Extract part of a filename shell script

Question

In bash I would like to extract part of many filenames and save that output to another file.

The files are formatted as coffee_{SOME NUMBERS I WANT}.freqdist.

#!/bin/sh
for f in $(find . -name 'coffee*.freqdist)

That code will find all the coffee_{SOME NUMBERS I WANT}.freqdist file. Now, how do I make an array containing just {SOME NUMBERS I WANT} and write that to file?

I know that to write to file one would end the line with the following.

  > log.txt

I'm missing the middle part though of how to filter the list of filenames.

Answer 1

You can do it natively in bash as follows:

filename=coffee_1234.freqdist
tmp=${filename#*_}
num=${tmp%.*}
echo "$num"

This is a pure bash solution. No external commands (like sed) are involved, so this is faster.

Append these numbers to a file using:

echo "$num" >> file

(You will need to delete/clear the file before you start your loop.)

Answer 2

If the intention is just to write the numbers to a file, you do not need find command:

ls coffee*.freqdist
coffee112.freqdist  coffee12.freqdist  coffee234.freqdist

The below should do it which can then be re-directed to a file:

$ ls coffee*.freqdist | sed 's/coffee\(.*\)\.freqdist/\1/'
112
12
234

Guru.