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  • Find last match with python regular expression

    - by SDD
    I wanto to match the last occurence of a simple pattern in a string, e.g. list = re.findall(r"\w+ AAAA \w+", "foo bar AAAA foo2 AAAA bar2) print "last match: ", list[len(list)-1] however, if the string is very long, a huge list of matches is generated. Is there a more direct way to match the second occurence of "AAAA" or should I use this workaround?

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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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  • Algorithm to match natural text in mail

    - by snøreven
    I need to separate natural, coherent text/sentences in emails from lists, signatures, greetings and so on before further processing. example: Hi tom, last monday we did bla bla, lore Lorem ipsum dolor sit amet, consectetur adipisici elit, sed eiusmod tempor incidunt ut labore et dolore magna aliqua. list item 2 list item 3 list item 3 Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquid x ea commodi consequat. Quis aute iure reprehenderit in voluptate velit regards, K. ---line-of-funny-characters-####### example inc. 33 evil street, london mobile: 00 234534/234345 Ideally the algorithm would match only the bold parts. Is there any recommended approach - or are there even existing algorithms for that problem? Should I try approximate regular expressions or more statistical stuff based on number of punctation marks, length and so on?

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  • Regular expressions and matching URLs with metacharacters

    - by James P.
    I'm having trouble finding a regular expression that matches the following String. Korben;http://feeds.feedburner.com/KorbensBlog-UpgradeYourMind?format=xml;1 One problem is escaping the question mark. Java's pattern matcher doesn't seem to accept \? as a valid escape sequence but it also fails to work with the tester at myregexp.com. Here's what I have so far: ([a-zA-Z0-9])+;http://([a-zA-Z0-9./-]+);[0-9]+ Any suggestions? Edit: The original intent was to match all URLs that could be found after the first semi colon.

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  • Replacing text node of HTML input in PHP

    - by Aman Kumar Jain
    Hi, I want to replace all the text nodes in a html text. I'll explain with an example: $html = " <div> <p> text2 text2 word text2 <span>abcd</span> text2 text2 word text2 <p> this is a long, very long statement with punctuations. </div> I want to replace "text2 text2 word text2" with "<span>text2 text2 word text2</span>" and "this is a long, very long statement with punctuations." with "<span>this is a long, very long statement with punctuations.</span>" What should be the regular expression for the same?

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  • RegularExpressionValidator - Windows ID Validation

    - by Albert
    I'd like to setup a RegularExpressionValidator to ensure users are entering valid windows IDs in a textbox. Specifically, I'd like to ensure it's any three capital letters (for our range of domains), followed by a backslash, followed by any number of letters and numbers. Does anyone know where I can find some examples of this type of validation...or can somebody whip one up for me? :)

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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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  • Why does this regular expression fail?

    - by Stephen
    I have a password validation script in PHP that checks a few different regular expressions, and throws a unique error message depending on which one fails. Here is an array of the regular expressions and the error messages that are thrown if the match fails: array( 'rule1' => array( '/^.*[\d].*$/i', 'Password must contain at least one number.' ), 'rule2' => array( '/^.*[a-z].*$/i', 'Password must contain at least one lowercase letter' ), 'rule3' => array( '/^.*[A-Z].*$/i', 'Password must contain at least one uppercase letter' ), 'rule4' => array( '/^.*[~!@#$%^&*()_+=].*$/i', 'Password must contain at least one special character [~!@#$%^&*()_+=]' ) ); For some reason, no matter what I pass through the validation, the "Special Characters" rule fails. I'm guessing it's a problem with the expression. If there's a better (or correct) way to write these expressions, I'm all ears!

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  • Java Regular Expression. Check if String contains ONLY Letters

    - by user69514
    How do I check if a String contains only letters in java? I want to write an if statement that will return false if there is a white space, a number, a symbol or anything else other than a-z A-Z. My string must contain ONLY letters. I thought I could do it this way, but I'm doing it wrong: if( ereg("[a-zA-Z]+", $myString)) return true; else return false;

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  • Matching several items inside one string with preg_match_all() and end characters

    - by nefo_x
    I have the following code: preg_match_all('/(.*) \((\d+)\) - ([\d\.\d]+)[,?]/U', "E-Book What I Didn't Learn At School... (2) - 3525.01, FREE Intro DVD/Vid (1) - 0.15", $match); var_dump($string, $match); and get the following ouput: array(4) { [0]=> array(1) { [0]=> string(54) "E-Book What I Didn't Learn At School... (2) - 3525.01," } [1]=> array(1) { [0]=> string(39) "E-Book What I Didn't Learn At School..." } [2]=> array(1) { [0]=> string(1) "2" } [3]=> array(1) { [0]=> string(7) "3525.01" } } which matches only one items... what i need is to get all items from such strings. when i've added "," sign to the end of the string - it worked fine. but that is non-sense in adding comma to each string. Any advice?

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  • Regular expression to match HTML table row ( <tr> ) NOT containing a specific value

    - by user1821136
    I'm using Notepad++ to clean up a long and messy HTML table. I'm trying to use regular expressions even if I'm a total noob. :) I need to remove all the table rows that doesn't contain a specific value (may I call that substring?). After having all the file contents unwrapped, I've been able to use the following regular expression to select, one by one, every table row with all its contents: <tr>.+?</tr> How can I improve the regular expression in order to select and replace only table rows containing, somewhere inside a part of them, that defined substring? I don't know if this does matter but the structure of every table row is the following (I've put there every HTML tag, the dots stand for standard content/values) <tr> <td> ... </td> <td> ... </td> <td> <a sfref="..." href="...">!! SUBSTRING I HAVE TO MATCH HERE !!</a> </td> <td> <img /> </td> <td> ... </td> <td> ... </td> <td> ... </td> <td> ... </td> </tr> Thanks in advance for your help!

