shell - Unix : saving a sed command in a variable, to echo it, then execute it -


im new unix scripts, sorry mistake may :)

i execute "sed" command.
first echo command executed.
so, save command in variable.
then, execute command.

what did (simplyfied version of .sh script) :

command="sed -i 's/foo/bar/g' myfile"; echo "command: \"$command\"" ; $command ;

i got error : 

command: "sed -i 's/foo/bar/g' myfile" sed: -e expression #1, char 1: unknown command: `''

how should escape single quotes ? (or may use double quotes ?)

(what found on google on error did not help)

define convenience function echo given command, run it.

verbosely_do () {   printf 'command: %s\n' "$*";  # printf, not echo, because $@ might contain switches echo   "$@"; }  verbosely_do sed -i 's/foo/bar/g' myfile

this give you:

command: sed -i s/foo/bar/g myfile

and perform sed(1) command.


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