如何“释放” bash的`printf“％q”的输出

Question

In bash, you can use printf "%q" to escape the special characters in a string. I've added a line break in the following examples for clarity's sake:

$ printf "%q\n" "foo"
foo
$ printf "%q\n" 'foo$bar'
foo\$bar
$ printf "%q\n" "foo    bar"  # Tab entered with Ctrl+V Tab
$'foo\tbar'

You can supply the -v option to printf to stick the output into a variable, rather than echoing to stdout.

Now what if I want to echo the original, unescaped string back to stdout? If I just do a simple echo , it includes all the meta/control characters; echo -e gets me slightly further, but not to a fully unescaped state.

Answer 1

Use printf and eval to "unescape" the string (it is perfectly safe, since it is quoted by bash already). echo is, in general, dangerous with arbitrary user input.

eval printf '%s\\n' "$var"

The argument to printf is %s\\\\n with double backslashes even within the single quotes, since eval will take it as one argument and then strip it (evaluate it) as is, so it effectively becomes something like: printf %s\\\\n $'12\\n34 56' which is a perfectly safe and valid command (the variable even includes a newline and space, as you can see).