Bash script: how to insert an external variable into a compound command?

Question

bash

There is the following script:

#!/bin/bash
find ./ -type f -name '*.png' | xargs -P 8 -I {} sh -c 'cwebp -q 75 $1 -o "${1%.png}.webp"' _ {} \;

In this form it works fine, but if I try to add a parameter when calling the script (like this: scriptname 85), then everything breaks:

#!/bin/bash
quality=${1:-75}
find ./ -type f -name '*.png' | xargs -P 8 -I {} sh -c 'cwebp -q $quality $1 -o "${1%.png}.webp"' _ {} \;
'                                                                ^^^^^^^^                  '

How can the command be refactored so that the external parameter is inserted?

Answer 1

I apologize for answering myself, but after suffering for another hour, thanks to the Internet, I nevertheless solved the problem - maybe not too elegant, but the main thing that works:

#!/bin/bash
quality=${1:-75}
part1='cwebp -q '
part2=' $1 -o "${1%.png}.webp"'
cmd=$part1$quality$part2
find ./ -type f -name '*.png' | xargs -P 8 -I {} sh -c "$cmd" _ {} \;

And you can call the script either with or without a parameter - by default, the quality is set to 75.
For example, let's create an alias "webp-convert" for the script, then the call will look like this:

$ webp-convert # качество 75
$ webp-convert 90 # качество 90

Answer 2

double quotes - in single not intepol