Confusion about bash command vs. variable substitution

Question

I've been writing one-line bash scripts on the command line since 1989. These usually have the form:

for name in 1 2 3; do wc $name.txt; done

Now I'm trying to do some image manipulation with GIMP, using a script that takes two filenames for arguments. Here's something that gives me a valid command line:

/Applications/GIMP.app/Contents/MacOS/GIMP -b '(script-fu-overlay "png0004.tif" "png0000.tif") (script-fu-overlay "png0004.tif" "png0001.tif") (script-fu-overlay "png0004.tif" "png0002.tif") (script-fu-overlay "png0004.tif" "png0003.tif")'

It was generated with the help of a typical one-liner:

X=\'$(for name in 0 1 2 3; do echo \(script-fu-overlay \"png0004.tif\" \"png000$name.tif\"\) ; done)\'

But...and here is my question:

echo /Applications/GIMP.app/Contents/MacOS/GIMP -b $X

gives me the exact command line above, but when I run

/Applications/GIMP.app/Contents/MacOS/GIMP -b $X

I get errors about not being able to open filenames that have quotes around them, but when I run

/Applications/GIMP.app/Contents/MacOS/GIMP -b "$X"

everything works. My real goal is to understand whether it's possible for me to write a one-liner without resorting to variables at all. Something like:

/Applications/GIMP.app/Contents/MacOS/GIMP -b \'$(for name in 0 1 2 3; do echo \(script-fu-overlay \"png0004.tif\" \"png000$name.tif\"\) ; done)\'

But that fails in apparently the same way that

/Applications/GIMP.app/Contents/MacOS/GIMP -b $X

fails.

Answer 1

Yes you can do it on the command line without resorting to shell variables.

/Applications/GIMP.app/Contents/MacOS/GIMP -b "$(for name in 0 1 2 3; do echo \(script-fu-overlay \"png0004.tif\" \"png000$name.tif\"\) ; done)"

Note that we need to do away with the escaped single quotes since we are not doing an eval here. Only thing that is needed are the double quotes around the command expansion $(...) to prevent word split from happening PLUS pass the result as one single argument to gimp.

Answer 2

What about this:

$ /Applications/GIMP.app/Contents/MacOS/GIMP -b \
"$(printf '%s\n' '(script-fu-overlay "png0004.tif" "png'{0..0004}'.png")')"

Answer 3

man bash says:

The order of expansions is: brace expansion; tilde expansion, parameter and variable expansion, arithmetic expansion, and command substitution (done in a left-to-right fashion); word splitting; and pathname expansion [...] After these expansions are performed, quote characters present in the original word are removed unless they have been quoted themselves (quote removal).

Thus, only quotes from the original word are removed. But those quotes of yours appear only after the expansion and they won't get removed.

In the given case, you can leave them away, as your file names don't need any quoting, but you'll run into problems with file names with whitespaces or other special characters. And neither quotes nor backslashes will help you as they are not considered to be quotes at this stage of processing.

Edit to serve the "one step" request:

But what if you have a bunch of files with whitespace like foo file foo and bar file bar and want to cat some of them?

Your command line would read

cat "foo file foo" "bar file bar"

but if there is more, you'll want to script it.

echo -n cat; for name in foo bar; do echo -n \ \"$name file $name\"; done

Will produce exactly the line you would enter, but

$(echo -n cat; for name in foo bar; do echo -n \ \"$name file $name\"; done)

will fail for the reason explained above. Enclosing everything in double quotes (as did help in your gimp case) will not work either, because the new command needs to be separated in words, but not within the quotes.

So what to do to make quoting work as if entered directly in the command line? Pass your string as command to the shell:

bash -c "$(echo -n cat; for name in foo bar; do echo -n \ \"$name file $name\"; done)"

That should be the most general approach for a one-liner executing a generated command.