
Old 02-13-2012
Is the second line of your data typed correctly? I think we can guess that the date format is American mm/dd/yyyy but that line is different.

What Operating System and version do you have? Do you have the GNU "date" command?
How do you define "during the last two days"? Do you mean 48 hours from the time it is at the moment you run the script, or is this something to do with the date?

Are you trying to run the script as a one-off for today only or is this a regular requirement?
# 3  
Old 02-13-2012
I'm using AIX 6.1. yes I have the "date" command

the date format is American.

I mean by "during the last two days" from the the moment I run the script.

The purpose of the script to list every IDs of employees who logged in during the last two days.

Thaaanks Smilie
# 4  
Old 02-13-2012
MySQL
below script will provide you the expected output. It will work only for the same year.

Code:
#! /bin/bash

da_date="`date +%d`"
da_month=`date "+%m"`

 prev_date=0
 prev_mon=0
da_year=`date "+%Y"`
if [ "$da_date" -gt "1" ];then
  da_dat=`expr $da_date - 2`
  da_dat="""$da_dat"
else
  da_dat=1
  prev_mon=`expr $da_month - 1`
  prev_date=`cal $prev_mon $da_year | xargs -n1 | tail -1`
  prev_date="""$prev_date"
fi

while read line1
do
line=`echo $line1 | grep -v Employee |  awk '{if($2=="M") print $4}'`
if [ ! -z "$line" ];then
op_da=`echo $line | awk -F/ '{print $2}'`
op_mo=`echo $line | awk -F/ '{print $1}'`
op_year=`echo $line | awk -F/ '{print $3}'`

if [ \( "$op_da" -gt "$da_dat" \) -a \( "$op_mo" -eq "$da_month" \) -a \( "$op_year" -eq "$da_year" \) ];then
 echo $line1 | awk '{print $1}'
fi

#If date is 1
if [ \( "$da_dat" -eq "1" \) -a \( \
 \( \( "$op_da" -eq "$prev_date" \) -a \( "$op_mo" -eq "$prev_mon" \) -a "$op_year" -eq "$da_year" \) -o \
\( \( "$op_da" -eq "1" \) -a  \( "$op_mo" -eq "$da_month" \) -a \( "$op_year" -eq "$da_year" \) \) \) ];then

 echo $line1 | awk '{print $1}'
fi

fi
done < inp6

Input: inp6
 
210125 M ABC 02/13/2012 08:07
451235 F EFG 02/012/2012 16:53
241567 M GHI 02/12/2012 14:03
314584 M JKL 02/02/2012 12:43
Output
210125
241567


Thanks,
Kalai
This User Gave Thanks to kalpeer For This Post:
Sara_84
# 5  
Old 02-13-2012
If Perl is an option, then -

Code:
$
$ cat myfile.txt
 Employee  Gender    NAME      Last Login
    ID                                       Time
-------------------------------------------------
210125       M         ABC         02/03/2012 08:07
451235       F         EFG         02/02/2012 16:53
241567       M         GHI         02/01/2012 14:03
314584       M         JKL         02/02/2012 12:43
123456       M         MNO         02/12/2012 00:00
654321       F         PQR         02/12/2012 00:00
$
$ perl -lne 'BEGIN {use Time::Local; $prior = time-2*24*60*60}
             if (m|^(\d+)\s+([MF])\s+\w+\s+(\d+)/(\d+)/(\d+)\s+(\d+):(\d+)$|){
               ($id, $g, $mo, $d, $y, $h, $mi) = ($1, $2, $3, $4, $5, $6, $7);
               $y -= 1900; $mo--;
               $ts = timelocal(0,$mi,$h,$d,$mo,$y);
               print $id if $ts > $prior and $g eq "M";
             }' myfile.txt
123456
$
$

tyler_durden
This User Gave Thanks to durden_tyler For This Post:
Sara_84
# 6  
Old 02-14-2012
Quote:
Originally Posted by kalpeer
below script will provide you the expected output. It will work only for the same year.

Code:
#! /bin/bash
 
da_date="`date +%d`"
da_month=`date "+%m"`
 
 prev_date=0
 prev_mon=0
da_year=`date "+%Y"`
if [ "$da_date" -gt "1" ];then
  da_dat=`expr $da_date - 2`
  da_dat="""$da_dat"
else
  da_dat=1
  prev_mon=`expr $da_month - 1`
  prev_date=`cal $prev_mon $da_year | xargs -n1 | tail -1`
  prev_date="""$prev_date"
fi
 
while read line1
do
line=`echo $line1 | grep -v Employee |  awk '{if($2=="M") print $4}'`
if [ ! -z "$line" ];then
op_da=`echo $line | awk -F/ '{print $2}'`
op_mo=`echo $line | awk -F/ '{print $1}'`
op_year=`echo $line | awk -F/ '{print $3}'`
 
if [ \( "$op_da" -gt "$da_dat" \) -a \( "$op_mo" -eq "$da_month" \) -a \( "$op_year" -eq "$da_year" \) ];then
 echo $line1 | awk '{print $1}'
fi
 
#If date is 1
if [ \( "$da_dat" -eq "1" \) -a \( \
 \( \( "$op_da" -eq "$prev_date" \) -a \( "$op_mo" -eq "$prev_mon" \) -a "$op_year" -eq "$da_year" \) -o \
\( \( "$op_da" -eq "1" \) -a  \( "$op_mo" -eq "$da_month" \) -a \( "$op_year" -eq "$da_year" \) \) \) ];then
 
 echo $line1 | awk '{print $1}'
fi
 
fi
done < inp6

Input: inp6
 
210125 M ABC 02/13/2012 08:07
451235 F EFG 02/012/2012 16:53
241567 M GHI 02/12/2012 14:03
314584 M JKL 02/02/2012 12:43
Output
210125
241567


Thanks,
Kalai
Thaaanks Kalai.. if you could please make comments on the code , so I can modify it.. the code is great but didnt work with me

---------- Post updated at 01:28 AM ---------- Previous update was at 01:27 AM ----------

Quote:
Originally Posted by durden_tyler
If Perl is an option, then -

Code:
$
$ cat myfile.txt
Employee  Gender    NAME      Last Login
   ID                                       Time
-------------------------------------------------
210125       M         ABC         02/03/2012 08:07
451235       F         EFG         02/02/2012 16:53
241567       M         GHI         02/01/2012 14:03
314584       M         JKL         02/02/2012 12:43
123456       M         MNO         02/12/2012 00:00
654321       F         PQR         02/12/2012 00:00
$
$ perl -lne 'BEGIN {use Time::Local; $prior = time-2*24*60*60}
            if (m|^(\d+)\s+([MF])\s+\w+\s+(\d+)/(\d+)/(\d+)\s+(\d+):(\d+)$|){
              ($id, $g, $mo, $d, $y, $h, $mi) = ($1, $2, $3, $4, $5, $6, $7);
              $y -= 1900; $mo--;
              $ts = timelocal(0,$mi,$h,$d,$mo,$y);
              print $id if $ts > $prior and $g eq "M";
            }' myfile.txt
123456
$
$

tyler_durden
thanks but unfortunately perl is not an option
# 7  
Old 02-14-2012
please provide me the sample input with more record
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