Search Results

Search found 3956 results on 159 pages for 'regex cookbook'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 100/159 | < Previous Page | 96 97 98 99 100 101 102 103 104 105 106 107  | Next Page >

  • 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.

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • Use matching value of a RegExp to name the output file.

    - by fx42
    I have this file "file.txt" which I want to split into many smaller ones. Each line of the file has an id field which looks like "id:1" for a line belonging to id 1. For each id in the file, I like to create a file named idid.txt and put all lines that belong to this id in that file. My brute force bash script solution reads as follows. count=1 while [ $count -lt 19945 ] do cat file.txt | grep "id:$count " >> ./sets/id$count.txt count='expr $count + 1' done Now this is very inefficient as I have do read through the file about 20.000 times. Is there a way to do the same operation with only one pass through the file? - What I'm probably asking for is a way to use the value that matches for a regular expression to name the associated output file.

    Read the article

  • Extract multiple values from one column in MySql

    - by Neil
    I've noticed that MySql has an extensive search capacity, allowing both wildcards and regular expressions. However, I'm in somewhat in a bind since I'm trying to extract multiple values from a single string in my select query. For example, if I had the text "<span>Test</span> this <span>query</span>", perhaps using regular expressions I could find and extract values "Test" or "query", but in my case, I have potentially n such strings to extract. And since I can't define n columns in my select statement, that means I'm stuck. Is there anyway I could have a list of values (ideally separated by commas) of any text contained with span tags? In other words, if I ran this query, I would get "Test,query" as the value of spanlist: select <insert logic here> as spanlist from HtmlPages ...

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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!

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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)?

    Read the article

  • Change Number Format

    - by gsembilan
    I have a lot lines contains XXXXXXXXX number format. I want change number XXXXXXXXX to XX.XXX.XXX.X XXXXXXXXX = 9 digit random number Anyone can help me? Thanks in advance

    Read the article

  • 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.

    Read the article

  • 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 ({.*?})

    Read the article

  • 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!

    Read the article

  • python: multiline regular expression

    - by facha
    Hi, everyone I have a piece of text and I've got to parse usernames and hashes out of it. Right now I'm doing it with two regular expressions. Could I do it with just one multiline regular expression? #!/usr/bin/env python import re test_str = """ Hello, UserName. Please read this looooooooooooooooong text. hash Now, write down this hash: fdaf9399jef9qw0j. Then keep reading this loooooooooong text. Hello, UserName2. Please read this looooooooooooooooong text. hash Now, write down this hash: gtwnhton340gjr2g. Then keep reading this loooooooooong text. """ logins = re.findall('Hello, (?P<login>.+).',test_str) hashes = re.findall('hash: (?P<hash>.+).',test_str)

    Read the article

  • 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());

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 96 97 98 99 100 101 102 103 104 105 106 107  | Next Page >