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  • How can I find all attributes with single quotes in a Sublime Text 2 document and replace with double quotes?

    - by Brandon Durham
    I'm feeling particularly nit-picky today. I'm working in some HTML docs that have single quotes around all attribute values through the docs, like this: <div class='classone classtwo'> I'd love to be able to do a find-and-replace in each doc and replace with double quotes, like this: <div class="classone classtwo"> Many elements in the document will have multiple attributes: <div class='classone classtwo' data-scripts='lazyload'> And some will have the correct double quotes: <div class='classone classtwo' data-scripts="lazyload"> What's the best way to replace all single quotes wrapping values with double?

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  • Use Regular expression with fileinput

    - by chrissygormley
    Hello, I am trying to replace a variable stored in another file using regular expression. The code I have tried is: r = re.compile(r"self\.uid\s*=\s*('\w{12})'") for line in fileinput.input(['file.py'], inplace=True): print line.replace(r.match(line), sys.argv[1]), The format of the variable in the file is: self.uid = '027FC8EBC2D1' I am trying to pass in a parameter in this format and use regular expression to verify that the sys.argv[1] is correct format and to find the variable stored in this file and replace it with the new variable. Can anyone help. Thanks for the help.

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  • Remove leading whitespaces using variable length lookbehind in RegExp

    - by Shizhidi
    Hello, I'm wondering if variable length lookbehind assertions are supported in JavaScript's RegExp engine? For example, I'm trying to match the string "variable length" in the string "[a lot of whitespaces and/or tabs]variable length lookbehind", and I have something like this but it does not go well in various RegExp testers: ^(?<=[ \t]+).+(?= lookbehind) If it's an illegal pattern, what would be a good workaround to it? Thanks!

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  • Ruby: backslash all non-alphanumeric characters in a string

    - by HBlend
    I have a script where I need to take a user's password and then run a command line using it. I need to backslash all (could be more then one) non-alphanumeric characters in the password. I have tried several things at this point including the below but getting no where. This has to be easy, just missing it. Tried these and several others: password = password.gsub(/(\W)/, '\\1') password = password.gsub(/(\W)/, '\\\1') password = password.gsub(/(\W)/, '\\\\1')

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  • classic asp ignore comma within speach marks CSV

    - by user2968227
    I Have a CSV File that looks like this 1,HELLO,ENGLISH 2,HELLO1,ENGLISH 3,HELLO2,ENGLISH 4,HELLO3,ENGLISH 5,HELLO4,ENGLISH 6,HELLO5,ENGLISH 7,HELLO6,ENGLISH 8,"HELLO7, HELLO7 ...",ENGLISH 9,HELLO7,ENGLISH 10,HELLO7,ENGLISH I want to step loop through the lines and write to a table using split classic asp function by comma. When Speech marks are present to ignore the comma within those speech marks and take the string. Please help. <% dim csv_to_import,counter,line,fso,objFile csv_to_import="uploads/testLang.csv" set fso = createobject("scripting.filesystemobject") set objFile = fso.opentextfile(server.mappath(csv_to_import)) str_imported_data="<table cellpadding='3' cellspacing='1' border='1'>" Do Until objFile.AtEndOfStream line = split(objFile.ReadLine,",") str_imported_data=str_imported_data&"<tr>" total_records=ubound(line) for i=0 to total_records if i>0 then str_imported_data=str_imported_data&"<td>"&line(i)&"</td>" else str_imported_data=str_imported_data&"<th>"&line(i)&"</th>" end if next str_imported_data=str_imported_data&"</tr>" & chr(13) Loop str_imported_data=str_imported_data&"<caption>Total Number of Records: "&total_records&"</caption></table>" objFile.Close response.Write str_imported_data %>

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  • what is the return value of BeautifulSoup.find ?

    - by prosseek
    I run to get some value as score. score = soup.find('div', attrs={'class' : 'summarycount'}) I run 'print score' to get as follows. <div class=\"summarycount\">524</div> I need to extract the number part. I used re module but failed. m = re.search("[^\d]+(\d+)", score) TypeError: expected string or buffer function search in re.py at line 142 return _compile(pattern, flags).search(string) What's the return type of the find function? How to get the number from the score variable? Is there any easy way to let BeautifulSoup to return the value(in this case 524) itself?

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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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  • How to use regular expressions to pull a substring? (screen scraping)

    - by Diego
    Hey guys, i'm really trying to understand regular expressions while scraping a site, i've been using it in my code enough to pull the following, but am stuck here. I need to quickly grab this: http://www.example.com/online/store/TitleDetail?detail&sku=123456789 from this: ('<a href="javascript:if(handleDoubleClick(this.id)){window.location=\'http://www.example.com/online/store/TitleDetail?detail&sku=123456789\';}" id="getTitleDetails_123456789">\r\n\t\t\t \tcheck store inventory\r\n\t\t\t </a>', 1) This is where I got confused. any ideas?

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  • using regular expression in Java

    - by Mrityunjay
    Hi, i need to check a string that should contain only ABCDEFG characters, in any sequence and with only 7 characters. Please let me know the correct way of using regular expression. as corrently i am using String abs = "ABPID"; if(!Pattern.matches("[[ABCDEFG]", abs)) System.out.println("Error"); i am using the following code which works when i use the String abcdefg but for other cases it fails. please help me out.

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