editing the name of a variable in bash

Question

So basically i have a program that searches each file in a directory and creates a variable which consists of the name of the file and a number. The problem is that instead of creating the variable with the file's name and the number it also includes the extension of the file which i dont want.

Expected output:

./my_script.sh inputs outputs 1
InputFile=test1 NumThreads=1
OutputFile=test1-1.txt

My output:

./my_script.sh inputs outputs 1
InputFile=inputs/test1.txt NumThreads=1
OutputFile=inputs/test1.txt-1.txt

Program:

#!/bin/bash

Word1="${1}"
Word2="${2}"
Num="${3}"
for file in $(ls ${Word1}/*.txt)
do
    for i in $(seq 1 ${Num})
    do
        echo "InputFile="${file} "NumThreads="${i}
        out="${file}-${Num}".txt
        echo "OutputFile="${out}
    done
done

Please paste your script first at shellcheck.net and try to implement the recommendations made there. — Cyrus
– Cyrus, Commented Nov 15, 2020 at 20:56
I am not sure If I got you problem. For what ist Word2? Maybe this link helps to cut parts of your output: stackoverflow.com/questions/16623835/… — Maik
– Maik, Commented Nov 15, 2020 at 20:56

e-malito · Accepted Answer · 2020-11-15 21:11:16Z

3

Since you already know the file extension, you can extract the filebase directly using basename as follow:

filebase=$(basename -- "$file" .txt)

and then return

out="${filebase}-${Num}".txt

If you don't know the extension, you can extract it first like that:

extension=".${file##*.}"

So you can extract the filebase in one line as follows:

filebase=$(basename -- "$file" .${file##*.})

You can check out this answer for ways to extract the different path parts.

The mysterious ${} syntax which allows you to modify a variable is called shell parameter extensions.

answered Nov 15, 2020 at 21:11

e-malito

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1 Comment

glenn jackman Over a year ago

As you say, the extension is known, use the ${file%.txt} form to strip it off, instead of relying on basename.

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