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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • Match string which doesn't start with

    - by Pinky
    I have a string that looks like this: var str = "Hello world, &nbsp;hello &gt;world, hello world!"; ... and I'd like to replace all the hellos with e.g. bye and world with earth, except the words that start with &nbsp or &gt. Those should be ignored. So the result should be: bye earth, &nbsp;hello &gt;world, bye earth! Tried to this with str.replace(/(?!\&nbsp;)hello/gi,'bye')); But it doesn't work.

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  • RegExp to validate a formula (string/boolean/numeric expression)?

    - by JSteve
    I have used regExp quit a bit of times but still far from being an expert. This time I want to validate a formula (or math expression) by regExp. The difficult part here is to validate proper starting and ending parentheses with in the formula. I believe, there would be some sample on the web but I could not find it. Can somebody post a link for such example? or help me by some other means?

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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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  • JavaScript regular expression literal persists between function calls

    - by Charles Anderson
    I have this piece of code: function func1(text) { var pattern = /([\s\S]*?)(\<\?(?:attrib |if |else-if |else|end-if|search |for |end-for)[\s\S]*?\?\>)/g; var result; while (result = pattern.exec(text)) { if (some condition) { throw new Error('failed'); } ... } } This works, unless the throw statement is executed. In that case, the next time I call the function, the exec() call starts where it left off, even though I am supplying it with a new value of 'text'. I can fix it by writing var pattern = new RegExp('.....'); instead, but I don't understand why the first version is failing. How is the regular expression persisting between function calls? (This is happening in the latest versions of Firefox and Chrome.) Edit Complete test case: <!DOCTYPE HTML> <html> <head> <meta http-equiv="Content-type" content="text/html;charset=UTF-8"> <title>Test Page</title> <style type='text/css'> body { font-family: sans-serif; } #log p { margin: 0; padding: 0; } </style> <script type='text/javascript'> function func1(text, count) { var pattern = /(one|two|three|four|five|six|seven|eight)/g; log("func1"); var result; while (result = pattern.exec(text)) { log("result[0] = " + result[0] + ", pattern.index = " + pattern.index); if (--count <= 0) { throw "Error"; } } } function go() { try { func1("one two three four five six seven eight", 3); } catch (e) { } try { func1("one two three four five six seven eight", 2); } catch (e) { } try { func1("one two three four five six seven eight", 99); } catch (e) { } try { func1("one two three four five six seven eight", 2); } catch (e) { } } function log(msg) { var log = document.getElementById('log'); var p = document.createElement('p'); p.innerHTML = msg; log.appendChild(p); } </script> </head> <body><div> <input type='button' id='btnGo' value='Go' onclick='go();'> <hr> <div id='log'></div> </div></body> </html> The regular expression continues with 'four' as of the second call on FF and Chrome, not on IE7 or Opera.

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  • awk or sed: Best way to grab [this text]

    - by Parand
    I'm trying to parse various info from log files, some of which is placed within square brackets. For example: Tue, 06 Nov 2007 10:04:11 INFO processor:receive: [someuserid], [somemessage] msgtype=[T] What's an elegant way to grab 'someuserid' from these lines, using sed, awk, or other unix utility?

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  • php Dollar amount Regular Expression

    - by Thildemar
    I am have completed javascript validation of a form using Regular Expressions and am now working on redundant verification server-side using PHP. I have copied this regular expression from my jscript code that finds dollar values, and reformed it to a PHP friendly format: /\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/ Specifically: if (preg_match("/\$?((\d{1,3}(,\d{3})*)|(\d+))(\.\d{2})?$/", $_POST["cost"])){} While the expression works great in javascript I get : Warning: preg_match() [function.preg-match]: Compilation failed: nothing to repeat at offset 1 when I run it in PHP. Anyone have a clue why this error is coming up?

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  • PHP & Regular expression: keyword just occurs once

    - by lauthiamkok
    Hi, how can I make sure a certain keyword just occurs once in the input with regular expression? I think there is some mistakes in the expression below as I can repeat the same keywords, if (!preg_match('/\b(.php?){1}\b/', $cfg_path)) { $error = true; echo '<error elementid="cfg_path" message="PATH - make sure you have a \'.php?\' in the path."/>'; } I just want this to be true, form.php?category=something or form.php? but not this, form.php?.php?category=something or form.php?.php? please let me know how to fix it. thanks.

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