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  • 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)

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  • Java Matcher groups: Understanding The difference between "(?:X|Y)" and "(?:X)|(?:Y)"

    - by user358795
    Can anyone explain: Why the two patterns used below give different results? (answered below) Why the 2nd example gives a group count of 1 but says the start and end of group 1 is -1? public void testGroups() throws Exception { String TEST_STRING = "After Yes is group 1 End"; { Pattern p; Matcher m; String pattern="(?:Yes|No)(.*)End"; p=Pattern.compile(pattern); m=p.matcher(TEST_STRING); boolean f=m.find(); int count=m.groupCount(); int start=m.start(1); int end=m.end(1); System.out.println("Pattern=" + pattern + "\t Found=" + f + " Group count=" + count + " Start of group 1=" + start + " End of group 1=" + end ); } { Pattern p; Matcher m; String pattern="(?:Yes)|(?:No)(.*)End"; p=Pattern.compile(pattern); m=p.matcher(TEST_STRING); boolean f=m.find(); int count=m.groupCount(); int start=m.start(1); int end=m.end(1); System.out.println("Pattern=" + pattern + "\t Found=" + f + " Group count=" + count + " Start of group 1=" + start + " End of group 1=" + end ); } } Which gives the following output: Pattern=(?:Yes|No)(.*)End Found=true Group count=1 Start of group 1=9 End of group 1=21 Pattern=(?:Yes)|(?:No)(.*)End Found=true Group count=1 Start of group 1=-1 End of group 1=-1

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  • Regular expression to match any table tag

    - by keeg
    I'm trying to write a regular expression to see if a string contains any of the typical table tags: <table></table> <td></td> <th></th> <tr></tr> <thead></thead> <tfoot></tfoot> <tbody></tbody> Along with tags that may contain other attributes e.g: <table border="1"> I've come up with this so far, however, it matches <br /> tag and I'm not sure why: /<\/?[table|td|th|tr|tfoot|thead|tbody]{1,}>?/ http://www.rexfiddle.net/20Xtqka

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  • How Do I Remove The First 4 Characters From A String If It Matches A Pattern In Ruby

    - by James
    I have the following string: "h3. My Title Goes Here" I basically want to remove the first 4 characters from the string so that I just get back: "My Title Goes Here". The thing is I am iterating over an array of strings and not all have the h3. part in front so I can't just ditch the first 4 characters blindly. I have checked the docs and the closest think I could find was chomp, but that only works for the end of a string. Right now I am doing this: "h3. My Title Goes Here".reverse.chomp(" .3h").reverse This gives me my desired output, but there has to be a better way right? I mean I don't want to reverse a string twice for no reason. I am new to programming so I might have missed something obvious, but I didn't see the opposite of chomp anywhere in the docs. Is there another method that will work? Thanks!

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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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  • 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 ...

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  • PHP regular expression for positive number with 0 or 2 decimal places

    - by Peter
    Hi I am trying to use the following regular expression to check whether a string is a positive number with either zero decimal places, or 2: ^\d+(\.(\d{2}))?$ When I try to match this using preg_match, I get the error: Warning: preg_match(): No ending delimiter '^' found in /Library/WebServer/Documents/lib/forms.php on line 862 What am I doing wrong?

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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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  • Mod rewrite with multiple query strings

    - by Boris
    Hi, I'm a complete n00b when it comes to regular expressions. I need these redirects: (1) www.mysite.com/products.php?id=001&product=Product-Name&source=Source-Name should become -> www.mysite.com/Source-Name/001-Product-Name (2) www.mysite.com/stores.php?id=002&name=Store-Name should become -> www.mysite.com/002-Store-Name Any help much appreciated :)

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  • Can a repeated piece of regular expression create multiple groups? Such as this example...

