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  • Mod_rewrite works on local, not on remote, version?

    - by TylerT
    I have this site. Let's call it htp://www.mysite.com I have a rewrite rule to change htp://www.mysite.com/?q=words%20etc/0/10 into http://www.mysite.com/words%20etc/0/10 (or http://www.mysite.com//0/10 or http://www.mysite.com/0/10) .htaccess:ErrorDocument 404 htp://www.mysite.com/404.html options +FollowSymlinks rewriteEngine on rewriteCond %{REQUEST_URI} !-f rewriteCond %{REQUEST_URI} !-d rewriteCond %{REQUEST_URI} !index\.php rewriteRule ^/?([^/]+?)?/?([0-9]+?)/([0-9]+?)$ index.php/%{THE_REQUEST} [NC] Now, this works on my local apache 2.2.11 server, no errors. However on my host's apache 1.3.41 server, I get the following error: [Sat Mar 5 21:42:14 2011] [alert] [client [ip]] /home/_/public_html/mysite.com/.htaccess: RewriteRule: cannot compile regular expression '^/?([^/]+?)?/?([0-9]+?)/([0-9]+?)$'\n I imagine it's something quirky about the apache version as other sites on this host use mod_rewrite without a hitch. I've tried removing the +followSymlinks line, even the rewrite engine line. I haven't tried removing the conditions cause I don't think I should have to, I'm probably wrong.

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  • Regexs in Ruby getting filename

    - by user1290757
    i am extracting file names of html files using line: filename = File.basename(input_filename, ".*") which currently prints full file name excluding .html extension All files are stored in the form of http^x.x.edu^1^2 all file names begin with http^ and contain edu^ what i want is to extract 2 (which changes) but it is always the second element after .edu I have attempted destructive gsub! but i m weak with regular expressions.

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  • Extract string that is delimited with constant and ends with two numbers (numbers have to be included)

    - by Edmon
    I have a text that contains string of a following structure: text I do not care about, persons name followed by two IDs. I know that: a person's name is always preceded by XYZ code and that is always followed by two, space separated numbers. Name is not always just a last name and first name. It can be multiple last or first names (think Latin american names). So, I am looking to extract string that follows the constant XYZ code and that is always terminated by two separate numbers. You can say that my delimiter is XYZ and two numbers, but numbers need to be part of the extracted value as well. From blah, blah XYZ names, names 122322 344322 blah blah I want to extract: names, names 122322 344322 Would someone please advise on the regular expression for this that would work with Python's re package.

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  • I am using a regex snippet query string path

    - by Shelby Poston
    Using the following to load images base on two ids one is the and bookid and the out is the client. My folder structures is this. root path = flipbooks subfolders under flipbooks are books and clients in subfolder books I have and .net page title tablet. the tablet code behind checks the bookid of client and render a the tablet page with images in a flipbook fashion. because we have over 15000 records and flipbooks already created and stored in the database. I don't move the client folder under the books subfolders. I need the code below to get to the client subfolder in the query string and help to change this would be helpful. The result now is http://www.somewebsite.com/books/client/images/someimage1.jpg[^] I need the results to be http://www.somewebsite.com/client/images/someimage1.jpg[^]. I tried moving the tablet.aspx file to the root flipbooks and it works but i have provide a user name and password each time. This need to be access by the public and my root is protected. Don't want to have to change permission. I am trying to remove the /books function getParameterByName(name) { var results = RegExp('[?&]' + name + '=([^&]*)').exec(window.location.search); return results ? decodeURIComponent(results[1].replace(/\+/g, ' ')) : null; } Thanks Mission Critical

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  • In C/C++ mode in Emacs, change face of code in #if 0...#endif block to comment face

    - by pogopop77
    I'm trying to add functionality found in some other code editors to my Emacs configuration, whereby C/C++ code within #if 0...#endif blocks is automatically set to the comment face/font. Based on my testing, cpp-highlight-mode does something like what I want, but requires user action. It seems like tying into the font-lock functionality is the correct option to make the behavior automatic. I have successfully followed examples in the GNU documentation to change the face of single-line regular expressions. For example: (add-hook 'c-mode-common-hook (lambda () (font-lock-add-keywords nil '(("\\<\\(FIXME\\|TODO\\|HACK\\|fixme\\|todo\\|hack\\)" 1 font-lock-warning-face t))))) works fine to highlight debug related keywords anywhere in a file. However, I am having problems matching #if 0...#endif as a multiline regular expression. I found some useful information in this post (How to compose region like ""), that suggested that Emacs must be told specifically to allow for multiline matches. But this code: (add-hook 'c-mode-common-hook (lambda () '(progn (setq font-lock-multiline t) (font-lock-add-keywords nil '(("#if 0\\(.\\|\n\\)*?#endif" 1 font-lock-comment-face t)))))) still does not work for me. Perhaps my regular expression is wrong (though it appears to work using M-x re-builder), I've messed up my syntax, or I'm following the wrong approach entirely. I'm using Aquamacs 2.1 (which is based on GNU Emacs 23.2.50.1) on OS X 10.6.5, if that makes a difference. Any assistance would be appreciated!

