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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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  • How to detect identical part(s) inside string?

    - by Horace Ho
    I try to break down the http://stackoverflow.com/questions/2711961/decoding-algorithm-wanted question into smaller questions. This is Part I. Question: two strings: s1 and s2 part of s1 is identical to part of s2 space is separator how to extract the identical part(s)? example 1: s1 = "12 November 2010 - 1 visitor" s2 = "6 July 2010 - 100 visitors" the identical parts are "2010", "-", "1" and "visitor" example 2: s1 = "Welcome, John!" s2 = "Welcome, Peter!" the identical parts are "Welcome," and "!" Python and Ruby preferred. 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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  • php - get content from second pair of quotes in string

    - by Aaron Turecki
    I'm trying to get the contents of the second quotes and only the second quotes from a string. Right now I'm able to get the contents of all three quotes. What am I doing wrong? Is it possible to just print the second value in the output array? Text 2014-06-02 11:48:41.519 -0700 Information 94 NICOLE Client "[WebDirect] (207.230.229.204) [207.230.229.204]" opening database "FMServer_Sample" as "Admin". PHP if (preg_match_all('~(["\'])([^"\']+)\1~', $line, $matches)) $database_names = $matches[2]; print_r($database); Output [WebDirect] (207.230.229.204) [207.230.229.204], FMServer_Sample, Admin

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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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  • 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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  • Which is more efficient regular expression?

    - by Vagnerr
    I'm parsing some big log files and have some very simple string matches for example if(m/Some String Pattern/o){ #Do something } It seems simple enough but in fact most of the matches I have could be against the start of the line, but the match would be "longer" for example if(m/^Initial static string that matches Some String Pattern/o){ #Do something } Obviously this is a longer regular expression and so more work to match. However I can use the start of line anchor which would allow an expression to be discarded as a failed match sooner. It is my hunch that the latter would be more efficient. Can any one back me up/shoot me down :-)

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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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  • preg_replace only replaces first occurrence then skips to next line

    - by Dom
    Got a problem where preg_replace only replaces the first match it finds then jumps to the next line and skips the remaining parts on the same line that I also want to be replaced. What I do is that I read a CSS file that sometimes have multiple "url(media/pic.gif)" on a row and replace "media/pic.gif" (the file is then saved as a copy with the replaced parts). The content of the CSS file is put into the variable $resource_content: $resource_content = preg_replace('#(url\((\'|")?)(.*)((\'|")?\))#i', '${1}'.url::base(FALSE).'${3}'.'${4}', $resource_content); Does anyone know a solution for why it only replaces the first match per line?

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  • Optimizing python link matching regular expression

    - by Matt
    I have a regular expression, links = re.compile('<a(.+?)href=(?:"|\')?((?:https?://|/)[^\'"]+)(?:"|\')?(.*?)>(.+?)</a>',re.I).findall(data) to find links in some html, it is taking a long time on certain html, any optimization advice? One that it chokes on is http://freeyourmindonline.net/Blog/

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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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  • 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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  • 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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  • 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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  • How to process this string via regular expression

    - by iiduce
    my string style like this: expression1/field1+expression2*expression3+expression4/field2*expression5*expression6/field3 a real style mybe like this: computer/(100)+web*mail+explorer/(200)*bbs*solution/(300) "+" and "*" represent operator "computer","web"...represent expression (100),(200) represent field num . field num may not exist. I want process the string to this: /(100)+web*+explorer/(200)bbs/(300) rules like this: if expression length is more than 3 and its field is not (200), then add brackets to it.

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  • How to get N random string from a {a1|a2|a3} format string?

    - by Pentium10
    Take this string as input: string s="planets {Sun|Mercury|Venus|Earth|Mars|Jupiter|Saturn|Uranus|Neptune}" How would I choose randomly N from the set, then join them with comma. The set is defined between {} and options are separated with | pipe. The order is maintained. Some output could be: string output1="planets Sun, Venus"; string output2="planets Neptune"; string output3="planets Earth, Saturn, Uranus, Neptune"; string output4="planets Uranus, Saturn";// bad example, order is not correct Java 1.5

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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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  • How to find if dataTable contains column which name starts with abc

    - by VilemRousi
    In my program I have a dataTable and I´d like to know if is there a column which name starts with abc. For example I have a DataTable and its name is abcdef. I like to find this column using something like this: DataTable.Columns.Constains(ColumnName.StartWith(abc)) Because I know only part of the column name, I cannot use a Contains method. Is there any simple way how to do that? Thanks a lot.

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  • How can I match a match a null byte (0x00) in the Visual Studio binary editor with a find using a re

    - by Paul K
    Open a file in the Visual Studio binary editor that contains a null byte (0x00), then use the Quick Find feature (Ctrl +F) to find null bytes. I would have thought I could use a regular expression such as \x00 to match null bytes but it doesn't work. Searching for any other hex value using this method works fine. Is this a VS bug, 'feature', or am I just missing something? Is there a work around?

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