setting variable from string in a file using sed and regex

Question

UPDATE Edited the script slightly for a more specific regex, still same problem.

Clarifying following questions, this is my exact script I want to get working (POC for a bigger script), test.txt is in the same directory as my test.sh file and I'm running the script cd'ed in the same directory.

Also, whilst tinkering, I think this might be related to carriage returns, but I still can't figure out what's going wrong.

END UPDATE

I've looked around here to paste together a bash file I can run to extract a version number from a specific file, and do stuff with that version number via the variable. I have the rest of the script working if I just supply the version via an argument $1, but it'd be neat if it could just determine the version number directly from the file in the folder I want to manipulate.

Here's my current script

#!/bin/sh version=$(sed -ne "s/\\Version *: *\([0-9a-zA-Z\.\-]+\)*/\1/p" test.txt) printf 'detected version %s for test.txt\n' "$version" 

Which returns me this value in terminal

 for test.txtion 0.6 

It seems to overwrite part of the string I'm trying to print. If I just take this part:

sed -ne "s/\\Version *: *\([0-9a-zA-Z\.\-]+\)*/\1/p" test.txt 

It works for example if there's a line in the file containing this line

Version: 0.6 

I get this:

root@webserver [/home/username]# sed -ne "s/\\Version *: *\([0-9a-zA-Z\.\-]+\)*/\1/p" test.txt 0.6 root@webserver [/home/username]# 

Any ideas on what's wrong for the script?

Answer 1

I found the reason this wasn't working, and also the solution. As @eike commented in my OP, the issue was the carriage return \r from DOS overwriting the output. More problematically for me, the \r got added to my variable for some reason.

Removing the \r manually wasn't an option because that file gets updated regularly.

After a bit of searching, I found this to fix my problem and print out the string correctly:

#!/bin/bash version=$(sed 's/\r$//' test.txt | sed -ne "s/\\Version *: *\([0-9a-zA-Z\.\-]+\)*/\1/p") printf 'detected version %s for test.txt\n' "$version" 

What it does here is first remove the \r from the text and pass that to the regex part. and I get the expected output:

detected version 0.6 for test.txt

I'm sure there's a way to do it from within the regex, to ignore the \r, but this got the job done at least!