Re: string escaping in bash

[Pier Paolo Grassi]

In zsh this could be solved like this:

local -A conversions=(n $'\n' t $'\t' \\ \\)
example='\a\n\c\\\t\e'
echo -E - ${example//(#b)\\(?)/${conversions[$match[1]]:-\\$match[1]}}

but I don't know if this can be expressed with bash syntax, maybe someone
can help me translate it?

Pier Paolo Grassi


Il giorno ven 12 mar 2021 alle ore 22:48 Alex fxmbsw7 Ratchev <
fxmbsw7@gmail.com> ha scritto:

> i think i cannot succeed
> the most was the v=nonexist// version
>
> thank you
>
> On Fri, Mar 12, 2021 at 10:41 PM Alex fxmbsw7 Ratchev <fxmbsw7@gmail.com>
> wrote:
>
> > thank you
> >
> > i may post a new version soon whenever i succeed it, i just gotta think
> > more about it
> >
> > thank you again
> >
> > On Fri, Mar 12, 2021 at 10:37 PM Lawrence Velázquez <vq@larryv.me>
> wrote:
> >
> >> > On Mar 12, 2021, at 3:25 PM, Alex fxmbsw7 Ratchev <fxmbsw7@gmail.com>
> >> wrote:
> >> >
> >> > this is without keep over and does the stupid non edge cases
> >> > maybe you can give me better test cases ?
> >> >
> >> > i=-1 res=( ) list=( '\one' '\\Two' '\\\Three' '\\\\four' '\\\\\five' )
> >> > right=( '\one' '\Two' '\\Three' '\\four' '\\\five' ) ; for str in
> >> > "${list[@]}" ; do str=${str//\\n/$'\n'} str=${str//\\t/$'\t'}
> >> > res[++i]=${str//\\\\/\\} ; printf '%d %s %s\n' $i "${res[i]}"
> >> "${right[i]}"
> >> > ; done
> >>
> >>         bash-5.1$ str='\\n'
> >>         bash-5.1$ printf '%s' "$str" | od -A n -t a -t x1
> >>                    \   \   n
> >>                    5c  5c  6e
> >>
> >> This should become '\n' (0x5c 0x6e).
> >>
> >>         bash-5.1$ str=${str//\\n/$'\n'}
> >>         bash-5.1$ str=${str//\\t/$'\t'}
> >>         bash-5.1$ str=${str//\\\\/\\}
> >>         bash-5.1$ printf '%s' "$str" | od -A n -t a -t x1
> >>                    \  nl
> >>                    5c  0a
> >>
> >> --
> >> vq
> >>
> >>
>