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# 1  
Old 08-02-2008
hard question

I have a directory containing a series of files of the format:
A2008001231000.L2

I only care about the 6-8 digits, so the files are effectively:
?????---*.L2

I have files that range from ?????001*.L2 to ?????366*.L2

It should be noted these three digits represent the julian day of the files. My goal is to be able to define two variables: start_day and end_day. How can I list off all the files between start_day and end_day. NOTE: Since the julian days must be 3 digits long, it complicates matters.
# 2  
Old 08-02-2008
If you think this is hard, you must be using the wrong language. Here it is in ksh. What you call "julian" I will call "day of year". Look up "julian day" on wikipedia. This script uses my datecalc script to allow the user to enter the endpoints as yyyymmdd. datecalc is posted to this site and our search function can find it for you.

One thing about ksh... a number with a leading zero is assumed to be octal. This is why I keep stripping leading zeros off numbers before doing arithmetic.

Code:
$ ls -1 files
A2008001231000.L2
A2008021231000.L2
A2008041231000.L2
A2008061231000.L2
A2008081231000.L2
A2008101231000.L2
$
$
$
$ ./findit
enter start yyyymmdd - 20080101
enter end yyyymmdd - 20080220
A2008001231000.L2 is in range
A2008021231000.L2 is in range
A2008041231000.L2 is in range
$
$
$
$ cat findit
#! /usr/bin/ksh

read start?"enter start yyyymmdd - "
y1=${start%????}
d1=${start#??????}
temp=${start%$d1}
m1=${temp#$y1}
m1=${m1#0}
d1=${d1#0}
doy1=$(($(datecalc -a $y1 $m1 $d1 - $y1 1 1) + 1))
#echo $start $y1 $m1 $d1 $doy1

read end?"enter end yyyymmdd - "
y2=${end%????}
d2=${end#??????}
temp=${end%$d2}
m2=${temp#$y2}
m2=${m2#0}
d2=${d2#0}
doy2=$(($(datecalc -a $y2 $m2 $d2 - $y2 1 1) + 1))
#echo $start $y2 $m2 $d2 $doy2

cd files
ls | while read name ; do
        temp=${name#?}
        temp1=${temp#????}
        f_year=${temp%$temp1}
        temp2=${temp1#???}
        f_doy=${temp1%$temp2}
        f_doy=${f_doy##*(0)}
#       echo $name $f_year $f_doy
        if ((y1<=f_year && doy1<=f_doy && f_year<=y2 && f_doy<=doy2)) ; then
                echo $name is in range
        fi
done
exit 0
$

# 3  
Old 08-02-2008
I see you're in mktime() paradise.
From the man mktime:
Code:
           struct tm {
               int tm_sec;         /* seconds */
               int tm_min;         /* minutes */
               int tm_hour;        /* hours */
               int tm_mday;        /* day of the month */
               int tm_mon;         /* month */
               int tm_year;        /* year */
               int tm_wday;        /* day of the week */
               int tm_yday;        /* day in the year */
               int tm_isdst;       /* daylight saving time */
           };

 tm_yday
              The number of days since January 1, in the range 0 to 365.

Well, I couldn't get mktime() to work directly, but I still used tm_yday.
Code:
Tsunami julian_days # perl test2.pl 
Enter start date [yyyy/mm/dd]: 20080101
Enter end date [yyyy/mm/dd]: 20080220
A2008041231000.L2
A2008021231000.L2
A2008001231000.L2
Tsunami julian_days # perl test2.pl 
Enter start date [yyyy/mm/dd]: 2008/01/01  
Enter end date [yyyy/mm/dd]: 2008/02/20
A2008041231000.L2
A2008021231000.L2
A2008001231000.L2
Tsunami julian_days #

and the code:

Code:
#!/usr/bin/perl -w

use POSIX;

$| = 1;

my $input;
my (@date_s, @date_e);

opendir(DIR, ".") || die "can't opendir: $!";
@dots = grep { ! /^\./ && /\.L2$/ } readdir(DIR);
closedir DIR;

print "Enter start date [yyyy/mm/dd]: ";
($input = <STDIN>) =~ /(\d{4})\/*-*(\d{2})\/*-*(\d{2})/;
$date_s[0] = mktime (0, 0, 0, $3, ($2-1), ($1-1900), 0, 0);
$date_s[2] = $1;
print "Enter end date [yyyy/mm/dd]: ";
($input = <STDIN>) =~ /(\d{4})\/*-*(\d{2})\/*-*(\d{2})/;
$date_e[0] = mktime (0, 0, 0, $3, ($2-1), ($1-1900), 0, 0);
$date_e[2] = $1;
(undef, undef, undef, undef, undef, undef, undef, $date_s[1] ,undef) = localtime($date_s[0]);
(undef, undef, undef, undef, undef, undef, undef, $date_e[1] ,undef) = localtime($date_e[0]);

foreach my $i (@dots)
{
        $i =~ /^A(\d{4})(\d{3})/;
        print $i . "\n" if (($1 >= $date_s[2] && $1 <= $date_e[2]) && (($2+0) >= $date_s[1] && ($2+0) <= $date_e[1]));  
}

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