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  • js regexp problem

    - by Alexander
    I have a searching system that splits the keyword into chunks and searches for it in a string like this: var regexp_school = new RegExp("(?=.*" + split_keywords[0] + ")(?=.*" + split_keywords[1] + ")(?=.*" + split_keywords[2] + ").*", "i"); I would like to modify this so that so that I would only search for it in the beginning of the words. For example if the string is: "Bbe be eb ebb beb" And the keyword is: "be eb" Then I want only these to hit "be ebb eb" In other words I want to combine the above regexp with this one: var regexp_school = new RegExp("^" + split_keywords[0], "i"); But I'm not sure how the syntax would look like. I'm also using the split fuction to split the keywords, but I dont want to set a length since I dont know how many words there are in the keyword string. split_keywords = school_keyword.split(" ", 3); If I leave the 3 out, will it have dynamic lenght or just lenght of 1? I tried doing a alert(split_keywords.lenght); But didnt get a desired response

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  • regular expression

    - by xyz
    I need regular expression to match braces correct e.g for every open one close one abc{abc{bc}xyz} I need it get all it from {abc{bc}xyz} not get {abc{bc} I tried this ({.*?})

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  • php preg_replace, regexp

    - by Michael
    I'm trying to extract the postal codes from yell.com using php and preg_replace. I successfully extracted the postal code but only along with the address. Here is an example $URL = "http://www.yell.com/ucs/UcsSearchAction.do?scrambleSeed=17824062&keywords=shop&layout=&companyName=&location=London&searchType=advance&broaderLocation=&clarifyIndex=0&clarifyOptions=CLOTHES+SHOPS|CLOTHES+SHOPS+-+LADIES|&ooa=&M=&ssm=1&lCOption32=RES|CLOTHES+SHOPS+-+LADIES&bandedclarifyResults=1"; //get yell.com page in a string $htmlContent = $baseClass-getContent($URL); //get postal code along with the address $result2 = preg_match_all("/(.*)/", $htmlContent, $matches); print_r($matches); The above code ouputs something like Array ( [0] = Array ( [0] = 7, Royal Parade, Chislehurst, Kent BR7 6NR [1] = 55, Monmouth St, London, WC2H 9DG .... the problem that I have is that I don't know how to extract the the postal code because it doesn't have an exact number of digits (sometimes it has 6 digits and sometimes has only 5 times). Basically I should extract the lasted 2 words from each array . Thank you in advance for any help !

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  • How do you validate a URL with a regular expression in Python?

    - by Zachary Spencer
    I'm building a Google App Engine app, and I have a class to represent an RSS Feed. I have a method called setUrl which is part of the feed class. It accepts a url as an input. I'm trying to use the re python module to validate off of the RFC 3986 Reg-ex (http://www.ietf.org/rfc/rfc3986.txt) Below is a snipped which should work, right? I'm incredibly new to Python and have been beating my head against this for the past 3 days. p = re.compile('^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\?([^#]*))?(#(.*))?') m = p.match(url) if m: self.url = url return url

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  • Are .NET's regular expressions Turing complete?

    - by Robert
    Regular expressions are often pointed to as the classical example of a language that is not Turning complete. For example "regular expressions" is given in as the answer to this SO question looking for languages that are not Turing complete. In my, perhaps somewhat basic, understanding of the notion of Turning completeness, this means that regular expressions cannot be used check for patterns that are "balanced". Balanced meaning have an equal number of opening characters as closing characters. This is because to do this would require you to have some kind of state, to allow you to match the opening and closing characters. However the .NET implementation of regular expressions introduces the notion of a balanced group. This construct is designed to let you backtrack and see if a previous group was matched. This means that a .NET regular expressions: ^(?<p>a)*(?<-p>b)*(?(p)(?!))$ Could match a pattern that: ab aabb aaabbb aaaabbbb ... etc. ... Does this means .NET's regular expressions are Turing complete? Or are there other things that are missing that would be required for the language to be Turing complete?

