Cannot assign output of sed to a variable

Question

I want to store this output of ls:

file1
file2
file3

in to a variable with value:

/../file1 /../file2 /../file3

This is my .sh file:

PREFIX="/../"
OUTPUT_INLINE=$(ls | tr "\n" " ")
OUTPUT="${OUTPUT_INLINE}" | sed "s|\<|$PREFIX|g"
echo "${OUTPUT_INLINE}"
echo "${OUTPUT}"

Output is:

file1 file2 file3

It means variable OUTPUT contains nothing.

Even if I do:

echo "${OUTPUT_INLINE}" | sed "s|\<|$PREFIX|g"

I will get:

/../file1 /../file2 /../file3

What is wrong here ?

Unix filenames can contain any character, including whitespace or newline. https://stackoverflow.com/questions/20796200/how-to-loop-over-files-in-directory-and-change-path-and-add-suffix-to-filename — G. C., Dec 15 '19 at 18:28

score 1 · Answer 1 · answered Dec 15 '19 at 12:12

OUTPUT="${OUTPUT_INLINE}" | sed "s|\<|$PREFIX|g"

That pipes OUTPUT="${OUTPUT_INLINE}" into sed "s|\<|$PREFIX|g", which doesn’t do anything. I think you meant:

OUTPUT=$(printf '%s' "${OUTPUT_INLINE}" | sed "s|\<|$PREFIX|g")

but there’s lots of fragility here around different delimiter types, and you should be able to avoid all that:

PREFIX="/../"

for filename in *; do
    printf '%s%s ' "$PREFIX" "$filename"
done

G. C. · Accepted Answer · 2019-12-15T18:08:34.280

1

You are assigning OUTPUT variable this command

"${OUTPUT_INLINE}" | sed "s|\<|$PREFIX|g"

Which means nothing. Do as you are already doing with OUTPUT_INLINE variable to assign the output of command.

OUTPUT=$(echo -n "${OUTPUT_INLINE}" | sed "s|\<|$PREFIX|g")

edited Dec 15 '19 at 18:08

answered Dec 15 '19 at 12:13

G. C.

2 Answers2