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Regex (regular expressions)
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Topic: Regex (regular expressions) (Read 6266 times)
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BlakPilar
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Regex (regular expressions)
«
on:
October 23, 2011, 02:19:00 pm »
Does anyone know regex well? I need to be able to select everything between a double quote and either another double quote, the end of a line, or a two character sequence ("->" for example). What I have now kind of works, except the highlighting only stops after the '>' and sometimes it selects everything for a while after the '>'.
Here's what it is now: (by the way, for the '->' part, I completely guessed lol)
Code:
[Select]
@"""[^>]+""|""[^>]+|""[^>]+'->'"
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Michael_Lee
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Re: Regex (regular expressions)
«
Reply #1 on:
October 23, 2011, 03:31:09 pm »
Try this?
Code:
[Select]
\"(.+?)(\"|$|->)
I tested it here:
http://regexpal.com/
but I'm not certain if there are any syntax differences between that and C# (which is what I think you're using?)
If you want only what's between the quotation marks/->/newline, capture only the
first matching group
(see the first answer).
«
Last Edit: October 23, 2011, 03:40:23 pm by Michael_Lee
»
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BlakPilar
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Re: Regex (regular expressions)
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Reply #2 on:
October 23, 2011, 03:39:52 pm »
Ahh, yes, that works. The only thing is it highlights '->' too... Oh well, good enough for now! Thank you!
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Michael_Lee
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Re: Regex (regular expressions)
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Reply #3 on:
October 23, 2011, 03:41:24 pm »
No problem.
Have you tried looking at the stackoverflow link and tried capturing the first matching group instead of capturing the entire regex? That might solve the highlighted '->' problem.
«
Last Edit: October 23, 2011, 03:41:45 pm by Michael_Lee
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BlakPilar
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Re: Regex (regular expressions)
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Reply #4 on:
October 23, 2011, 03:51:12 pm »
Oops, didn't see that part lol. I'll look at it and see what I can do, thanks again.
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C0deH4cker
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Re: Regex (regular expressions)
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Reply #5 on:
October 23, 2011, 04:54:00 pm »
can anyone point me to a really good regex tutorial that you KNOW is good?
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Deep Toaster
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Re: Regex (regular expressions)
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Reply #6 on:
October 23, 2011, 04:54:59 pm »
http://regular-expressions.info/
Everything is there, complete with examples.
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C0deH4cker
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Re: Regex (regular expressions)
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Reply #7 on:
October 23, 2011, 04:57:03 pm »
alright, thanks! +1
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BlakPilar
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Re: Regex (regular expressions)
«
Reply #8 on:
April 05, 2012, 12:00:44 pm »
So, I have another regex question... I looked at the tutorial and reference guide, but I can't seem to find out how to do what I want.
I want to be able to match everything between two single quotes ('...') but only if there is a single character between them, or a backslash then another character. Basically, C-style character declarations. This is what I have right now:
\'.+?\'
. It doesn't capture
''
which is what I wanted, but it captures anything that contains one or more character between the single quotes (I know it's because of the +'s "rule"). I mean, I suppose I could do something like
\'[a-zA-Z0-9 _\?!@#$%^&\*\(\)\[\]]\'
but that would be cheating and wouldn't capture everything...
«
Last Edit: April 05, 2012, 12:01:17 pm by BlakPilar
»
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cooliojazz
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Re: Regex (regular expressions)
«
Reply #9 on:
April 05, 2012, 04:47:05 pm »
How about /'([^\\]|\\.)'/ does that do what you want?
«
Last Edit: April 05, 2012, 04:47:17 pm by cooliojazz
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BlakPilar
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Re: Regex (regular expressions)
«
Reply #10 on:
April 05, 2012, 05:58:12 pm »
I'm using C# (so .NET) and I had to change that to
\'([^\\]|\\.)\'
, but it says there's an unterminated [] pair, which is weird... I tried changing it to
\'(.|\\.)\'
(I don't understand the need for a character class
). My test search string is
'gh' ' ' 'j' ';' ' fsdf' ''
and when I did the changed version my result was
' ', ' ', ' ', ' '
EDIT: Wow, nevermind... I added an @ before the edited regex thing and changed my string so there were commas between the character examples and it worked fine. Thank you!
«
Last Edit: April 05, 2012, 06:03:42 pm by BlakPilar
»
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cooliojazz
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Re: Regex (regular expressions)
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Reply #11 on:
April 05, 2012, 06:16:36 pm »
The character class is so it will capture \ characters, without it, it will return just \ if you say have '\a' instead of what it should, which is \a. The problem is that you probably didn't re-escape everything... =P if you wanted to actually put it in a string it would need to be "\'([^\\\\]|\\\\.)\'". (Do you really need to escape single quotes in strings defined with double quotes?)
EDIT, also I'm curious, what does the @ do?
«
Last Edit: April 05, 2012, 06:20:13 pm by cooliojazz
»
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