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  • 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

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  • Regular Expression for username

    - by neobie
    I need help on regular expression on the condition (4) below: Begin with a-z End with a-z0-9 allow 3 special characters like ._- The characters in (3) must be followed by alphanumeric characters, and it cannot be followed by any characters in (3) themselves. Not sure how to do this. Any help is appreciated, with the sample and some explanations.

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  • Help with this reg. exp. in PHP

    - by Jonathan
    Hi, i don't know about regular expressions, I asked here for one that: gets either anything up to the first parenthesis/colon or the first word inside the first parenthesis. This was the answer: preg_match('/(?:^[^(:]+|(?<=^\\()[^\\s)]+)/', $var, $match); I need an improvement, I need to get either anything up to the first parenthesis/colon/quotation marks or the first word inside the first parenthesis. So if I have something like: $var = 'story "The Town in Hell"s Backyard'; // I get this: $match = 'story'; $var = "screenplay (based on)"; // I get this: $match = 'screenplay'; $var = "(play)"; // I get this: $match = 'play'; $var = "original screen"; // I get this: $match = 'original screen'; Thanks!

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  • parse string with regular exression

    - by llamerr
    I trying to parse this string: $right = '34601)S(1,6)[2] - 34601)(11)[2] + 34601)(3)[2,4]'; with following regexp: const word = '(\d{3}\d{2}\)S{0,1}\([^\)]*\)S{0,1}\[[^\]]*\])'; preg_match('/'.word.'{1}(?:\s{1}([+-]{1})\s{1}'.word.'){0,}/', $right, $matches); print_r($matches); i want to return array like this: Array ( [0] => 34601)S(1,6)[2] - 34601)(11)[2] + 34601)(3)[2,4] [1] => 34601)S(1,6)[2] [2] => - [3] => 34601)(11)[2] [4] => + [5] => 34601)(3)[2,4] ) but i return only following: Array ( [0] => 34601)S(1,6)[2] - 34601)(11)[2] + 34601)(3)[2,4] [1] => 34601)S(1,6)[2] [2] => + [3] => 34601)(3)[2,4] ) i think, its becouse of [^)]* or [^]]* in the word, but how i should correct regexp for matching this in another way? i tryied to specify it: \d+(?:[,#]\d+){0,} so word become const word = '(\d{3}\d{2}\)S{0,1}\(\d+(?:[,#]\d+){0,}\)S{0,1}\[\d+(?:[,#]\d+){0,}\])'; but it gives nothing

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  • Perl Regular expression remove double tabs, line breaks, white spaces

    - by Scoox
    Hi guys, I want to write a perl script that removes double tabs, line breaks and white spaces. What I have so far is: $txt=~s/\r//gs; $txt=~s/ +/ /gs; $txt=~s/\t+/\t/gs; $txt=~s/[\t\n]*\n/\n/gs; $txt=~s/\n+/\n/gs; But, 1. It's not beautiful. Should be possible to do that with far less regexps. 2. It just doesn't work and I really do not know why. It leaves some double tabs, white spaces and empty lines (i.e. lines with only a tab or whitespace) I could solve it with a while, but that is very slow and ugly. Any suggestions?

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  • Square Brackets in Python Regular Expressions (re.sub)

