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Regex to match only till first occurence of class match
You were missing ? Your regex would be (?i)(.*?)case[^a-zd]*(d+)(.*) You can toggle case insensitive match using (?i) in regex

Categories : Regex

Match last occurence with regex
You can use in greedy quantifier with a reduced character class (assuming you have no tags between you <br>): <br>([^<]*)<br>s*$ or <br>((?:[^<]+|<(?!br>))*)<br>s*$ to allow tags inside. Since the string you search is Tizi Ouzou without <br> you can extract the first capturing group.

Categories : Python

Get first occurence of match in Regex
A way to do that is to use a lazy quantifier with dotall option: Regex regex = new Regex(@"^.*?(?>dog|mouse)"); Another way is to do that; Regex regex = new Regex(@"^(?>[^dm]*+|d++(?!og)|m++(?!ouse))*(?>dog|mouse)"); it is longer but more efficient. The idea is to avoid lazy quantifier that is slow because it tests on each characters to see what follows. Here i describe the begining as "all that is not a d or a m OR some d not followed by og OR some m not followed by ouse zero or more times. (?>..) is an atomic group, this is to avoid that the regex engine backtrack, it is a kind of 'all or nothing', more informations here ++ is a possessive quantifier that avoid backtracks too.

Categories : C#

RegEx pattern to match between nth and mth occurence of a word
K is so under-appreciated grep -oP '^(.*?one){2}K.*?(?=one)' <<< "There is one wherever you one and here is the one , nowhere is the one" and here is the

Categories : Regex

loop goes to next occurence after one match
my @attendees = ('company'); my @whatever; open ( my $fh, '<', "file.txt") or die "could not open file.txt: $!"; while ( <$fh> ) { chomp $_; push @whatever, $_; } foreach my $attendee ( @attendees ) { foreach my $thing ( @whatever ) { if ($thing =~ /$attendee/i) { print $attendee, " ", $thing, " "; } } } print STDERR "FINISHED! "; Perhaps this does what you want it to, but I must admit that I'm not quite sure.

Categories : Arrays

How to stop at first occurence of match?
You can try this received date[ws]*(?!due date)[ws]*[^d](d{1,4}/d{1,4}/d{1,4}) Group 1 captures date OR /received date.*?(?![ ]+due date).*?[^d](d{1,4}/d{1,4}/d{1,4})/s

Categories : Regex

Regex: Get after last occurence
try this out: grep -Po '(?<=()[^)]*(?=@)|(?<=@)[^)]*(?=))' test with your example: kent$ echo "Something (foo@bar.com@baz.com) something..."|grep -Po '(?<=()[^)]*(?=@)|(?<=@)[^)]*(?=))' foo@bar.com baz.com

Categories : Regex

How to search for the second occurence of ' or " in regex?
Well, from your updates, it seems that you can use something a bit like this: str = "adfa odf 'aso'"; if(/(?:'[^']+'|"[^"]+")$/.test(str)){ res = str.replace(/(?:'[^']+'|"[^"]+")$/, "!string!"); alert(res); } JSFiddle.

Categories : Javascript

Regex to extract lines after the last occurence
You need to make sure that the characters you are repeating cannot include output or last. You can do this with a negative lookahead at each position: (?<=output )w+((?:(?!output|last)[^])*)last First we make sure to start after an output (like you did in your own attempt). Then we match the word that's following (because you don't want it in your captured group). Then comes the interesting part: at every position, we check that there is neither output nor last with (?!output|last). Then we match an arbitrary character with [^]. And then we repeat, until we find last. Instead of [^] you could also use [sS] or . and the dotall option. Working demo.

Categories : Java

Span-wrapping multiple occurence of cout's "<<" using regex
Your problem was that the semicolon in your regex finds also the semicolon in the &lt; special char. Fixed by using the following regex: /(couts*(&lt;&lt;.*?)[^(&lt;)];)/g jsFiddle Demo

Categories : Javascript

grep & regex in terminal - stop at first occurence of character
.* matches greedily. Use .*? (and an extended regular expression) instead: grep -ohE 'translatedString:.*?]' ArticleV.m If you don't like the -E flag, egrep works the same way: egrep -oh 'translatedString:.*?]' ArticleV.m

Categories : Regex

Regex to match paths that don't match a specific pattern: Express Router
The following regex will match any path except those starting with /foo/ app.get(/^/([^f][^o][^o]|.{1,2}|.{4,})/.*$/, routes.index); I assume that this is a standard javascript regex.

