Search Results

Search found 10005 results on 401 pages for 'regex trouble'.

Page 122/401 | < Previous Page | 118 119 120 121 122 123 124 125 126 127 128 129  | Next Page >

  • Regexp look-behind to match internet speeds

    - by Sandman
    So the user may search for "10 mbit" after which I want to capture the "10" so I can use it in a speed-search rather than a string-search. This isn't a problem, the below regexp does this fine: if (preg_match("/(\d+)\smbit/", $string)){ ... } But, the user may search for something like "10/10 mbit" or "10-100 mbit". I don't want to match those with the above regexp - they should be handled in another fashion. So I would like a regexp that matches "10 mbit" if the number is all-numeric as a whole word (i.e. contained by whitespace, newline or lineend/linestart) Using lookbehind, I did this: if (preg_match("#(?<!/)(\d+)\s+mbit#i", $string)){ Just to catch those that doesn't have "/" before them, but this matched true for this string: "10/10 mbit" so I'm obviously doing something wrong here, but what?

    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

  • 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!

    Read the article

  • 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?

    Read the article

  • Regular Expression :match string containing only non repeating words

    - by nash
    I have this situation(Java code): 1) a string such as : "A wild adventure" should match. 2) a string with adjacent repeated words: "A wild wild adventure" shouldn't match. With this regular expression: .* \b(\w+)\b\s*\1\b.* i can match strings containing adjacent repeated words. How to reverse the situation i.e how to match strings which do not contain adjacent repeat words

    Read the article

  • 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 :)

    Read the article

  • How do I write this URL in Django?

    - by alex
    (r'^/(?P<the_param>[a-zA-z0-9_-]+)/$','myproject.myapp.views.myview'), How can I change this so that "the_param" accepts a URL(encoded) as a parameter? So, I want to pass a URL to it. mydomain.com/http%3A//google.com

    Read the article

  • 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.

    Read the article

  • Extract a sentence out of sentences separated by delimitors

    - by Laura
    Below is a sample line I have extracted from a website: below a satisfactory level; &quot;an off year for tennis&quot;; &quot;his performance was off&quot; The output displays as: below a satisfactory level; "an off year for tennis"; "his performance was off" I want to get only the first sentence "below a satisfactory level"; Here is the code I have tried after exploring many stackoverflow posts: $data=explode('; ',$str); echo $data[0]; But somehow it is not working. Thanks in advance.

    Read the article

  • 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

    Read the article

  • What is the RFC complicant and working regular expression to check if a string is a valid URL

