.~s/ is just the replace operator. For example ~s/old/new/ replaces 'old' with 'new'.
Perl has these auto variables $1, $2 etc that correspond to brackets () you put in the reg exp. I guess what's missing in your code above is the $2. that's what's removing your Y.
(,
Sun 20 Apr 2008, 6:30,
archived)
Perl has these auto variables $1, $2 etc that correspond to brackets () you put in the reg exp. I guess what's missing in your code above is the $2. that's what's removing your Y.
Ok, thanksI really appreciate your help.
I don't know if I can adapt the code sample you've given me, but the rephrasing is really helpful, I'm much clearer now about what I need to achieve.
Thanks!
(I don't know if I can adapt the code sample you've given me, but the rephrasing is really helpful, I'm much clearer now about what I need to achieve.
Thanks!
mofaha ┐( ˘_˘)┌ ʅ(́◡◝)ʃ,
Sun 20 Apr 2008, 6:35,
archived)
*points below*
StickFigureNinja is worried he may be on the list,
Sun 20 Apr 2008, 6:38,
archived)
yeah. i came up with (on my own)newString = oldString.replace(/\.(?!\.|\s)/g, ". ");
seems to work. English goes: Replace a "." which is not followed by a "." or a " "
but you know. i'm no expert.
(seems to work. English goes: Replace a "." which is not followed by a "." or a " "
but you know. i'm no expert.
connor,
Sun 20 Apr 2008, 6:41,
archived)
Ahhh, i see, you used the (? ) lookahead optionwhich checks following characters without operating on them, very clever, why didnt I think of that.
COUGH
(COUGH
StickFigureNinja is worried he may be on the list,
Sun 20 Apr 2008, 6:43,
archived)