    - by Yousui
    Hi guys, I'm using RUBY 's regular expression to deal with text such as ${1:aaa|bbbb} ${233:aaa | bbbb | ccc ccccc } ${34: aaa | bbbb | cccccccc |d} ${343: aaa | bbbb | cccccccc |dddddd ddddddddd} ${3443:a aa|bbbb|cccccccc|d} ${353:aa a| b b b b | c c c c c c c c | dddddd} I want to get the trimed text between each pipe line. For example, for the first line of my upper example, I want to get the result aaa and bbbb, for the second line, I want aaa, bbbb and ccc ccccc. Now I have wrote a piece of regular expression and a piece of ruby code to test it: array = "${33:aaa|bbbb|cccccccc}".scan(/\$\{\s*(\d+)\s*:(\s*[^\|]+\s*)(?:\|(\s*[^\|]+\s*))+\}/) puts array Now my problem is the (?:\|(\s*[^\|]+\s*))+ part can't create multiple groups. I don't know how to solve this problem, because the number of text I need in each line is variable. Anyone can help? Great thanks.

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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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  • Need a regular expression for an Irish phone number

    - by Eoghan O'Brien
    I need to validate an Irish phone number but I don't want to make it too user unfriendly, many people are used to writing there phone number with brackets wrapping their area code followed by 5 to 7 digits for their number, some add spaces between the area code or mobile operator. The format of Irish landline numbers is an area code of between 1 and 4 digits and a number of between 5 to 8 digits. e.g. (021) 9876543 (01)9876543 01 9876543 (0402)39385 I'm looking for a regular expression for Javascript/PHP. Thanks.

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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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  • Flex 3 Regular Expression Problem

    - by Tommy
    I've written a url validator for a project I am working on. For my requirements it works great, except when the last part for the url goes longer than 22 characters it breaks. My expression: /((https?):\/\/)([^\s.]+.)+([^\s.]+)(:\d+\/\S+)/i It expects input that looks like "http(s)://hostname:port/location". When I give it the input: https://demo10:443/111112222233333444445 it works, but if I pass the input https://demo10:443/1111122222333334444455 it breaks. You can test it out easily at http://ryanswanson.com/regexp/#start. Oddly, I can't reproduce the problem with just the relevant (I would think) part /(:\d+\/\S+)/i. I can have as many characters after the required / and it works great. Any ideas or known bugs?

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  • Delete all characters in a multline string up to a given pattern

    - by biffabacon
    Using Python I need to delete all charaters in a multiline string up to the first occurrence of a given pattern. In Perl this can be done using regular expressions with something like: #remove all chars up to first occurrence of cat or dog or rat $pattern = 'cat|dog|rat' $pagetext =~ s/(.*?)($pattern)/$2/xms; What's the best way to do it in Python?

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  • 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.

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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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  • Dreamweaver regular expression substitution followed by number

    - by mark
    Hi. I'm using Dreamweaver to update copyright dates across my site. I want to preserve the existing spacing (or lack thereof) between years. Examples: © 2002-2008 should update to © 2002-2009 © 2003 - 2008 should update to © 2003 - 2009 This is the regular expression I'm using to accomplish this in Dreamweaver's find & replace function Find: ©\s*(\d{4}\s*-\s*)\d{3}[^9] Replace: © $1 2009 Here's the PROBLEM: This expression works, but has that that extra space between the hyphen and 2009. If I write the replace expression without the space, as © $12009 then dreamweaver looks for the 12,009th substitution in the find expression, and, not finding one, prints $12009. Any ideas?

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  • Django - urls.py - Filenames with a hash/pound (#) sign?

    - by miya
    I'm using django and realized that when the filename that the user wants to access (let's say a photo) has the pound sign, the entry in the url.py does not match. Any ideas? url(r'^static/(?P<path>.*)$', 'django.views.static.serve', {'document_root': MEDIA_ROOT}, it just says: "/home/user/project/static/upload/images/hello" does not exist when actually the name of the file is: hello#world.jpg Thanks, Nico

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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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  • 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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  • 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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  • 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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