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  • Help with this reg. exp. in PHP

    - by Jonathan
    Hi, i don't know about regular expressions, I asked here for one that: gets either anything up to the first parenthesis/colon or the first word inside the first parenthesis. This was the answer: preg_match('/(?:^[^(:]+|(?<=^\\()[^\\s)]+)/', $var, $match); I need an improvement, I need to get either anything up to the first parenthesis/colon/quotation marks or the first word inside the first parenthesis. So if I have something like: $var = 'story "The Town in Hell"s Backyard'; // I get this: $match = 'story'; $var = "screenplay (based on)"; // I get this: $match = 'screenplay'; $var = "(play)"; // I get this: $match = 'play'; $var = "original screen"; // I get this: $match = 'original screen'; Thanks!

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  • Python RegExp exception

    - by Jasie
    How do I split on all nonalphanumeric characters, EXCEPT the apostrophe? re.split('\W+',text) works, but will also split on apostrophes. How do I add an exception to this rule? Thanks!

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  • What is the RFC complicant and working regular expression to check if a string is a valid URL

    - by bestis
    There is question by the almost the same name already: What is the best regular expression to check if a string is a valid URL I don't understand this stackoverflow. It seems like I need reputation to comment an answer. As I don't have it, I don't know how to tell/ask that the proposed solution doesn't seem to work. So I'm forced to make a new question and ask for the solution this way? But that regexp seems to fail in input which has IPv6 address in it: For example facebook's IPv6 address: http://2620:0:1cfe:face:b00c::3/ Also link to localhost fails: http://::1/ Or is PHP to blame? /** * Validate URL - RFC 3987 (IRI) * * http://stackoverflow.com/questions/161738/what-is-the-best-regular-expression-to-check-if-a-string-is-a-valid-url * * @param string $str_url * @return boolean */ function is_url($str_url) { // RFC 3987 For absolute IRIs (internationalized): // @todo FIXME - Has bugs in IPv6 (http://2620:0:1cfe:face:b00c::3/) fails return (bool) preg_match('/^[a-z](?:[-a-z0-9\+\.])*:(?:\/\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:])*@)?(?:\[(?:(?:(?:[0-9a-f]{1,4}:){6}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|::(?:[0-9a-f]{1,4}:){5}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){4}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:[0-9a-f]{1,4}:[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){3}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,2}[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){2}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,3}[0-9a-f]{1,4})?::[0-9a-f]{1,4}:(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,4}[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,5}[0-9a-f]{1,4})?::[0-9a-f]{1,4}|(?:(?:[0-9a-f]{1,4}:){0,6}[0-9a-f]{1,4})?::)|v[0-9a-f]+[-a-z0-9\._~!\$&\'\(\)\*\+,;=:]+)\]|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3}|(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=@])*)(?::[0-9]*)?(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*|\/(?:(?:(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))+)(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*)?|(?:(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))+)(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*|(?!(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])))(?:\?(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])|[\x{E000}-\x{F8FF}\x{F0000}-\x{FFFFD}|\x{100000}-\x{10FFFD}\/\?])*)?(?:\#(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])|[\/\?])*)?$/iu',$str_url); } Here is the test for it: $urls=array('http://www.example.org/','http://www.example.org:80/','example.org','ftp://user:[email protected]/','http://example.org/?cat=5&test=joo','http://www.fi/?cat=5&amp;test=joo','http://::1/','http://2620:0:1cfe:face:b00c::3/','http://2620:0:1cfe:face:b00c::3:80/'); foreach ($urls as $a) { echo $a."\n"; $a=is_url($a); var_dump($a); } And that outputs: > `http://www.example.org/` bool(true) > `http://www.example.org:80/` bool(true) > example.org bool(false) > `ftp://user:[email protected]/` > bool(true) > `http://example.org/?cat=5&test=joo` > bool(true) > `http://www.fi/?cat=5&amp;test=joo` > bool(true) `http://::1/` bool(false) > `http://2620:0:1cfe:face:b00c::3/` > bool(false) > `http://2620:0:1cfe:face:b00c::3:80/` > bool(false) And it also seems that stackoverflow's code is miss behaving on those :) So what is the RFC compilicant and working regexp? ps. If you close this, please then tell me how this situation should be handled? I don't think that the answer is, just earn your reputation. Who wants to do that if they cannot even tell that some proposed solution isn't working correctly. pps. "we're sorry, but as a spam prevention mechanism, new users can only post a maximum of one hyperlink. Earn more than 10 reputation to post more hyperlinks.". Oh C'mon, I'm fine with plain text :D

