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  • regexp target last main li in list

    - by veilig
    I need to target the starting tag of the last top level LI in a list that may or may-not contain sublists in various positions - without using CSS or Javascript. Is there a simple/elegant regexp that can help with this? I'm no guru w/ them, but it appears the need for greedy/non-greedy selectors when I'm selecting all the middle text (.*) / (.+) changes as nested lists are added and moved around in the list - and this is throwing me off. $pattern = '/^(<ul>.*)<li>(.+<\/li><\/ul>)$/'; $replacement = '$1<li id="lastLi">$3'; Perhaps there is an easier approach?? converting to XML to target the LI and then convert back? ie: Single Element <ul> <li>TARGET</li> </ul> Multiple Elements <ul> <li>foo</li> <li>TARGET</li> </ul> Nested Lists before end <ul> <li> foo <ul> <li>bar</li> </ul> <li> <li>TARGET</li> </ul> Nested List at end <ul> <li>foo</li> <li> TARGET <ul> <li>bar</li> </ul> </li> </ul>

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  • Simple PHP form Validation and the validation symbols

    - by Cool Hand Luke UK
    Hi have some forms that I want to use some basic php validation (regular expressions) on, how do you go about doing it? I have just general text input, usernames, passwords and date to validate. I would also like to know how to check for empty input boxes. I have looked on the interenet for this stuff but I haven't found any good tutorials. Thanks

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  • [Python] OR in regular expression?

    - by www.yegorov-p.ru
    Hello. I have text file with several thousands lines. I want to parse this file into database and decided to write a regexp. Here's part of file: blablabla checked=12 unchecked=1 blablabla unchecked=13 blablabla checked=14 As a result, I would like to get something like (12,1) (0,13) (14,0) Is it possible?

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  • String pattern matching in Javascript

    - by kwokwai
    Hi all, I am doing some self learning about Patern Matching in Javascript. I got a simple input text field in a HTML web page, and I have done some Javascript to capture the string and check if there are any strange characters other than numbers and characters in the string. But I am not sure if it is correct. Only numbers, characters or a mixture of numbers and characters are allowed. var pattern = /^[a-z]+|[A-Z]+|[0-9]+$/; And I have another question about Pattern Matching in Javascript, what does the percentage symbol mean in Pattern matching. For example: var pattern = '/[A-Z0-9._%-]+@[A-Z0-9.-]+\.[A-Z]{2,4}/';

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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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  • Why does this regular expression for sed break inside Makefile?

    - by jcrocholl
    I'm using GNU Make 3.81, and I have the following rule in my Makefile: jslint : java org.mozilla.javascript.tools.shell.Main jslint.js mango.js \ | sed 's/Lint at line \([0-9]\+\) character \([0-9]\+\)/mango.js:\1:\2/' This works fine if I enter it directly on the command line, but the regular expression does not match if I run it with "make jslint". However, it works if I replace \+ with \{1,\} in the Makefile: jslint : java org.mozilla.javascript.tools.shell.Main jslint.js mango.js \ | sed 's/Lint at line \([0-9]\{1,\}\) character \([0-9]\{1,\}\)/mango.js:\1:\2/' Is there some special meaning to \+ in Makefiles, or is this a bug?

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  • Trying to create a Reg Ex for the following patterns

    - by Travis
    Here are the patterns: Red,Green (and so on...) Red (+5.00),Green (+6.00) (and so on...) Red (+5.00,+10.00),Green (+6.00,+20.00) (and so on...) Red (+5.00),Green (and so on...) Each attribute ("Red,"Green") can have 0, 1, or 2 modifiers (shown as "+5.00,+10.00", etc.). I need to capture each of the attributes and their modifiers as a single string (i.e. "Red (+5.00,+10.00)", "Green (+6.00,+20.00)". Help?

