filter and get the latest order number

Shell Programming and Scripting
1

hello,

how can I filter and get the latest order number (last five digits) below:

input file:
johnmm00001
maryyy00121
johnm100222
johnmm00003
maryyy00122

output file:
johnmm00003
maryyy00122
johnm100222

2

Whats the criteria to get the latest order number ?

3

the bigger number of last 5 digits (which is the invoice number for customer)..and I just need to get latest one...for a month

4

Input File :
cat /tmp/s
johnmm00001
maryyy00121
johnm100222
johnmm00003
maryyy00122

sort -k1.7n,12n /tmp/s
Output :
johnmm00001
johnmm00003
maryyy00121
maryyy00122
johnm100222

Assumptions :There are only 6 alphabets before the numeric digits.

Please let me know if this works. Experts comment on this

5

yes...but I need to delete all old invoice number from the same customer:

expect output:
johnmm00003
maryyy00122
johnm100222

6

happy.pl

#!/usr/bin/perl
my $no=1;
my $str;
while( <> )
{
        chomp;
        if(/([0-9]{5})$/)
        {
                $hsh{ $_ }=$& ;
        }
}
foreach $char(sort { $hsh{$b} <=> $hsh{$a} } keys %hsh )
{
      $char =~ /([0-9]{5})$/ ;
      if( $str ne "$`" ) {
      $str=$`;
      print "$char\n" if( $no++ <= 3);
      }
}

$ happy.pl file
johnm100222
maryyy00122
johnmm00003

Replace 3 in above code by the number of invoices you require

7 
awk '{name = value = $0
      gsub(/[0-9]/,"",name)
      gsub(/[a-zA-Z]/,"",value)
      if ( value > xv[name] ) x[name] = $0; xv[name]=value }
     END { for ( name in x ) print x[name] }' ~/txt