查找文件名大于的文件

Question

If I have 10 files called 01-a.txt, 02-a.txt,...10-a.txt - how can I find the files where the number is greater than 5? I would like a general solution, and I would be putting the contents of all files into one file using something like

cat *.txt > bigfile.txt

I can get files with numbers using

ls *[0-9]*

but can't seem go beyond that.

thanks.

Answer 1

You may use seq for that, but it works only if all files have same naming convention:

seq -f "%02g-a.txt" 6 10
06-a.txt
07-a.txt
08-a.txt
09-a.txt
10-a.txt

Ie:

cat `seq -f "%02g-a.txt" 6 10` > bigfile.txt

Answer 2

It will cat all files named as "< numeric_value >-< rest >" and having this < numeric_value > greater than $LIM .

Even if they are written with one single digit (like 5 ), with two digits (like 05 ), or more...

And even if the < rest > are different among the files.

LIM=5
for file in $(ls);
do
   number=$(echo $file | cut -f1 -d'-');
   [ $number -gt $LIM ] && cat $file >> bigfile.txt;
done

Answer 3

Assuming the folder contains only these files.

This would list all files where the number is > 5

ls [0-9]* | awk -F '-' '{if ($1 > 5) print $0}'

Answer 4

ls *.txt | perl -n -e '$f = $_; $f =~ s/\D//g; $f > 5 and print'