Categories : Regex

Regex to match single new line. Regex to match double new line
To match exactly N repetitions of the same character you need lookaheads and lookbehinds (see Match exactly N repetitions of the same character). Since javascript doesn't support the latter, a pure regexp solution seems to be impossible. You'll have to use a helper function, for example: > x = "...a...aa...aaa...aaaa...a...aa" "...a...aa...aaa...aaaa...a...aa" > x.replace(/a+/g, function($0) { return $0.length == 2 ? '@@' : $0; }) "...a...@@...aaa...aaaa...a...@@"

Categories : Javascript

Regex that match if the match contains special word
You're kind of on the right track with lookahead assertions: {{START}}(?:(?!{{END}})[sS])*specialword(?:(?!{{END}})[sS])*{{END}} Explanation: {{START}} # Match {{START}} (?: # Match... (?!{{END}}) # ...as long as we haven't reached {{END}} yet: [sS] # any character )* # any number of times. specialword # Match "specialword" (?: # Match (as before)... (?!{{END}}) # whatever follows, unless it's {{END}} [sS] )* {{END}} # Then finally match {{END}}

Categories : Regex

Regex - Find the match that is inside a match
You can try this regex: /href=[^>]+.pdf/ regex101 demo Most of the time, when you can avoid .* or .+ (or their lazy versions), it's better :) Also, don't forget to escape periods.

Categories : PHP

Java regex: need one regex to match all the formats specified
Try using a reluctant quantifier: _year:.*?s. .replaceAll("_year:.*?\s", "_year:Y ") System.out .println("utc-hour_of_year:2013-07-30T17 dsfsdgfsgf utc-week_of_year:2013-W31 dsfsdgfsdgf" .replaceAll("_year:.*?\s", "_year:Y ")); utc-hour_of_year:Y dsfsdgfsgf utc-week_of_year:Y dsfsdgfsdgf

Categories : Java

regex not returning match but there is clearly a match
You need to escape the dollar sign. start = '>$' end = '</td>' AnnualDiv = re.search('%s(.*)%s' % (start, end), s).group(1) The reason is that the $ is a special character in regex. (It matches the end of a string or before the newline.) This will set AnnualDiv to the string '0.48'. If you want to add the $, you can do it using this: AnnualDiv = "$%s" % re.search('%s(.*)%s' % (start, end), s).group(1)

Categories : Python

Regex.Match() won't match a substring
Try removing ^ and $: Regex regex = new Regex(@"[ABCEGHJKLMNPRSTVXY]{1}d{1}[A-Z]{1} *d{1}[A-Z]{1}d{1}", RegexOptions.None); ^ : The match must start at the beginning of the string or line. $ : The match must occur at the end of the string or before at the end of the line or string. If you want to match only in word boundaries you can use  as suggested by Mike Strobel: Regex regex = new Regex(@"[ABCEGHJKLMNPRSTVXY]{1}d{1}[A-Z]{1} *d{1}[A-Z]{1}d{1}", RegexOptions.None);

Categories : C#

Javascript regex to match a regex
A regular expression to match a regular expression is //((?![*+?])(?:[^ [/\]|\.|[(?:[^ ]\]|\.)*])+)/((?:g(?:im?|mi?)?|i(?:gm?|mg?)?|m(?:gi?|ig?)?)?)/ To break it down, / matches a literal / (?![*+?]) is necessary because /* starts a comment, not a regular expression. [^ [/\] matches any non-escape sequence character and non-start of character group [...] matches a character group which can contain an un-escaped /. \. matches a prefix of an escape sequence + is necessary because // is a line comment, not a regular expression. (?:g...)? matches any combination of non-repeating regular expression flags. So ugly. This doesn't attempt to pair parentheses, or check that repetition modifiers are not applied to themselves, but filters out most of the other ways that regular expressions

Categories : Javascript

Looking for non-zero property TOs: Can I match a Description with number property, but use a regex match?
It is known that integer types has to be passed as integers in the description rendering the usage of regular expressions useless unfortunately. I do not have a QTP installation at hand right now, but to investigate it further, what happens if you use Print Browser("myBrowser").WebElement("height:=11").ChildObjects.Count and Print Browser("myBrowser").WebElement("height:=^[1-9][0-9]*$").ChildObjects.Count Where "myBrowser" is your browser definition of course.

Categories : Regex

Match BOL and EOL with std::regex
libstdc++ has no full support for regex (you can check it here). I'm tried to compile this code with clang 3.2 with libc++-3.2 and result is "true". Use libc++, or boost. Especially libstdc++ regex implementation status 8 Regular expressions 28.1 General N 28.2 Definitions N 28.3 Requirements N 28.4 Header <regex> synopsis N 28.5 Namespace std::regex_constants Y 28.6 Class regex_error Y 28.7 Class template regex_traits Partial 28.8 Class template basic_regex Partial 28.9 Class template sub_match Partial 28.10 Class template match_results Partial 28.11 Regular expression algorithms N 28.12 Regular expression Iterators N 28.13 Modified ECMAScript regular expression grammar N

Categories : C++

Regex to match [] but not []
http://rubular.com/r/16q3jSPHN0 [^\](?:]?([(.+?)])) should work for most cases. Edit: Seems like this will not match [test][test], as Rory pointed out. For that, I can't really think of a good solution without using multiple regexps, but if you want just one then try this: http://rubular.com/r/QBqFAbqW9E (?:[^\](?:]?([(.+?)]))|((?:]?([(.+?)])))\) Match groups will be populated in the first 3 if it a block with escaped brackets occurs after a regular block, and the last 3 if the opposite occurs. Match 1 1. 2. 3. [test] 4. [test] 5. test Match 2 1. [test] 2. test 3. 4. 5.

