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  • Regular expression to retrieve everything before first slash

    - by alex
    I need a regular expression to basically get the first part of a string, before the first slash (). For example in the following: C:\MyFolder\MyFile.zip The part I need is "C:" Another example: somebucketname\MyFolder\MyFile.zip I would need "somebucketname" I also need a regular expression to retrieve the "right hand" part of it, so everything after the first slash (excluding the slash.) For example somebucketname\MyFolder\MyFile.zip would return MyFolder\MyFile.zip.

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  • Confusion in RegExp Reluctant quantifier? Java

    - by Dusk
    Hi, Could anyone please tell me the reason of getting an output as: ab for the following RegExp code using Relcutant quantifier? Pattern p = Pattern.compile("abc*?"); Matcher m = p.matcher("abcfoo"); while(m.find()) System.out.println(m.group()); // ab and getting empty indices for the following code? Pattern p = Pattern.compile(".*?"); Matcher m = p.matcher("abcfoo"); while(m.find()) System.out.println(m.group());

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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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  • Django - urls.py - Filenames with a hash/pound (#) sign?

    - by miya
    I'm using django and realized that when the filename that the user wants to access (let's say a photo) has the pound sign, the entry in the url.py does not match. Any ideas? url(r'^static/(?P<path>.*)$', 'django.views.static.serve', {'document_root': MEDIA_ROOT}, it just says: "/home/user/project/static/upload/images/hello" does not exist when actually the name of the file is: hello#world.jpg Thanks, Nico

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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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  • Regular expressions in python unicode

    - by Remy
    I need to remove all the html tags from a given webpage data. I tried this using regular expressions: import urllib2 import re page = urllib2.urlopen("http://www.frugalrules.com") from bs4 import BeautifulSoup, NavigableString, Comment soup = BeautifulSoup(page) link = soup.find('link', type='application/rss+xml') print link['href'] rss = urllib2.urlopen(link['href']).read() souprss = BeautifulSoup(rss) description_tag = souprss.find_all('description') content_tag = souprss.find_all('content:encoded') print re.sub('<[^>]*>', '', content_tag) But the syntax of the re.sub is: re.sub(pattern, repl, string, count=0) So, I modified the code as (instead of the print statement above): for row in content_tag: print re.sub(ur"<[^>]*>",'',row,re.UNICODE But it gives the following error: Traceback (most recent call last): File "C:\beautifulsoup4-4.3.2\collocation.py", line 20, in <module> print re.sub(ur"<[^>]*>",'',row,re.UNICODE) File "C:\Python27\lib\re.py", line 151, in sub return _compile(pattern, flags).sub(repl, string, count) TypeError: expected string or buffer What am I doing wrong?

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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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  • 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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  • 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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  • jquery sortable with regexp

    - by Chris Lively
    I am trying to figure out the right regexp to match on list item id's. For example: <ul id="MyList" class="connectedSortable"> <li id="id=1-32">Item 1</li> <li id="id=2_23">Item 2</li> <li id="id=3">Item 3</li> <li id="id=4">Item 4</li> <li id="id=5">Item 5</li> <li id="id=6">Item 6</li> </ul> On the serialize method, I want it to pull everything after the equal sign (=) $(function () { $("#MyList, #OtherList").sortable({ connectWith: '.connectedSortable', update: function () { $("#MyListOrder").val($("#MyList").sortable('serialize', { regexp: '/(.+)[=](.+)/)' })); } }).disableSelection(); }); I tried the above, but that didn't quite work. My regexp expression is wrong and I don't know what it should be. Ideas?

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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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  • javascript split() array contains

    - by Mahesha999
    While learning JavaScript, I did not get why the output when we print the array returned of the Sting.split() method (with regular expression as an argument) is as explained below. var colorString = "red,blue,green,yellow"; var colors = colorString.split(/[^\,]+/); document.write(colors); //this print 7 times comma: ,,,,,,, However when I print individual element of the array colors, it prints an empty string, three commas and an empty string: document.write(colors[0]); //empty string document.write(colors[1]); //, document.write(colors[2]); //, document.write(colors[3]); //, document.write(colors[4]); //empty string document.write(colors[5]); //undefined document.write(colors[6]); //undefined Then, why printing the array directly gives seven commas. Though I think its correct to have three commas in the second output, I did not get why there is a starting (at index 0) and ending empty string (at index 4). Please explain I am screwed up here.

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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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  • 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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  • 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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  • 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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