Re: [Help-bash] Transform strings with special characters so that the strings don't need to be quoted?

[Eric Blake]

On 03/25/2015 02:57 PM, Peng Yu wrote:
>> printf -v escaped_var %q "$var"

Fork-free.

> 
> I don't see this usage in the manual.

It's there in 'man bash':

       printf [-v var] format [arguments]
...
              %q     causes  printf  to output the corresponding
argument in a
                     format that can be reused as shell input.

and also in 'help printf'

      -v var    assign the output to shell variable VAR rather than
                display it on the standard output
...
      %q        quote the argument in a way that can be reused as shell input



> 
>> shquote() {
>>   local q=\'
>>   printf "'%s'" "${1//"$q"/$q\\$q$q}"
>> }
>> escaped_var=$(shquote "$var")

Requires a fork for the command substitution.

> 
> This is not robust with other characters that need to be escaped.

How so?  Single quote is the only character that needs to be
special-cased when using single quoting.  What characters are you
thinking of?

Or are you trying to escape a string to be suitable for use within ""
instead of within '', as was done here?  If so, then you have a bit more
work to do (\"$` are the four characters that need escaping), but it is
still doable.  But why bother, when single quoting is easier?

-- 
Eric Blake   eblake redhat com    +1-919-301-3266
Libvirt virtualization library http://libvirt.org

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