echo that shell-escapes arguments [duplicate]
With bash, the printf builtin has an additional format specifier %q, which prints the corresponding argument in a friendly way:
In addition to the standard printf(1) formats,
%bcauses printf to expand backslash escape sequences in the corresponding argument (except that\cterminates output, backslashes in\',\", and\?are not removed, and octal escapes beginning with\0may contain up to four digits), and%qcauses printf to output the corresponding argument in a format that can be reused as shell input.
So you can do something like this:
printf %q "$VARIABLE"printf %q "$(my_command)"to get the contents of a variable or a command's output in a format which is safe to pass in as input again (i.e. spaces escaped). For example:
$ printf "%q\n" "foo bar"foo\ bar(I added a newline just so it'll be pretty in an interactive shell.)