Re: string escaping in bash

[Peng Yu]

On Sat, Mar 13, 2021 at 6:18 AM Koichi Murase <myoga.murase@gmail.com>
wrote:

> 2021年3月12日(金) 23:06 Peng Yu <pengyu.ut@gmail.com>:
> > I wondering if there is a simple but robust way to implement string
> escaping.
> >
> > Specifically, the string "\n" (a slash and the letter "n") should be
> [...] All other
> > characters and their preceding slash (if there is) should remain as
> > is.
>
> I think this is `unescaping' rather than `escaping'.


You are right. It should be unescape. My function name reflects this. But
the original question was wrong in this aspect.

>
>
> > Does anybody have a robust way to implement this in bash?
>
> function unescape-helpbash202103-47 {
>   local s=$1
>   s=${s//z/z0}


This idea is great. It can be a general approach to solve the problem of
custom unescaping without the support of regex replacement.

>
>   s=${s//'\\'/z1}
>   s=${s//'\n'/$'\n'}
>   s=${s//'\t'/$'\t'}
>   s=${s//z1/'\'}
>   s=${s//z0/z}
>   result=$s
> }
> unescape-helpbash202103-47 '\\n'; x=$result
> declare -p x | cat -v
>
> Result:
>
> declare -- x="\\n"
>
> --
> Koichi
>
-- 
Regards,
Peng