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  • How to export the matches only in a pattern search in vim?

    - by Mert Nuhoglu
    Is there a way to grab and export the match part only in a pattern search without changing the current file? For example, from a file containing: 57","0","37","","http://www.thisamericanlife.org/Radio_Episode.aspx?episode=175" 58","0","37","","http://www.thisamericanlife.org/Radio_Episode.aspx?episode=170" I want to export a new file containing: http://www.thisamericanlife.org/Radio_Episode.aspx?episode=175 http://www.thisamericanlife.org/Radio_Episode.aspx?episode=170 I can do this by using substitution like this: :s/.\{-}\(http:\/\/.\{-}\)".\{-}/\1/g :%w>>data But the substitution command changes the current file. Is there a way to do this without changing the current file?

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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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  • 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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  • Writing a PHP web crawler using cron

    - by Horse
    Hi all I have written myself a web crawler using simplehtmldom, and have got the crawl process working quite nicely. It crawls the start page, adds all links into a database table, sets a session pointer, and meta refreshes the page to carry onto the next page. That keeps going until it runs out of links That works fine however obviously the crawl time for larger websites is pretty tedious. I wanted to be able to speed things up a bit though, and possibly make it a cron job. Any ideas on making it as quick and efficient as possible other than setting the memory limit / execution time higher?

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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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  • Jakarta Regexp 1.5 Backreferences?

    - by Matt Smith
    Why does this match: String str = "099.9 102.2" + (char) 0x0D; RE re = new RE("^([0-9]{3}.[0-9]) ([0-9]{3}.[0-9])\r$"); System.out.println(re.match(str)); But this does not: String str = "099.9 102.2" + (char) 0x0D; RE re = new RE("^([0-9]{3}.[0-9]) \1\r$"); System.out.println(re.match(str)); The back references don't seem to be working... What am I missing?

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  • Weird error using preg_match and unicode

    - by Thorpe Obazee
    if (preg_match('(\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+)', '2010/02/14/this-is-something')) { // do stuff } The above code works. However this one doesn't. if (preg_match('/\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+/u', '2010/02/14/this-is-something')) { // do stuff } Maybe someone could shed some light as to why the one below doesn't work. This is the error that is being produced: A PHP Error was encountered Severity: Warning Message: preg_match() [function.preg-match]: Unknown modifier '\'

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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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  • 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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  • Regular expression match, extracting only wanted segments of string

    - by Ben Carey
    I am trying to extract three segments from a string. As I am not particularly good with regular expressions, I think what I have done could probably be done better... I would like to extract the bold parts of the following string: SOMETEXT: ANYTHING_HERE (Old=ANYTHING_HERE, New=ANYTHING_HERE) Some examples could be: ABC: Some_Field (Old=,New=123) ABC: Some_Field (Old=ABCde,New=1234) ABC: Some_Field (Old=Hello World,New=Bye Bye World) So the above would return the following matches: $matches[0] = 'Some_Field'; $matches[1] = ''; $matches[2] = '123'; So far I have the following code: preg_match_all('/^([a-z]*\:(\s?)+)(.+)(\s?)+\(old=(.+)\,(\s?)+new=(.+)\)/i',$string,$matches); The issue with the above is that it returns a match for each separate segment of the string. I do not know how to ensure the string is the correct format using a regular expression without catching and storing the match if that makes sense? So, my question, if not already clear, how I can retrieve just the segments that I want from the above string?

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  • 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

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  • regexp for detect that the url doesn´t end with an extension

    - by devnieL
    Hello. I'm using this regular expression for detect if an url ends with a jpg : var exp = /(\b(https?|ftp|file):\/\/[-A-Z0-9+&@#\/%?=~_|!:,.;]*[-A-Z0-9+&@#\/%=~_|]*^\.jpg)/ig; it detects the url : e.g. http://www.blabla.com/sdsd.jpg but now i want to detect that the url doesn't ends with an jpg extension, i try with this : var exp = /(\b(https?|ftp|file):\/\/[-A-Z0-9+&@#\/%?=~_|!:,.;]*[-A-Z0-9+&@#\/%=~_|]*[^\.jpg]\b)/ig; but only get http://www.blabla.com/sdsd then i used this : var exp = /(\b(https?|ftp|file):\/\/[-A-Z0-9+&@#\/%?=~_|!:,.;]*[-A-Z0-9+&@#\/%=~_|]*[^\.jpg]$)/ig; it works if the url is alone, but dont work if the text is e.g. : http://www.blabla.com/sdsd.jpg text

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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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  • string substitution regular expression not working in tcl

    - by Puneet Mittal
    i am trying to replace all the special characters including white space, hyphen, etc, to underscore, from a string variable in tcl. I wrote the code below but it doesn't seem to be working. set varname $origVar puts "Variable Name :>> $varname" if {$varname != ""} { regsub -all {[\s-\]\[$^?+*()|\\%&#]} $varname "_" $newVar } puts "New Variable :>> $newVar" one issue is that, instead of replacing the string in $varname, it is replacing the data inside $origVar. No idea why, and also i read the example code (for proper syntax) in my tcl book and according to that it should be something like this regsub -all {[\s-][$^?+*()|\\%&#]} $varname "_" newVar so i used the same syntax but it didn't work and gave the same result as modifying the $origVar instead of required $varname value.

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  • What regular expression(s) would I use to remove escaped html from large sets of data.

    - by Elizabeth Buckwalter
    Our database is filled with articles retrieved from RSS feeds. I was unsure of what data I would be getting, and how much filtering was already setup (WP-O-Matic Wordpress plugin using the SimplePie library). This plugin does some basic encoding before insertion using Wordpress's built in post insert function which also does some filtering. I've figured out most of the filters before insertion, but now I have whacko data that I need to remove. This is an example of whacko data that I have data in one field which the content I want in the front, but this part removed which is at the end: <img src="http://feeds.feedburner.com/~ff/SoundOnTheSound?i=xFxEpT2Add0:xFbIkwGc-fk:V_sGLiPBpWU" border="0"></img> <img src="http://feeds.feedburner.com/~ff/SoundOnTheSound?d=qj6IDK7rITs" border="0"></img> &lt;img src=&quot;http://feeds.feedburner.com/~ff/SoundOnTheSound?i=xFxEpT2Add0:xFbIkwGc-fk:D7DqB2pKExk&quot; Notice how some of the images are escape and some aren't. I believe this has to do with the last part being cut off so as to be unrecognizable as an html tag, which then caused it to be html endcoded. Another field has only this which is now filtered before insertion, but I have to get rid of the others: &lt;img src=&quot;http://farm3.static.flickr.com/2183/2289902369_1d95bcdb85.jpg&quot; alt=&quot;post_img&quot; width=&quot;80&quot; (all examples are on one line, but broken up for readability) Question: What is the best way to work with the above escaped html (or portion of an html tag)? I can do it in Perl, PHP, SQL, Ruby, and even Python. I believe Perl to be the best at text parsing, so that's why I used the Perl tag. And PHP times out on large database operations, so that's pretty much out unless I wanted to do batch processing and what not. PS One of the nice things about using Wordpress's insert post function, is that if you use php's strip_tags function to strip out all html, insert post function will insert <p> at the paragraph points. Let me know if there's anything more that I can answer. Some article that didn't quite answer my questions. (http://stackoverflow.com/questions/2016751/remove-text-from-within-a-database-text-field) (http://stackoverflow.com/questions/462831/regular-expression-to-escape-html-ampersands-while-respecting-cdata)

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