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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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  • Need help parsing HTML with a regex in python

    - by laspal
    Hi, My string is mystring = "<tr><td><span class='para'><b>Total Amount : </b>INR (Indian Rupees) 100.00</span></td></tr>" My problem here is I have to search and get the total amount test = re.search("(Indian Rupees)(\d{2})(?:\D|$)", mystring) but my test give me None. How can I get the values and values can be 10.00, 100.00, 1000.00 Thanks

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  • Regular Expression with Names and Emails

    - by Nina
    I am having a problem with regular expressions at the moment. What I'm trying to do is that for each line through the iteration, it checks for this type of pattern: Lastname, Firstname If it finds the name, then it will take the first letter of the first name, and the first six letters of the lastname and form it as an email. I have the following: $checklast = "[A-z],"; $checkfirst = "[A-z]"; if (ereg($checklast, $parts[1])||ereg($checkfirst, $parts[2])){ $first = preg_replace($checkfirst, $checkfirst{1,1}, $parts[2]); print "<a href='mailto:[email protected];'> $parts[$i] </a>"; } This one obviously broke the code. But I was initially attempting to find only the first letter of the firstname and then after that the first six letters of the lastname followed by the @email.com This didn't work out too well. I'm not sure what to do at this point. Any help is much appreciated.

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  • How Do You Parse Column Data ?

    - by discwiz
    I am trying to parse a file generated by LGA Tracon that lists the position data for aircraft over a given time frame. The data of interest starts with TRACKING DATA and ends with SST and there are thousands of entries per file. The system generating the file, Common ARTS, is very rigid in its formatting and we can expect the column spacing to be consistent. Any help would be greatly appreciated. Thanks, Here is an image to preserve the exact formatting Here is a reduced text file. link text

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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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  • 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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  • PHP: URL detection (regexp) includes line breaks

    - by marco92w
    I want to have a function which gets a text as the input and gives back the text with URLs made to HTML links as the output. My draft is as follows: function autoLink($text) { return preg_replace('/https?:\/\/[\S]+/i', '<a href="\0">\0</a>', $text); } But this doesn't work properly. For the input text which contains ... http://www.google.de/ ... I get the following output: <a href="http://www.google.de/<br">http://www.google.de/<br</a> /> Why does it include the line breaks? How could I limit it to the real URL? Thanks in advance!

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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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  • Dealing with regular expressions, Python

    - by Gusto
    I want to remove some symbols from a string using a regular expression, for example: == (that occur both at the beginning and at the end of a line), * (at the beginning of a line ONLY). def some_func(): clean = re.sub(r'= {2,}', '', clean) #Removes 2 or more occurrences of = at the beg and at the end of a line. clean = re.sub(r'^\* {1,}', '', clean) #Removes 1 or more occurrences of * at the beginning of a line. What's wrong with my code? It seems like expressions are wrong. How do I remove a character/symbol if it's at the beginning or at the end of the line (with one or more occurrences)?

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  • Using `rack-rewrite` to Remove the Month and Date from a Permlink

    - by Bryan Veloso
    I've started the process of moving my blog to Octopress, but unfortunately, a limitation of Jekyll doesn't allow me to use abbreviated month names for my permalinks. Therefore I'm looking to just get rid of the month and day bits altogether. I'ved read in this article that you can use rack-rewrite to take care of the redirection, since I am using Heroku to host this. So how would I turn: This: example.com/journal/2012/jan/03/post-of-the-day/ Into this: example.com/journal/2012/post-of-the-day/ Extra points: If I had another rule that redirected /blog/ to /journal/, would that rule still adhere to the above one as well? So from: This: example.com/blog/2012/jan/03/post-of-the-day/ To this: example.com/journal/2012/jan/03/post-of-the-day/ And finally to: example.com/journal/2012/post-of-the-day/ Thanks for the assistance in advance. :)

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  • A more elegant way to parse a string with ruby regular expression using variable grouping?

