On 23/01/19 07:37, Alexander Kapshuk wrote: > On Wed, Jan 23, 2019 at 9:05 AM Paul Colquhoun > <paul...@andor.dropbear.id.au> wrote: >> >> On Wednesday, 23 January 2019 5:52:57 PM AEDT Alexander Kapshuk wrote: >>> On Wed, Jan 23, 2019 at 5:20 AM Adam Carter <adamcart...@gmail.com> wrote: >>>>>> François-Xavier >>>>> >>>>> My bad, it should be: >>>>> >>>>> sed 's/0*\([0-9][0-9]*\)/\1/g' >>>>> >>>>> (tests are indeed needed!) >>>> >>>> Many thanks François. This is almost right, but it is also stripping zeros >>>> that follow a letter, and I only want it to strip zeros that are >>>> proceeded by a period. There are no leading zeros in the first octet of >>>> the IP so that case does not need to be handled. >>>> >>>> Does the \1 refer to what's in the ()'s? So anything that one would wont >>>> to carry through should be inside the ()'s and anything that's outside is >>>> stripped, right? >>> Would something like to do the trick? >>> echo 198.088.062.01 | sed 's/\.0/./g' >>> 198.88.62.1 >> >> In a word, no. >> >> echo 198.088.0.01 | sed 's/\.0/./g' >> 198.88..1 >> >> >> -- >> Reverend Paul Colquhoun, ULC. http://andor.dropbear.id.au/ >> Asking for technical help in newsgroups? Read this first: >> http://catb.org/~esr/faqs/smart-questions.html#intro >> >> >> >> > > How about this one? > > echo '198.088.0.01 > 198.088.062.01' | sed 's/\.0\([0-9][0-9]*\)/.\1/g' > 198.88.0.1 > 198.88.62.1 >
I've just done a bit of digging, and would this work to match an octet? [0-9][0-9]?[0-9]? I know ? normally matches a single character, but apparently in this syntax it means "0 or 1 occurrence of the preceding expression". So that will detect a number consisting of at most three digits. I thought there must be a "detect a single optional character" operator ... :-) Cheers, Wol