    - by user1479984
    I'm migrating wiki pages from the FlexWiki engine to the FOSwiki engine using Python regular expressions to handle the differences between the two engines' markup languages. The FlexWiki markup and the FOSwiki markup, for reference. Most of the conversion works very well, except when I try to convert the renamed links. Both wikis support renamed links in their markup. For example, Flexwiki uses: "Link To Wikipedia":[http://www.wikipedia.org/] FOSwiki uses: [[http://www.wikipedia.org/][Link To Wikipedia]] both of which produce something that looks like I'm using the regular expression renameLink = re.compile ("\"(?P<linkName>[^\"]+)\":\[(?P<linkTarget>[^\[\]]+)\]") to parse out the link elements from the FlexWiki markup, which after running through something like "Link Name":[LinkTarget] is reliably producing groups <linkName> = Link Name <linkTarget = LinkTarget My issue occurs when I try to use re.sub to insert the parsed content into the FOSwiki markup. My experience with regular expressions isn't anything to write home about, but I'm under the impression that, given the groups <linkName> = Link Name <linkTarget = LinkTarget a line like line = renameLink.sub ( "[[\g<linkTarget>][\g<linkName>]]" , line ) should produce [[LinkTarget][Link Name]] However, in the output to the text files I'm getting [[LinkTarget [[Link Name]] which breaks the renamed links. After a little bit of fiddling I managed a workaround, where line = renameLink.sub ( "[[\g<linkTarget>][ [\g<linkName>]]" , line ) produces [[LinkTarget][ [[Link Name]] which, when displayed in FOSwiki looks like <[[Link Name> <--- Which WORKS, but isn't very pretty. I've also tried line = renameLink.sub ( "[[\g<linkTarget>]" + "[\g<linkName>]]" , line ) which is producing [[linkTarget [[linkName]] There are probably thousands of instances of these renamed links in the pages I'm trying to convert, so fixing it by hand isn't any good. For the record I've run the script under Python 2.5.4 and Python 2.7.3, and gotten the same results. Am I missing something really obvious with the syntax? Or is there an easy workaround?

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  • Removing whitespace in Java string?

    - by waitinforatrain
    Hi guys, I'm writing a parser for some LISP files. I'm trying to get rid of leading whitespace in a string. The string contents are along the lines of: :FUNCTION (LAMBDA (DELTA PLASMA-IN-0) (IF (OR (>= #61=(+ (* 1 DELTA) PLASMA-IN-0) 100) (<= #61# 0)) PLASMA-IN-0 #61#)) The tabs are all printed as 4 spaces in the file, so I want to get rid of these leading tabs. I tried to do this: string.replaceAll("\\s{4}", " ") - but it had no effect at all on the string. Does anyone know what I'm doing wrong? Is it because it is a multi-line string? Thanks

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  • How do I process the largest match first in PHP?

    - by animuson
    Ok, so I tried searching around first but I didn't exactly know how to word this question or a search phrase. Let me explain. I have data that looks like this: <!-- data:start --> <!-- 0:start --> <!-- 0:start -->0,9<!-- 0:stop --> <!-- 1:start -->0,0<!-- 1:stop --> <!-- 2:start -->9,0<!-- 2:stop --> <!-- 3:start -->9,9<!-- 3:stop --> <!-- 4:start -->0,9<!-- 4:stop --> <!-- 0:stop --> <!-- 1:start --> <!-- 0:start -->1,5<!-- 0:stop --> <!-- 1:start -->1,6<!-- 1:stop --> <!-- 2:start -->3,6<!-- 2:stop --> <!-- 3:start -->3,8<!-- 3:stop --> <!-- 4:start -->4,8<!-- 4:stop --> <!-- 1:stop --> <!-- 2:start --> <!-- 0:start -->0,7<!-- 0:stop --> <!-- 1:start -->1,7<!-- 1:stop --> <!-- 2:stop --> <!-- data:stop --> So it's basically a bunch of points. Here is the code I'm currently using to try and parse it so that it would create an array like so: Array ( 0 => Array ( 0 => "0,9", 1 => "0,0", 2 => "9,0", 3 => "9,9", 4 => "0,9" ), 1 => Array ( 0 => "1,5", 1 => "1,6", 2 => "3,6", 3 => "3,8", 4 => "4,8" ), 2 => Array ( 0 => "0,7", 1 => "1,7" ) ) However, it is returning an array that looks like this: Array ( 0 => "0,9", 1 => "0,0", 2 => "9,0" ) Viewing the larger array that I have on my screen, you see that it's setting the first instance of that variable when matching. So how do I get it to find the widest match first and then process the insides. Here is the function I am currently using: function explosion($text) { $number = preg_match_all("/(<!-- ([\w]+):start -->)\n?(.*?)\n?(<!-- \\2:stop -->)/s", $text, $matches, PREG_SET_ORDER); if ($number == 0) return $text; else unset($item); foreach ($matches as $item) if (empty($data[$item[2]])) $data[$item[2]] = $this->explosion($item[3]); return $data; } I'm sure it will be something stupid and simple that I've overlooked, but that just makes it an easy answer for you I suppose.

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  • Retain Delimiters when Splitting String

    - by JoeC
    Edit: OK, I can't read, thanks to Col. Shrapnel for the help. If anyone comes here looking for the same thing to be answered... print_r(preg_split('/([\!|\?|\.|\!\?])/', $string, null, PREG_SPLIT_DELIM_CAPTURE)); Is there any way to split a string on a set of delimiters, and retain the position and character(s) of the delimiter after the split? For example, using delimiters of ! ? . !? turning this: $string = 'Hello. A question? How strange! Maybe even surreal!? Who knows.'; into this array('Hello', '.', 'A question', '?', 'How strange', '!', 'Maybe even surreal', '!?', 'Who knows', '.'); Currently I'm trying to use print_r(preg_split('/([\!|\?|\.|\!\?])/', $string)); to capture the delimiters as a subpattern, but I'm not having much luck.