    - by i0n
    At the moment I have a regular expression that looks like this: ^(cat|dog|bird){1}(cat|dog|bird)?(cat|dog|bird)?$ It matches at least 1, and at most 3 instances of a long list of words and makes the matching words for each group available via the corresponding variable. Is there a way to revise this so that I can return the result for each word in the string without specifying the number of groups beforehand? ^(cat|dog|bird)+$ works but only returns the last match separately , because there is only one group.

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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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  • How to replace plain URLs with links?

    - by Sergio del Amo
    I am using the function below to match URLs inside a given text and replace them for HTML links. The regular expression is working great, but currently I am only replacing the first match. How I can replace all the URL? I guess I should be using the exec command, but I did not really figure how to do it. function replaceURLWithHTMLLinks(text) { var exp = /(\b(https?|ftp|file):\/\/[-A-Z0-9+&@#\/%?=~_|!:,.;]*[-A-Z0-9+&@#\/%=~_|])/i; return text.replace(exp,"<a href='$1'>$1</a>"); }

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  • How to replace only part of the match with python re.sub

    - by Arty
    I need to match two cases by one reg expression and do replacement 'long.file.name.jpg' - 'long.file.name_suff.jpg' 'long.file.name_a.jpg' - 'long.file.name_suff.jpg' I'm trying to do the following re.sub('(\_a)?\.[^\.]*$' , '_suff.',"long.file.name.jpg") But this is cut the extension '.jpg' and I'm getting long.file.name_suff. instead of long.file.name_suff.jpg I understand that this is because of [^.]*$ part, but I can't exclude it, because I have to find last occurance of '_a' to replace or last '.' Is there a way to replace only part of the match?

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  • Regular Expression Pattern for C# with matches

    - by Sumit Gupta
    I am working on project where I need to find Frequency from a given text. I wrote a Regular expression that try to detect frequency, however I am stuck with how C# handle it and how exactly I use it in my software My regular experssion is (\d*)(([,\.]?\s*((k|m)?hz)*)|(\s*((k|m)?hz)*))$ And I am trying to find value from 23,2 Hz 24,4Hz 25,0 Hzsadf 26 Hz 27Khz 28hzzhzhzhdhdwe 29 30.4Hz 31.8 Hz 4343.34.234 Khz 65SD Further Explanation: System needs to work for US and Belgium Culture hence, 23.2 (US) = 23,2 (Be) I try to find a Digit, followed by either khz,mhz,hz or space or , or . If it is , or . then it should have another Digit followed by khz, mhz, hz Any help is appericated.

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  • How to capture strings using * or ? with groups in python regular expressions

    - by user1334085
    When the regular expression has a capturing group followed by "*" or "?", there is no value captured. Instead if you use "+" for the same string, you can see the capture. I need to be able to capture the same value using "?" >>> str1='This string has 29 characters' >>> re.search(r'(\d+)*', str1).group(0) '' >>> re.search(r'(\d+)*', str1).group(1) >>> >>> re.search(r'(\d+)+', str1).group(0) '29' >>> re.search(r'(\d+)+', str1).group(1) '29' More specific question is added below for clarity: I have str1 and str2 below, and I want to use just one regexp which will match both. In case of str1, I also want to be able to capture the number of QSFP ports >>> str1='''4 48 48-port and 6 QSFP 10GigE Linecard 7548S-LC''' >>> str2='''4 48 48-port 10GigE Linecard 7548S-LC''' >>> When I do not use a metacharacter, the capture works: >>> re.search(r'^4\s+48\s+.*(?:(\d+)\s+QSFP).*-LC', str1, re.I|re.M).group(1) '6' >>> It works even when I use the "+" to indicate one occurrence: >>> re.search(r'^4\s+48\s+.*(?:(\d+)\s+QSFP)+.*-LC', str1, re.I|re.M).group(1) '6' >>> But when I use "?" to match for 0 or 1 occurrence, the capture fails even for str1: >>> re.search(r'^4\s+48\s+.*(?:(\d+)\s+QSFP)?.*-LC', str1, re.I|re.M).group(1) >>>

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