How do I use super-sed's Perl regex dot match? - regex

Question

I'm trying to use super-sed's Perl regex /S, but can't get it to work at all. This flag makes dots match newlines. This would be a very handy tool, if only I could understand how it's used! For example, I expect the following command will match and replace the pattern which spans across a newline to be replaced with Xs:
echo "[phone]\n(212) 567-9050" | ssed -R -e "s/78.*?5/x/S"
So, I am expecting this output:
(123) 456-XXXX
XXXXXXX67-9050
Instead I get (no match):
[phone]
[phone]

Answer 1

Ssed, like sed, works in a line-based manner. If you want to work on multiple lines at the same time, you have to fetch them first. One way to do that in sed (and ssed) is
:a $! { N; ba; }
Where :a is a jump label, N fetches the next line, ba jumps back to :a and the $! check sees to it that this only happens as long as there are more lines to read.
Once we have that, the other difficulty is to get the right number of Xs into the right places. Ssed, like sed, does not make this very convenient, and it requires some shuffling around with the hold buffer to get the substituted part isolated and ready for processing. I came up with the following:
$ ssed -R ':a $! { N; ba; }; h; s/(.*?78)(.*?5)(.*)/\2/S; s/./X/g; s/^/#/; x; G; s/(.*?78)(.*?5)(.*)\n#(.*)/\1\4\3/S' << EOF
> [phone]
> [phone]
> EOF
(123) 456-78XX
XXXXXXX67-9050
This works as follows:
:a $! { N; ba; } # read full input into pattern space
h # save a copy of it in the hold buffer
s/(.*?78)(.*?5)(.*)/\2/S # isolate the part to substitute
s/./X/g # replace non-newlines with X
s/^/#/ # Put an # as marker before the X's.
x # Swap hold buffer and pattern space
G # append hold buffer (now the X's) to
# the pattern space. The PS now contains
# the input followed by an # followed by
# the X's.
s/(.*?78)(.*?5)(.*)\n#(.*)/\1\4\3/S # Use the # marker (that we know to be
# the last # in the PS) to isolate the
# X's and the original regex to isolate
# the part we want to replace, then
# reassemble.
As you can see, this is about as messy in ssed as it would be in sed, so I still suggest that it might be saner to use Perl:
$ perl -0777 -pe 's/(?<=78)(.*?5)/$1=~s{[^\n]}{X}gr/se' << EOF
> [phone]
> [phone]
> EOF
(123) 456-78XX
XXXXXXX67-9050
Here, the -0777 option puts perl into slurp mode, which makes it read the whole input in one go rather than linewise, and the code is a simple substitution, where
(?<=78) is a lookbehind expression that matches an empty string if it is preceded by 78
/e enables us to use a perl expression in the replacement clause of s///, and
$1=~s{[^\n]}{X}gr takes the first capture and replaces all non-newline charaters in it with X, yielding the result of the substitution. This is then substituted into the string where (.*?5) was matched.

Answer 2

Noooo!!!! It's bad enough people are using sed for all sorts of wacky machinations but now there's super-sed for even more crazy rune combinations???
You don't tell us what sseds /S command does so I'm guessing it's for doing substitutions across multi-line blocks but sed is for simple substitutions on individual lines, that is all, and you should forget you ever heard about super-sed. For anything interesting related to manipulating text you should just use awk, e.g. with GNU awk for multi-char RS:
$ printf "[phone]\n(212) 567-9050\n" |
awk -v RS='78[^5]*5' -v ORS= '{print $0 gensub(/[^\n]/,"X","g",RT)}'
(123) 456-XXXX
XXXXXXX67-9050
or if you didn't want the 78 to be replaced:
$ printf "[phone]\n(212) 567-9050\n" |
awk -v RS='78[^5]*5' -v ORS= '{print $0 substr(RT,1,2) gensub(/[^\n]/,"X","g",substr(RT,3))}'
(123) 456-78XX
XXXXXXX67-9050
or:
$ printf "[phone]\n(212) 567-9050\n" |
awk -v RS='^$' -v ORS= 'match($0,/(.*78)([^5]*5)(.*)/,a){print a[1] gensub(/[^\n]/,"X","g",a[2]) a[3]}'
(123) 456-78XX
XXXXXXX67-9050
and if you don't like that for some reason then just use perl, it's got to be every bit as readily available as ssed, probably more so!