Need to pick Latest File From a Dir Using Shell Script - bash

Question

I am new to Shell Script and I got a requirement to pick the latest files from a dir using Shell script
Directory Name : FTPDIR
File In this Dir will be of
APC5502015VP072020121826.csv
APC5502015VP082020122314.csv
APC5502015VP092020121451.csv
CBC5502015VP092020122045.csv
CBC5502015VP102020122045.csv
S5502015VP072020121620.csv
S5502015VP072020122314.csv
S5502015VP092020122045.csv
Note: (Need to Pick one Latest from each Group)- Below is the out put which I need to get after executing the shell script
APC5502015VP092020121451.csv
CBC5502015VP102020122045.csv
S5502015VP092020122045.csv
Ex: In the latest File APC5502015VP092020121451.csv the no 092020121451 is the date part in the format : MMDDYYYYHHMM and string part is APC5502015VP (Length Not Fixed in String Part)
I need to pick those three files from the dir using shell script
Can you help me to resolve this?

Answer 1

It's going to be really problematic to do this safely in just bash. As Jonathan mentioned, "special" characters like spaces or newlines may bung up your script.
If we can assume that there won't be any of those, then we can do most of job in bash, without involving other tools.
# Make an associative array to record types, in the second loop...
declare -A a
for file in *.csv; do
# First, we convert the filenames into something that can be sorted.
# The next three lines account for your "unknown length" in the first part
# of the filename. We assume the date+time is the 12 chars before ".csv".
new="$(rev <<<"$file")"
new="${new:4:12}"
new="$(rev <<<"$new")"
new="${new:4:4}${new:0:2}${new:2:2}${new:8:4}"
len=$(( ${#file} - 16 ))
echo "$new ${file:0:$len} $file"
done | sort | while read date type file; do
# Next, we print only the first of each "type"...
if [[ ${a[$type]} -eq 0 ]]; then
a[$type]=1
echo "$file"
fi
# And stop once we have collected three types.
if [[ ${#a[*]} -ge 3 ]]; then
break
fi
done
As I say, this doesn't handle newlines in filenames.
Note also that this uses rev and sort, which are not built in to bash. The rev parts could be done internally, using more code, which might make them execute faster, but you'd only see a difference in very extreme cases. There's not much we can do about sort, since there isn't a built-in within bash.

Answer 2

This Perl script works on the given data. No doubt it could be improved.
#!/usr/bin/env perl
use strict;
use warnings;
my %bases;
while (<>)
{
chomp;
my $name = $_;
my($prefix, $mmdd, $yyyy, $hhmm) = ($name =~ m/(.*)(\d{4})(\d{4})(\d{4})\.csv/);
#print "$name = $prefix $yyyy $mmdd $hhmm\n";
my $stamp = "$yyyy$mmdd$hhmm";
if (!exists($bases{$prefix}) || ($stamp > $bases{$prefix}->{stamp}))
{
$bases{$prefix} = { name => $name, stamp => $stamp };
}
}
foreach my $prefix (sort keys %bases)
{
print "$bases{$prefix}->{name}\n";
}
Output:
APC5502015VP092020121451.csv
CBC5502015VP102020122045.csv
S5502015VP092020122045.csv

Answer 3

this is the awk solution:
cd FTPDIR
ls -1|awk -F"VP" '{split($2,a,".");if(a[1]>b[$1]){b[$1]=$2}}END{for(i in b)print i"VP"b[i]}'
Testted Below:
> cat temp
APC5502015VP072020121826.csv
APC5502015VP082020122314.csv
APC5502015VP092020121451.csv
CBC5502015VP092020122045.csv
CBC5502015VP102020122045.csv
S5502015VP072020121620.csv
S5502015VP072020122314.csv
S5502015VP092020122045.csv
> awk -F"VP" '{split($2,a,".");if(a[1]>b[$1]){b[$1]=$2}}END{for(i in b)print i"VP"b[i]}' temp
CBC5502015VP102020122045.csv
S5502015VP092020122045.csv
APC5502015VP092020121451.csv