How to pass every line of a file as options to a command?

Question

I'd like to write a script that reads a file and passes every line as options (or "option arguments") to a command, like this:

command -o "1st line" -o "2nd line" ... -o "last line" args

What's the simplest way of doing this?

Answer 1

# step 1, read the lines of the file into a shell array
mapfile -t lines < filename

# build up the command
cmd_ary=( command_name )
for elem in "${lines[@]}"; do
    cmd_ary+=( -o "$elem" )
done
cmd_ary+=( other args here )

# invoke the command
"${cmd_ary[@]}"

Answer 2

Here is one possibility:

$ cat tmp
1st line
2nd line
3rd line
4th line
$ command $(sed 's|.*|-o "&"|' tmp | tr '\n' ' ')

As glennjackman points out in the comments, word splitting can be circumvented by wrapping in eval, though the security implications of doing so should be appreciated:

$ eval "command $(sed 's|.*|-o "&"|' tmp | tr '\n' ' ')"

Edit: Combining my suggestion of using sed to assemble arguments with glenn jackman's mapfile/readarray approach gives the following concise form:

$ mapfile -t args < <(sed 's|.*|-o\n&|' tmp) && command "${args[@]}"

As a trivial demonstration, consider the aforementioned tmp file, the command grep, and the file text:

$ cat text
some text 1st line and
a 2nd nonmatching line
some more text 3rd line end
$ mapfile -t args < <(sed 's|.*|-e\n&|' tmp) && grep "${args[@]}" text
some text 1st line and
some more text 3rd line end
$ printf "%s\n" "${args[@]}"
-e
1st line
-e
2nd line
-e
3rd line
-e
4th line