Categories : Javascript

Regex to match this
You can try this pattern: ^(?:[^e ]+|Be|e(?!xception))+.php:d+$ or this pattern, if you don't need to check a specific line format: ^(?>[^e ]++|Be|e(?!xception))+$ Notice: If you need to select all consecutive lines in one block, you just need to remove from the character classes.

Categories : Regex

Regex match everything after
Why not using a mix of preg_match() and explode()?: $str = '/events/display/id/featured'; $pattern = '~/events/(?P<method>.*?)/(?P<parameter>.*)~'; preg_match($pattern, $str, $matches); // explode the params by '/' $matches['parameter'] = explode('/', $matches['parameter']); var_dump($matches); Output: array(5) { [0] => string(27) "/events/display/id/featured" 'method' => string(7) "display" [1] => string(7) "display" 'parameter' => array(2) { [0] => string(2) "id" [1] => string(8) "featured" } [2] => string(11) "id/featured" }

Categories : PHP

Given regex does not match to the end
You need to pass the global modifier. I'm not sure which programming language you are using, but the syntax often resembles the following: /$myregex/g For example, given the following text: Hello Adam, how are you? Hello Sarah, how are you? The regular expression /Hellos(.*),/g will match both Adam and Sarah.

Categories : Python

Match a^xb^x with regex
The \1 is a backreference and refers to the value of the group, not to the pattern as the recursion (?1) does in Perl. Unfortunately, Java regexes do not support recursion, but the pattern can be expressed using lookarounds and backrefs.

Categories : Java

Regex right match url with DOT at the end
The simplest fix is to require a non-punctuation character as the last character: /(^|[?s])(www.[^? ]+/[^/ ]*?[^? ]*[^?.,! ]|www.[^? ]*[^?.,! ])/g Note that I removed some of your backslash, because they were not necessary. JSFiddle. However, this is still by for not a robust URL pattern. So, why reinvent the wheel instead of just using some established URL pattern?

Categories : Javascript

Java Regex does not match
You can use this pattern to check whether an M character appears as at the beginning of the string: if (line.matches("M.*")) But for something this simple, you can just use this: if (line.length() > 0 && line.charAt(0) == 'M')

Categories : Java

regex (?!) and (?=) and null match
(?=foo) matches any position that's followed by foo, and (?!foo) matches any position that's not followed by foo. So, since every position is followed by the empty string — there's an empty substring between any two characters, and there's an empty substring at the start and end of any string — (?=) matches any position at all, and (?!) doesn't match any position at all.

Categories : Regex

How match a paragraph using regex
You can split on double-newline like this: paragraphs = re.split(r" ", DATA) Edit: To capture the paragraphs as matches, so you can get their start and end points, do this: for match in re.finditer(r'(?s)((?:[^ ][ ]?)+)', DATA): print match.start(), match.end() # Prints: # 0 214 # 215 298 # 299 589

Categories : Python

Check if a given regex will match anything
This doesn't exactly answer your question, but hopefully explains a little why a simple answer is hard to come by: First, the term 'regex' is a bit murky, so just to clarify, we have: "Strict" regular expressions, which are equivalent to deterministic finite automatons (DFAs). Perl-compatible regular expressions (PCREs), which add a bunch of bells and whistles such as lookaheads, backreferences, etc. These are implemented in other languages too, such as Python and Java. Actual Perl regular expressions, which can get even more crazy, including arbitrary Perl code, via the ?{...} construct. I think this problem is solvable for strict regular expressions. You just construct the corresponding DFA and search that graph to see if there's any path to a non-accept state. But that doesn't help

Categories : Regex

How to match "not followed by some literals" using regex?
If you are using a perl-compatible regex engine you should be able to use a zero-width negative lookbehind assertion like this... (?<!hello|hi|bye) ( An example using R (with perl-compatability switched on)... string <- c( "hello (" , "hi (" , "bye (" , "Hello (" , "Anything (" ) grepl( pattern = "(?<!hello|hi|bye) \(" , string , perl = TRUE ) [1] FALSE FALSE FALSE TRUE TRUE We can be a bit more precise like so.... ^.+(?<!^hello|^hi|^bye)s( Matching the start of the string, then optionally any characters, but not hello, hi or bye at the start of a string, then a space, then an open parentheses.

