Re: string escaping in bash

[Peng Yu]

I don't think the chained solution will work robustly. Here is what I
figured out using a character by character-based approach. It should
be correct regarding all input.

But I am not sure whether its performance is the best.

        function tabnlunesc {
                declare __tabnlunesc__i
                declare -a __tabnlunesc__r
                
for((__tabnlunesc__i=0;__tabnlunesc__i<${#1}-1;++__tabnlunesc__i)); do
                        if [[ ${1:__tabnlunesc__i:1} = \\ ]]; then
                                if [[ ${1:__tabnlunesc__i+1:1} = n ]]; then
                                        __tabnlunesc__r[__tabnlunesc__i]=$'\n'
                                        ((++__tabnlunesc__i))
                                        continue
                                elif [[ ${1:__tabnlunesc__i+1:1} = t ]]; then
                                        __tabnlunesc__r[__tabnlunesc__i]=$'\t'
                                        ((++__tabnlunesc__i))
                                        continue
                                elif [[ ${1:__tabnlunesc__i+1:1} = \\ ]]; then
                                        __tabnlunesc__r[__tabnlunesc__i]='\'
                                        ((++__tabnlunesc__i))
                                        continue
                                fi
                        fi
                        __tabnlunesc__r[__tabnlunesc__i]=${1:__tabnlunesc__i:1}
                done
                ((__tabnlunesc__i==${#1})) || eval
__tabnlunesc__r[__tabnlunesc__i]=${1:__tabnlunesc__i:1}
                builtin printf %s "${__tabnlunesc__r[@]}"
        }

On 3/12/21, Alex fxmbsw7 Ratchev <fxmbsw7@gmail.com> wrote:
> i figured as much :
>
> str=a\\ttab\ and\ a\ newline\\nand\ \\backslashes\\$'\n'
> str=${str//\\\\/$'\1'} str=${str//\\n/$'\n'} str=${str//\\t/$'\t'}
> str=${str//$'\1'/\\} ; printf %s "$str"
> a tab and a newline
> and \backslashes\
>
> On Fri, Mar 12, 2021 at 5:40 PM Alex fxmbsw7 Ratchev <fxmbsw7@gmail.com>
> wrote:
>
>> i see
>> well manual chaining then
>>
>> var=b\\c\\tab\\t\\bnewline\\n\\\\n var=${var//\\n/$'\n'}
>> var=${var//\\t/$'\t'} vvar=${var//\\/\\\\} ; printf %s "$var"
>>
>> but that doesnt cover \\ backslashes yet, eg \\n for \n liteeral
>> no idea of your needs, maybe you can describe in more detail
>>
>> On Fri, Mar 12, 2021, 17:07 Peng Yu <pengyu.ut@gmail.com> wrote:
>>
>>> No. `printf %b '\a'` prints a bell character. But I still want a slash
>>> and the character "a". Basically, I only want to treat \\ \n and \t
>>> specially. All others should be treated literally.
>>>
>>> On 3/12/21, Alex fxmbsw7 Ratchev <fxmbsw7@gmail.com> wrote:
>>> > you may be looking for
>>> > # printf %b "$str"
>>> > to interpret the string by prrintf
>>> >
>>> > or mass chained singular statements, like you showed or similiar
>>> >
>>> > On Fri, Mar 12, 2021, 16:05 Peng Yu <pengyu.ut@gmail.com> wrote:
>>> >
>>> >> Hi,
>>> >>
>>> >> 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
>>> >> replaced as a newline character, the string "\t" (a slash and the
>>> >> letter "t") should be replaced as a tab character, and "\\" (two
>>> >> consecutive slashes should be replaced with a single slash. All other
>>> >> characters and their preceding slash (if there is) should remain as
>>> >> is.
>>> >>
>>> >> If I use a multi-string-replacement strategy, it will not be robust.
>>> >> For example, if I do it in the order 1) \\ -> \, 2) \n -> newline,
>>> >> \\n
>>> >> will not be replaced correctly. The result should be "\n" (a slash
>>> >> followed by the letter "n").
>>> >>
>>> >> $ x='\\n'; x=${x//\\\\/\\}; x=${x//\\n/$'\n'}; declare -p x
>>> >> declare -- x="
>>> >> "
>>> >>
>>> >> Does anybody have a robust way to implement this in bash?
>>> >>
>>> >> --
>>> >> Regards,
>>> >> Peng
>>> >>
>>> >>
>>> >
>>>
>>>
>>> --
>>> Regards,
>>> Peng
>>>
>>
>


-- 
Regards,
Peng