CODEWITHSUNDEEP
bash
user2416047
user2416047

Reputation: 100

Trying to use a variable in Single Quotes Bash

I'm trying to use the variable $y in this snippet below but keep running into the filename showing up as $y.pdf

ls -v *.jpg | tr '\n' ' ' | sed 's/$/\ $y.pdf/' | xargs convert

I've also tried the below:

ls -v *.jpg | tr '\n' ' ' | sed 's/$/\'"$y.pdf"'/' | xargs convert

ls -v *.jpg | tr '\n' ' ' | sed 's/$/\ {$y}.pdf/' | xargs convert

The top one usually fails, as sed is expecting something else. The bottom one just makes my filename {$y}.pdf.

Any ideas for me?

Thanks!

Upvotes: 0

Views: 53

Answers (1)

chepner
chepner

Reputation: 532418

Never mind the quoting problem; this is the wrong approach entirely. Just use a for loop to avoid all kinds of potential problems.

for f in *.jpg; do
    convert "$f" "${f%.jpg}.pdf"
done

${f%.jpg} expands to the name of the current file minus the .jpg extension.

To merge the files into one PDF is even simpler (I think):

convert *.jpg "$y.pdf"

Assuming ls -v outputs the file names in the correct order, and there aren't any of the usual concerns with parsing ls involved, use

ls -v *.jpg > input.txt
convert @input.txt "$y.pdf"

(There might be a way to avoid the use of a tempfile, maybe as simple as ls -v *.jpg | convert @- "$y.pdf", but I'm too lazy to figure out all the ways convert can be called.)

Upvotes: 1

Related Questions