    - by bestis
    There is question by the almost the same name already: What is the best regular expression to check if a string is a valid URL I don't understand this stackoverflow. It seems like I need reputation to comment an answer. As I don't have it, I don't know how to tell/ask that the proposed solution doesn't seem to work. So I'm forced to make a new question and ask for the solution this way? But that regexp seems to fail in input which has IPv6 address in it: For example facebook's IPv6 address: http://2620:0:1cfe:face:b00c::3/ Also link to localhost fails: http://::1/ Or is PHP to blame? /** * Validate URL - RFC 3987 (IRI) * * http://stackoverflow.com/questions/161738/what-is-the-best-regular-expression-to-check-if-a-string-is-a-valid-url * * @param string $str_url * @return boolean */ function is_url($str_url) { // RFC 3987 For absolute IRIs (internationalized): // @todo FIXME - Has bugs in IPv6 (http://2620:0:1cfe:face:b00c::3/) fails return (bool) preg_match('/^[a-z](?:[-a-z0-9\+\.])*:(?:\/\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:])*@)?(?:\[(?:(?:(?:[0-9a-f]{1,4}:){6}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|::(?:[0-9a-f]{1,4}:){5}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){4}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:[0-9a-f]{1,4}:[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){3}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,2}[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){2}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,3}[0-9a-f]{1,4})?::[0-9a-f]{1,4}:(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,4}[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,5}[0-9a-f]{1,4})?::[0-9a-f]{1,4}|(?:(?:[0-9a-f]{1,4}:){0,6}[0-9a-f]{1,4})?::)|v[0-9a-f]+[-a-z0-9\._~!\$&\'\(\)\*\+,;=:]+)\]|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3}|(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=@])*)(?::[0-9]*)?(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*|\/(?:(?:(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))+)(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*)?|(?:(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))+)(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*|(?!(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])))(?:\?(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])|[\x{E000}-\x{F8FF}\x{F0000}-\x{FFFFD}|\x{100000}-\x{10FFFD}\/\?])*)?(?:\#(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])|[\/\?])*)?$/iu',$str_url); } Here is the test for it: $urls=array('http://www.example.org/','http://www.example.org:80/','example.org','ftp://user:[email protected]/','http://example.org/?cat=5&test=joo','http://www.fi/?cat=5&amp;test=joo','http://::1/','http://2620:0:1cfe:face:b00c::3/','http://2620:0:1cfe:face:b00c::3:80/'); foreach ($urls as $a) { echo $a."\n"; $a=is_url($a); var_dump($a); } And that outputs: > `http://www.example.org/` bool(true) > `http://www.example.org:80/` bool(true) > example.org bool(false) > `ftp://user:[email protected]/` > bool(true) > `http://example.org/?cat=5&test=joo` > bool(true) > `http://www.fi/?cat=5&amp;test=joo` > bool(true) `http://::1/` bool(false) > `http://2620:0:1cfe:face:b00c::3/` > bool(false) > `http://2620:0:1cfe:face:b00c::3:80/` > bool(false) And it also seems that stackoverflow's code is miss behaving on those :) So what is the RFC compilicant and working regexp? ps. If you close this, please then tell me how this situation should be handled? I don't think that the answer is, just earn your reputation. Who wants to do that if they cannot even tell that some proposed solution isn't working correctly. pps. "we're sorry, but as a spam prevention mechanism, new users can only post a maximum of one hyperlink. Earn more than 10 reputation to post more hyperlinks.". Oh C'mon, I'm fine with plain text :D

    Read the article

  • Using s/// in an expression

    - by mikeY
    I got a headache looking for this: How do you use s/// in an expression as opposed to an assignment. To clarify what I mean, I'm looking for a perl equivalent of python's re.sub(...) when used in the following context: newstring = re.sub('ab', 'cd', oldstring) The only way I know how to do this in perl so far is: $oldstring =~ s/ab/cd/; $newstring = $oldstring; Note the extra assignment.

    Read the article

  • C# regular expression

    - by vert
    How would I write a regular expression (C#) which will check a given string to see if any of its characters are characters OTHER than the following: a-z A-Z Æ æ Å å Ø ø - ' Thanks!

    Read the article

  • How do you validate a URL with a regular expression in Python?

    - by Zachary Spencer
    I'm building a Google App Engine app, and I have a class to represent an RSS Feed. I have a method called setUrl which is part of the feed class. It accepts a url as an input. I'm trying to use the re python module to validate off of the RFC 3986 Reg-ex (http://www.ietf.org/rfc/rfc3986.txt) Below is a snipped which should work, right? I'm incredibly new to Python and have been beating my head against this for the past 3 days. p = re.compile('^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\?([^#]*))?(#(.*))?') m = p.match(url) if m: self.url = url return url

    Read the article

  • regular expression code

    - by Gaia Andreoletti
    Deal all, I need to find match between two tab delimited files files like this: File 1: ID1 1 65383896 65383896 G C PCNXL3 ID1 2 56788990 55678900 T A ACT1 ID1 1 56788990 55678900 T A PRO55 File 2 ID2 34 65383896 65383896 G C MET5 ID2 2 56788990 55678900 T A ACT1 ID2 2 56788990 55678900 T A HLA what I would like to do is to retrive the matching line between the two file. What I would like to match is everyting after the gene ID So far I have written this code but unfortunately perl keeps giving me the error: use of "Use of uninitialized value in pattern match (m//)" Could you please help me figure out where i am doing it wrong? Thank you in advance! use strict; open (INA, $ARGV[0]) || die "cannot to open gene file"; open (INB, $ARGV[1]) || die "cannot to open coding_annotated.var files"; my @sample1 = <INA>; my @sample2 = <INB>; foreach my $line (@sample1) { my @tab = split (/\t/, $line); my $chr = $tab[1]; my $start = $tab[2]; my $end = $tab[3]; my $ref = $tab[4]; my $alt = $tab[5]; my $name = $tab[6]; foreach my $item (@sample2){ my @fields = split (/\t/,$item); if ($fields[1]=~ m/$chr(.*)/ && $fields[2]=~ m/$start(.*)/ && $fields[4]=~ m/$ref(.*)/ && $fields[5]=~ m/$alt(.*)/&& $fields[6]=~ m/$name(.*)/){ print $line,"\n",$item; } } }

    Read the article

  • best REGEXP friendly Text Editors + most powerful REGEXP syntax?