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  • How would I create a VIM or Vi command to delete all text after a certain character for every line i

    - by Jason Down
    Scenario: I have a text file that has pipe (as in the "|" character) delimited data. Each field of data in the pipe delimited fields can be of variable length, so counting characters won't work (or using some sort of substring function... if that even exists in VIM). Is it possible, using VIM / Vi to delete all data from the second pipe to the end of the line for the entire file? There are approx 150,000 lines, so doing this manually would only be appealing to a masochist... e.g. Change the following lines from: 1111|random sized text 12345|more random data la la la|1111|abcde 2222|random sized text abcdefghijk|la la la la|2222|defgh 3333|random sized text|more random data|33333|ijklmnop to: 1111|random sized text 12345 2222|random sized text abcdefghijk 3333|random sized text I'm sure this can be done somehow... I hope. TIA UPDATE: I should have mentioned that I'm running this on Windows XP, so I don't have access to some of the mentioned *nix commands (CUT is not recognized on Windows).

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  • How to start matching and saving matched from exact point in a text

    - by yuliya
    I have a text and I write a parser for it using regular expressions and perl. I can match what I need with two empty lines (I use regexp), because there is a pattern that allows recognize blocks of text after two empty lines. But the problem is that the whole text has Introduction part and some text in the end I do not need. Here is a code which matches text when it finds two empty lines #!/usr/bin/perl use strict; use warnings; my $file = 'first'; open(my $fh, '<', $file); my $empty = 0; my $block_num = 1; open(OUT, '>', $block_num . '.txt'); while (my $line = <$fh>) { chomp ($line); if ($line =~ /^\s*$/) { $empty++; } elsif ($empty == 2) { close(OUT); open(OUT, '>', ++$block_num . '.txt'); $empty = 0; } else { $empty = 0;} print OUT "$line\n"; } close(OUT); This is example of the text I need (it's really small :)) this is file example I think that I need to iterate over the text till the moment it will find the word LOREM IPSUM with regexps this kind "/^LOREM IPSUM/", because it is the point from which needed text starts(and save the text in one file when i reach the word). And I need to finish iterating over the text when INDEX word is fount or save the text in separate file. How could I implement it. Should I use next function to proceed with lines or what? BR, Yuliya

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  • User will input some filter criteria -- how can I turn it into a regular expression for String.match

    - by envinyater
    I have a program where the user will enter a string such as PropertyA = "abc_*" and I need to have the asterisk match any character. In my code, I'm grabbing the property value and replacing PropertyA with the actual value. For instance, it could be abc_123. I also pull out the equality symbol into a variable. It should be able to cover this type of criteria PropertyB = 'cba' PropertyC != '*-this' valueFromHeader is the lefthand side and value is the righthand side. if (equality.equals("=")) { result = valueFromHeader.matches(value); } else if (equality.equals("!=")) { result = !valueFromHeader.matches(value); } EDIT: The existing code had this type of replacement for regular expressions final String ESC = "\\$1"; final String NON_ALPHA = "([^A-Za-z0-9@])"; final String WILD = "*"; final String WILD_RE_TEMP = "@"; final String WILD_RE = ".*"; value = value.replace(WILD, WILD_RE_TEMP); value = value.replaceAll(NON_ALPHA,ESC); value = value.replace(WILD_RE_TEMP, WILD_RE); It doesn't like the underscore here... abcSite_123 != abcSite_123 (evaluates to true) abcSite_123$1.matches("abcSite$1123") It doesn't like the underscore...

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  • regular expression: extract last 2 characters

    - by dotnet-practitioner
    what is the best way to extract last 2 characters of a string using regular expression. For example, I want to extract state code from the following "A_IL" I want to extract IL as string.. please provide me C# code on how to get it.. string fullexpression = "A_IL"; string StateCode = some regular expression code.... thanks

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  • Reading a line backwards

    - by Jimmy
    Hi, I'm using regular expression to count the total spaces in a line (first occurrence). match(/^\s*/)[0].length; However this reads it from the start to end, How can I read it from end to start. Thanks

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  • validation of special characters

    - by jpallavi
    I want to validate login name with special characters !@#S%^*()+_-?/<:"';. space using regular expression in ruby on rails. These special characters should not be acceptable. What is the code for that? Thanks, Pallavi

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