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  • regular expression and escaping

    - by pstanton
    Sorry if this has been asked, my search brought up many off topic posts. I'm trying to convert wildcards from a user defined search string (wildcard is "*") to postgresql like wildcard "%". I'd like to handle escaping so that "%" => "\%" and "\*" => "*" I know i could replace \* with something else prior to replacing * and then swap it back, but i'd prefer not to and instead only convert * using a pattern that selects it when not proceeded by \. String convertWildcard(String like) { like = like.replaceAll("%", "\\%"); like = like.replaceAll("\\*", "%"); return like; } Assert.assertEquals("%", convertWildcard("*")); Assert.assertEquals("\%", convertWildcard("%")); Assert.assertEquals("*", convertWildcard("\*")); // FAIL Assert.assertEquals("a%b", convertWildcard("a*b")); Assert.assertEquals("a\%b", convertWildcard("a%b")); Assert.assertEquals("a*b", convertWildcard("a\*b")); // FAIL ideas welcome.

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  • Regular Expression Help

    - by WizardsSleeve
    Hi There, Does anyone have a regurlar expression available which only accepts dates in the format dd/mm/yy but also has strict checking to make sure that the date is valid, including leap year support? I am coding in vb.net and am struggling to work this one out. Many Thanks

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  • How can I replace a line which contains only -------- by |||

    - by mimou
    I have something like: ------------------------------------------------------------------------ r2 | username | 2011-01-16 16:52:23 +0100 (Sun, 16 Jan 2011) | 1 line Changed paths: D /foo Removed foo ------------------------------------------------------------------------ r1 | username | 2011-01-16 16:51:03 +0100 (Sun, 16 Jan 2011) | 1 line Changed paths: A /foo created foo ------------------------------------------------------------------------ My target is to identify the file added by the "username" in a specific date. Thus, I need to have the combination (username, 16 Jan 2011, A) to insure that it is the right file ands then print foo. My idea is to: delete the white spaces change the newlines into | get rid of the --------------- and replace them with newlines but the problem is that I couldn't replace the ------- since they are mixed with other characters. ------------------------------------------------------------------------ |r2|username|2011-01-1616:52:23+0100(Sun,16Jan2011)|1line|Changedpaths:|D/foo|Removedfoo| ------------------------------------------------------------------------ |r1|username|2011-01-1616:51:03+0100(Sun,16Jan2011)|1line|Changedpaths:|A/foo|createdfoo| ------------------------------------------------------------------------ So I thought it would be a good idea to start by replacing the --------------- by a special character like ||| and then change this character by a newline using awk FS=||| OFS=\n Can anyone help me! thanks

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  • Alter Regular Expression to Return 2 Values Instead of 3 from userAgent String

    - by Jay
    I've taken a regular expression from jQuery to detect if a browser's engine is WebKit and gets it's version number, it returns 3 values extracted from the userAgent string: webkit/….…, webkit and ….… [“….…” being the version number]. I would like the regular expression to return just 2 values: webkit and ….…. I'm rubbish at regular expressions, so please can you give an explanation of the expression with your answer. The regular expression I'm currently working with and wish to improve is: /(webkit)[\/]([\w.]+)/. I appreciate all your help, thanks in advance!

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  • Confusion in RegExp Reluctant quantifier? Java

    - by Dusk
    Hi, Could anyone please tell me the reason of getting an output as: ab for the following RegExp code using Relcutant quantifier? Pattern p = Pattern.compile("abc*?"); Matcher m = p.matcher("abcfoo"); while(m.find()) System.out.println(m.group()); // ab and getting empty indices for the following code? Pattern p = Pattern.compile(".*?"); Matcher m = p.matcher("abcfoo"); while(m.find()) System.out.println(m.group());

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  • Please help on multiple match replacement

    - by duenguyen
    I have a perl code: my $s = "The+quick+brown+fox+jumps+over+the+lazy+dog+that+is+my+dog"; what I want is to replace every + with space and dog with cat i have this regular expression $s =~ s/+(.*)dog/ ${1}cat/g; But it only match first occurrence of + and last dog. Please help

