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# 1  
Old 12-22-2011
Perl script
Hi All,
I have a question here in perl script
Suppose I have a date 12/22/2011(mm/dd/yyyy) what i have to do is to find
the start date of the week in which that date lies and end date of that week. In this scenario the start date will be 12/19/2011 and the end date will be 12/25/2011Smilie
# 2  
Old 12-22-2011
here is your code :-) (
p.s. i am a novice in shell scripting.. so i dont know what perl script is...
)

Code:
datee="12/22/2011"
len=${#datee}
mm=${datee:0:2}
dd=${datee:3:2}
yy=${datee:6:4}
mydate=$(date -d 12/22/2011)
day=$( echo $mydate | awk -F' ' '{print $1}' )
 
startdate=0
 
var1="Sun"
var2="Mon"
var3="Tue"
var4="Wed"
var5="Thu"
var6="Fri"
var7="Sat"
no1=0
no2=1
no3=2
no4=3
no5=4
no6=5
no7=6
 
for i in $(seq 1 7)
do
        val=$( eval eval echo \$var$i )
        num=$( eval eval echo \$no$i )
        if [[ "$val" == "$day" ]]
        then
                number=`expr $dd - $num + 1`
                echo "weekstart: $mm/$number/$yy"
                date -d $mm/$number/$yy
                number2=`expr $dd + 7 - $num`
                echo "weekend: $mm/$number2/$yy"
                date -d $mm/$number2/$yy
        fi
done

output:
Code:
weekstart: 12/19/2011
Mon Dec 19 00:00:00 EST 2011
weekend: 12/25/2011
Sun Dec 25 00:00:00 EST 2011

# 3  
Old 12-22-2011
The +%u format specifier to the date command returns a numeric value for the day of the week ...

Alternatively the Perl localtime function returns in an array context has the day of the week at index 6.
This User Gave Thanks to Skrynesaver For This Post:
vivek d r
# 4  
Old 12-22-2011
Code:
date +"%u"

gives current days numeric value of the week... if i want to find that of 12/19/2011 what would be the syntax?
# 5  
Old 12-22-2011
If u have -d option, u can use it
Code:
date -d '12/22/11' +'%u'

This User Gave Thanks to siva shankar For This Post:
vivek d r
# 6  
Old 12-22-2011
oh okay thanks :-)
# 7  
Old 12-22-2011
@vivek_d_r: What would your program print if input date is 12/01/2011 in mm/dd/yyyy????

@parthmittal2007: Here's the program you asked for:
perl code:
  1. #! /usr/bin/perl -w
  2. use strict;
  3. use Time::Local;
  4.  
  5. my ($mt, $dt, $yr, $sec, $wk_st, $wk_nd, $st_date, $nd_date);
  6.  
  7. (@ARGV != 1) && die "Invalid parameters. Enter date in mm/dd/yyyy format. Exiting";
  8. ($ARGV[0] !~ /^(0[1-9]|1[012])\/(0[1-9]|[12][0-9]|3[01])\/[\d]{4}$/)
  9. && die "Invalid date format. Enter date in mm/dd/yyyy. Exiting";
  10.  
  11. $mt = substr $ARGV[0], 0, 2;
  12. $dt = substr $ARGV[0], 3, 2;
  13. $yr = substr $ARGV[0], 6, 4;
  14.  
  15. $sec = timelocal (0, 0, 0, $dt, $mt - 1, $yr);
  16. $wk_st = $sec - (((localtime ($sec))[6] - 1) * 86400);
  17. $wk_nd = $sec + ((6 - ((localtime ($sec))[6] - 1)) * 86400);
  18.  
  19. $st_date = localtime ($wk_st);
  20. $nd_date = localtime ($wk_nd);
  21.  
  22. print "Week Start date: $st_date\n";
  23. print "Week End date: $nd_date\n";

Code:
[root@hostname test]# ./test.pl 12/01/2011
Week Start date: Mon Nov 28 00:00:00 2011
Week End date: Sun Dec  4 00:00:00 2011

Last edited by balajesuri; 12-22-2011 at 07:11 AM..
This User Gave Thanks to balajesuri For This Post:
parthmittal2007
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