Remove extension in loop

Shell Programming and Scripting
1

Dear all
sorry for bothering you wityh dumb question but I am stucked with an issue.
Well, I am trying to loop over files in folder, make an operation and rewrite the output.
Example:

INPUT

file1.txt
file2.txt
file3.txt

My command (doesn't work!!)

for file in /path/to/*.txt
do command -input $file -output "${file%.*}"_sorted.txt
done

OUTPUT(desired!)

file1_sorted.txt
file2_sorted.txt
file3_sorted.txt

Any suggestion?

Best

Giuliano

2

In what way does it "doesn't work"? What do you actually get? The substitution appears to work here, in the BASH shell.

3

Yes sorry for misunderstanding..

Input

file1.txt
file2.txt
file3.txt

My command (doesn't work!!)

for file in /path/to/*.txt
do command -input $file -output "${file%.*}"_sorted.txt
done

My output

file1.txt_sorted.txt
file2.txt_sorted.txt
file3.txt_sorted.txt

desired output

file1_sorted.txt
file2_sorted.txt
file3_sorted.txt
4

What is your shell? What version, too?

5

/bin/bash
3.2.57(1)-release

6

Try ${file/.txt/}

7

no it doesn't work. It doesn't recognize the pattern.

8

Are you sure that it's interpreted by bash ? Or is it part of a script that has a #!sh shebang?

9

A shell that does not understand this would throw an error.
--
In case you have two or more extensions, e.g. file1.txt.txt , then you get

echo ${file%.*}
file1.txt

Perhaps you need

echo ${file%%.*}
file1
10

Sorry all my shell is

#!sh

Thanks made in Germany but your suggestion doesn't work!

11

Please run the script with the -x option set and post the output.

12

Really sorry guys the solution of MadeinGermany works fine...I had a my mistake!

Sorry for annoying you!

Thanks

Giuliano

1 Like
13

No trouble, though I'm left curious what the issue was :slight_smile:

14

my fault..

Usage is

command [options] <input> <ouput.prefix>

my command for this site was was

for file in /path/to/*.txt
do command -input $file -output "${file%.*}"_sorted.txt
done

My real command was

for file in /path/to/*.txt
do command $file -o "${file%.*}"_sorted.txt
done

That was because I though was more clear specify INPUT and OUTPUT for you.
BUT
My mistake was to think that

-o

was

output

when instead is

final output to stdout

well that was do not know how it matter..
hope that my awful explanation is clear.. better I go home now!

still sorry for my mistake next time i'll read carefully the options!

best

1 Like