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  • Nullability (Regular Expressions)

    - by danportin
    In Brzozowski's "Derivatives of Regular Expressions" and elsewhere, the function d(R) returning ? if a R is nullable, and Ø otherwise, includes clauses such as the following: d(R1 + R2) = d(R1) + d(R2) d(R1 · R2) = d(R1) ? d(R2) Clearly, if both R1 and R2 are nullable then (R1 · R2) is nullable, and if either R1 or R2 is nullable then (R1 + R2) is nullable. It is unclear to me what the above clauses are supposed to mean, however. My first thought, mapping (+), (·), or the Boolean operations to regular sets is nonsensical, since in the base case, d(a) = Ø (for all a ? S) d(?) = ? d(Ø) = Ø and ? is not a set (nor is the return type of d, which is a regular expression). Furthermore, this mapping isn't indicated, and there is a separate notation for it. I understand nullability, but I'm lost on the definition of the sum, product, and Boolean operations in the definition of d: how are ? or Ø returned from d(R1) ? d(R2), for instance, in the definition off d(R1 · R2)?

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  • Using regular expressions

    - by Tom
    What is wrong with this regexp? I need it to make $name to be letter-number only. Now it doens't seem to work at all. if (!preg_match("/^[A-Za-z0-9]$/",$name)) { $e[]="name must contain only letters or numbers"; }

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  • List files with two dots in their names using java regular expressions

    - by Nivas
    I was trying to match files in a directory that had two dots in their name, something like theme.default.properties I thought the pattern .\\..\\.. should be the required pattern [. matches any character and \. matches a dot] but it matches both oneTwo.txt and theme.default.properties I tried the following: [resources/themes has two files oneTwo.txt and theme.default.properties] 1. public static void loadThemes() { File themeDirectory = new File("resources/themes"); if(themeDirectory.exists()) { File[] themeFiles = themeDirectory.listFiles(); for(File themeFile : themeFiles) { if(themeFile.getName().matches(".\\..\\..")); { System.out.println(themeFile.getName()); } } } } This prints nothing and the following File[] themeFiles = themeDirectory.listFiles(new FilenameFilter() { public boolean accept(File dir, String name) { return name.matches(".\\..\\.."); } }); for (File file : themeFiles) { System.out.println(file.getName()); } prints both oneTwo.txt theme.default.properties I am unable to find why these two give different results and which pattern I should be using to match two dots... Can someone help?

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  • How to change source order of <div> in less steps/automatically?

    - by metal-gear-solid
    How can i do this task automate. i need to change source order of div, which has same id in above 100 pages. i created example This is default condition <div class="identification"> <div class="number">Number 1</div> </div> <div class="identification"> <div class="number">Number 2</div> </div> <div class="identification"> <div class="number">Number 3</div> </div> <div class="identification"> <div class="number">Number 4</div> </div> <div class="identification"> <div class="number">Number 5</div> </div> <div class="identification"> <div class="number">Number 6</div> </div> I need lik this <div class="identification"> <div class="number">Number 1</div> </div> <div class="identification"> <div class="number">Number 3</div> </div> <div class="identification"> <div class="number">Number 2</div> </div> <div class="identification"> <div class="number">Number 6</div> </div> <div class="identification"> <div class="number">Number 4</div> </div> <div class="identification"> <div class="number">Number 5</div> </div> Is the manual editing only option? I use dreamweaver.

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  • Switch statement for string matching in JavaScript

    - by yaya3
    How do I write a swtich for the following conditional? If the url contains "foo", then settings.base_url is "bar". The following is achieving the effect required but I've a feeling this would be more manageable in a switch: var doc_location = document.location.href; var url_strip = new RegExp("http:\/\/.*\/"); var base_url = url_strip.exec(doc_location) var base_url_string = base_url[0]; //BASE URL CASES // LOCAL if (base_url_string.indexOf('xxx.local') > -1) { settings = { "base_url" : "http://xxx.local/" }; } // DEV if (base_url_string.indexOf('xxx.dev.yyy.com') > -1) { settings = { "base_url" : "http://xxx.dev.yyy.com/xxx/" }; } Thanks

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