How to use awk's output on bash command line

Question

I have a file named "01 - Welcome To The Jungle.mp3", and I want to do eyeD3 -t "Welcome To the Jungle" 01 - Welcome To The Jungle.mp3 to modify the tag of the all the files in the folder. I've extracted from the file with awk: "Welcome To The Jungle" doing:

#!/bin/bash
for i in *.mp3
do
eyeD3 -t $(echo ${i} | awk -F' - ' '{print $2}' | awk -F'.' '{print $1}') ${i}
done

It doesn't work. Neither the whole "$(echo S{i}....)" nor the "${i}" seem to work for replacing the names of the respective files.

heemayl · Accepted Answer · 2016-11-06 05:29:37Z

1

You need to prevent word splitting on IFS (default: space, tab, newline) by shell, as your input filename contains space(s). The typical workaround is to use double quotes around the variable expansion.

Do:

for i in *.mp3; do eyeD3 -t "$(echo "$i")" | ...; done

You can leverage here string, <<<, to avoid the echo-ing:

for i in *.mp3; do eyeD3 -t <<<"$i" | ...; done

answered Nov 6, 2016 at 5:29

heemayl

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janos · Accepted Answer · 2016-11-06 07:41:07Z

0

You need to double-quote variables that may contain spaces, as @heemayl already pointed out.

Also, in this example, instead of using awk, it would be better to use native Bash pattern substitution to extract the title, for example:

for file in *.mp3; do
    title=${file%.mp3}
    title=${title#?????}
    eyeD3 -t "$title" "$file"
done

That is:

Remove .mp3 at the end
Remove the first 5 characters (the count prefix NN -)

answered Nov 6, 2016 at 7:41

janos

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2 Comments

JPJ Over a year ago

Thank you so much @janos! I didn't even know you could do that! And it looks so much more intuitive than the awk command!

JPJ Over a year ago

No, Thank you. I think I understand it now.

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