    - by John
    I am fluent with Microsoft Visual 2005 regular expressions and they are a big time saver. I seem to learn them best by having a vaguely organized cheat sheet thrown at me, at which point I read just a little and play with them until I understand what's going on. That learning approach has worked well for me, for now. I would really like to take this to the next level though. Basically -- What is the REGEXP convention that is generally regarded as the most open-ended and powerful? VS2005 Regexps seem kind of gimped, so maybe I'm a kid playing in a sandbox. Are there text editors out there that can perform a highlight all matches, list lines containing string, or some kind of powerful function like that in conjunction with the very strongest REGEXP language? If not I can just use multiple programs and a weird technique but I'd like to avoid that. I wonder if a stronger REGEXP language or a "stronger" regEXP writer might be able to have his search match all results on all lines even by clicking a "find next" by adding some simple criteria to the search. Anyway, please provide advice!

    Read the article

  • Find and Replace with Notepad++

    - by Levi
    I have a document that was converted from PDF to HTML for use on a company website to be referenced and indexed for search. I'm attempting to format the converted document to meet my needs and in doing so I am attempting to clean up some of the junk that was pulled over from when it was a PDF such as page numbers, headers, and footers. luckily all of these lines that need to be removed are in blocks of 4 lines unfortunately they are not exactly the same therefore cannot be removed with a simple literal replace. The lines contain numbers which are incremental as they correlate with the pages. How can I remove the following example from my html file. Title<br> 10<br> <hr> <A name=11></a>Footer<br> I've tried many different regular expression attempts but as my skill in that area is limited I can't find the proper syntax. I'm sure i'm missing something fairly easy as it would seem all I need is a wildcard replace for the two numbers in the code and the rest is literal. any help is apprciated

    Read the article

  • Help with Regular Expression

    - by shivesh
    Hello I need help with Regular Expression, I want to match each section (number and it's text - 2 groups), the text can be multi line, each section ends when another section starts (another number) or when .END is reached or EOF. Demo Expression: \(\d{1,3}\) ([\s\S]*?)(\.END|\(\d{1,3}\)) Input text: (1) some text some text some text some text some text some text (2) some text some textsome text (3) some textsome text some textsome textsome text (4) some text .END first group should match number (with brackets) and second group should match corresponded text.

    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

  • How can I replace a line which contains only -------- by |||

    - by mimou
    I have something like: ------------------------------------------------------------------------ r2 | username | 2011-01-16 16:52:23 +0100 (Sun, 16 Jan 2011) | 1 line Changed paths: D /foo Removed foo ------------------------------------------------------------------------ r1 | username | 2011-01-16 16:51:03 +0100 (Sun, 16 Jan 2011) | 1 line Changed paths: A /foo created foo ------------------------------------------------------------------------ My target is to identify the file added by the "username" in a specific date. Thus, I need to have the combination (username, 16 Jan 2011, A) to insure that it is the right file ands then print foo. My idea is to: delete the white spaces change the newlines into | get rid of the --------------- and replace them with newlines but the problem is that I couldn't replace the ------- since they are mixed with other characters. ------------------------------------------------------------------------ |r2|username|2011-01-1616:52:23+0100(Sun,16Jan2011)|1line|Changedpaths:|D/foo|Removedfoo| ------------------------------------------------------------------------ |r1|username|2011-01-1616:51:03+0100(Sun,16Jan2011)|1line|Changedpaths:|A/foo|createdfoo| ------------------------------------------------------------------------ So I thought it would be a good idea to start by replacing the --------------- by a special character like ||| and then change this character by a newline using awk FS=||| OFS=\n Can anyone help me! thanks

    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

< Previous Page | 118 119 120 121 122 123 124 125 126 127 128 129  | Next Page >