Categories : Regex

Regex Match string
I think something like this would work: ^C?R?U?D?$ To avoid matching empty strings, you can use a lookahead assertion: ^(?!$)C?R?U?D?$

Categories : PHP

Regex match UK postcode
Wrap your regex in ^ and $ to ensure that full string is matched: var re = /^(GIR[ ]?0AA|((AB|AL|B|BA|BB|BD|BH|BL|BN|BR|BS|BT|BX|CA|CB|CF|CH|CM|CO|CR|CT|CV|CW|DA|DD|DE|DG|DH|DL|DN|DT|DY|E|EC|EH|EN|EX|FK|FY|G|GL|GY|GU|HA|HD|HG|HP|HR|HS|HU|HX|IG|IM|IP|IV|JE|KA|KT|KW|KY|L|LA|LD|LE|LL|LN|LS|LU|M|ME|MK|ML|N|NE|NG|NN|NP|NR|NW|OL|OX|PA|PE|PH|PL|PO|PR|RG|RH|RM|S|SA|SE|SG|SK|SL|SM|SN|SO|SP|SR|SS|ST|SW|SY|TA|TD|TF|TN|TQ|TR|TS|TW|UB|W|WA|WC|WD|WF|WN|WR|WS|WV|YO|ZE)(d[dA-Z]?[ ]?d[ABD-HJLN-UW-Z]{2}))|BFPO[ ]?d{1,4})$/; console.log(re.test('WD4 9PL')); ^ matches beginning of the line $ matches end of the line Note, I've also wrapped it in (): /^abc|def$/ will match abc.... and ....def /^(abc|def)$/ will match only abc or def Example: > /abc/.test("abcd") true > /^abc$/.test("abcd") false

Categories : Javascript

Need Regex to match [#URL^Url Description^#]
Why don't you use String.Replace()? A regex would work, but it looks like the format is well defined and regexes are harder to read. string url = "[#URL^blah^#]"; string url_html = url.Replace("[#URL^", "<a href="http://www.somewhere.net">") .Replace("^#]", "</a>");

Categories : C#

Regex to match tags with <%= %>
That's a weird html tag if you ask me, but that should do it I think: /<([a-z]+)((?:[^<]|<%)*)(style)="([^"]*)"([^<]*)>/gi ^^^ regex101 demo [^<]* was replaced by (?:[^<]|<%)* which accepts either a non < character or <%.

Categories : Regex

match first character in a regex?
http://([^/:]*):?([0-9]*)(/.*) The first group is matching everything but : and now I added /, that's because the [^] operator means match everything but what's inside the group, everything else is just the same. Hope it helped!

Categories : Regex

regex match within parenthesis
If you print m, you'll see gregexpr(..., perl = TRUE) gives you the positions and lengths of matches for a) your full pattern including the leading and closing quotes and b) the captured (.*). Unfortunately for you, when m is used by regmatches, it use the positions and lengths of the former. There are two solutions I can think of. Pass your final output through sub: line <- 'VARIABLES = "First [T]" "Second [L]" "Third [1/T]"' m <- gregexpr('"(.*?)"', line, perl = TRUE) z <- regmatches(line, m)[[1]] sub('"(.*?)"', "\1", z) Or use substring using the positions and lengths of the captured expressions: start.pos <- attr(m[[1]], "capture.start") end.pos <- start.pos + attr(m[[1]], "capture.length") - 1L substring(line, start.pos, end.pos) To further your understandi

Categories : Regex

How to match `` in regex in PetitParserDart?
The parser returned by pattern only looks at a single character. Have a look at the tests for some examples. A first approximation of the regular expression w+ would be: word().neg() & word().plus() & word().not() However, this requires a non-word character at the beginning of the parsed string. You can avoid this problem by removing word().neg() and making sure that the caller starts at a valid place. The problem you describe is common when using parsing expression grammars. You can typically solve it by reordering the choices accordingly, or by using the logical predicates like and() and not(). For example the Smalltalk grammar defines the token true as follows: def('trueToken', _token('true') & word().not()); This avoids that the token parser accidentally consumes

Categories : Dart

Regex Gets Match Wrong
- when used within character class depicts a range So you are trying to match characters from space till . Move - to end or the beginning of character class or escape it - Now referring to ascii table you are specifying a range from ascii decimal value 32 till 46 which includes !,",#,$,%........ So,it should be ^[-a-zA-Z0-9 .,()/ / ]+$ ^ or ^[a-zA-Z0-9 .,()/ / -]+$ ^ or escape it ^[a-zA-Z0-9 -.,()/ / ]+$ ^

Categories : Dotnet



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