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  • Why does this Bash regex match return an Exit Status of "2"?

    - by PreservedMoose
    I'm writing a Bash script that needs to scan for the existence of non-ASCII characters in filenames. I'm using the POSIX bracket regex syntax to match the non-ASCII characters, but for some reason, when I test for the match in an if/then statement, the test always returns an Exit Status of 2, and never matches my test string. Here's the code in question: FILEREQ_SOURCEFILE="Filename–WithNonAScII-Charàcters-05sec_23.98.mov" REGEX_MATCH_NONASCII="[^[:ascii:]]" if [[ $FILEREQ_SOURCEFILE =~ $REGEX_MATCH_NONASCII ]]; then echo "Exit Status: $?" echo "Matched!" else echo "Exit Status: $?" echo "No Match" fi This code always returns: Exit Status: 2 No Match I've read and re-read the bash-hackers.org explanation of how regex matching works, as well as this previous question on SO regarding matching non-ASCII characters, but for the life of me, I can't get this to work. What am I missing here?

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  • How can I include a line until # but without the # when parsing 'sources.list' with regex?

    - by stwissel
    I want to parse my sources.list to extract the list of repositories. I have: ## Some comment deb http://some.vendor.com/ubuntu precise stable deb-src http://some.vendor.com/ubuntu precise stable deb http://some.othervendor.com/ubuntu precise experimental # my current favorite I want: http://some.vendor.com/ubuntu precise stable http://some.othervendor.com/ubuntu precise experimental So I need: only lines with deb at the beginning and until the end of the line or a # character but excluding it. So far I have: grep -o "^deb .*" But how to match # or LineEnd and excluding the #?

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  • How can I exclude a file in a folder from basic auth (regex help)?

    - by simon180
    Hi I have a folder on my site which contains admin files and I've added basic auth following a little unwanted attention. This works fine however a couple of the admin functions won't work through basic auth as they handle file uploads and so I want to exclude these files from the auth. It shouldn't have any security implications as any rogue user wouldn't be able to access the pages that could create a session to use these functions. I am using the following basic code to exclude a file: <FilesMatch "(index.php\/myadminfolder\/myurl\/myaction/someotherstuff?)$"> Satisfy Any Order allow,deny Allow from all Deny from none </FilesMatch> The URL exclusion is not working. The URL to exclude is in the form: index.php/directory/subdirectory/action/uniqueid/blah What is the correct URL string to add to FilesMatch to exclude any files that start with the pattern of index.php/directory/subdirectory/action - regardless of what comes after action? Thanks Simon

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  • Trouble recording unique regex output to array in perl

    - by Structure
    The goal of the following code sample is to read the contents of $target and assign all unique regex search results to an array. I have confirmed my regex statement works so I am simplifying that so as not to focus on it. When I execute the script I get a list of all the regex results, however, the results are not unique which leads me to believe that my manipulation of the array or my if (grep{$_ eq $1} @array) { check is causing a problem(s). #!/usr/bin/env perl $target = "string to search"; $inc = 0; $once = 1; while ($target =~ m/(regex)/g) { #While a regex result is returned if ($once) { #If $once is not equal to zero @array[$inc] = $1; #Set the first regex result equal to @array[0] $once = 0; #Set $once equal to zero so this is not executed more than once } else { if (grep{$_ eq $1 } @array ) { #From the second regex result, check to see if the result is already in the array #If so, do nothing } else { @array[$inc] = $1; #If it is not, then assign the regex search result to the next unused position in the array in any position. $inc++; #Increment to next unused array position. } } } print @array; exit 0;

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  • How I use RegExp in my Java program? [migrated]

    - by MIH1406
    I have the following string examples: 00001 1 12 123 00002 3 7 321 00003 99 23 332 00004 192 50 912 In a separate text file. Numbers are separated by tabs not spaces. I tried to read the file and print each line if it matches a given RegExp, but I could not find the suitable RegExp for these lines. private static void readFile() { String fileName = "processes.lst"; FileReader file = null; String result = ""; try { file = new FileReader(fileName); BufferedReader reader = new BufferedReader(file); String line = null; String regEx = "[0-9]\t[0-9]\t[0-9]\t[0-9]"; while((line = reader.readLine()) != null) { if(line.matches(regEx)) { result += "\n" + line; } } } catch(Exception e) { System.out.println(e.getMessage()); } finally { if(file != null) try { file.close(); } catch(Exception e) { System.out.println(e.getMessage()); } } System.out.println(result); } I ended up without any string being printed!!

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  • How to write regex in asp.net MVC 3 razor

    - by anirudha
    here is a small trick to write regex and exceptional code in MVC3. first trick is use @@ instead of @ it’s work don’ worry output goes @ not @@. second trick is that you can use <text></text> tag in MVC to use some code who give error because viewengine does not accept them. like var pattern = new RegExp(/^(("[\w-\s]+")|([\w-]+(?:\.[\w-]+)*)|("[\w-\s]+")([\w-]+(?:\.[\w-]+)*))(@((?:[\w-]+\.)*\w[\w-]{0,66})\.([a-z]{2,6}(?:\.[a-z]{2})?)$)|(@\[?((25[0-5]\.|2[0-4][0-9]\.|1[0-9]{2}\.|[0-9]{1,2}\.))((25[0-5]|2[0-4][0-9]|1[0-9]{2}|[0-9]{1,2})\.){2}(25[0-5]|2[0-4][0-9]|1[0-9]{2}|[0-9]{1,2})\]?$)/i);   this regex not work because it’s use @ so replace them to @@ e var pattern = new RegExp(/^(("[\w-\s]+")|([\w-]+(?:\.[\w-]+)*)|("[\w-\s]+")([\w-]+(?:\.[\w-]+)*))(@@((?:[\w-]+\.)*\w[\w-]{0,66})\.([a-z]{2,6}(?:\.[a-z]{2})?)$)|(@@\[?((25[0-5]\.|2[0-4][0-9]\.|1[0-9]{2}\.|[0-9]{1,2}\.))((25[0-5]|2[0-4][0-9]|1[0-9]{2}|[0-9]{1,2})\.){2}(25[0-5]|2[0-4][0-9]|1[0-9]{2}|[0-9]{1,2})\]?$)/i); now it will work. second way is that use text tag in MVC to make them work like <text> function isValidEmailAddress(emailAddress) { var pattern = new RegExp(/^(("[\w-\s]+")|([\w-]+(?:\.[\w-]+)*)|("[\w-\s]+")([\w-]+(?:\.[\w-]+)*))(@((?:[\w-]+\.)*\w[\w-]{0,66})\.([a-z]{2,6}(?:\.[a-z]{2})?)$)|(@\[?((25[0-5]\.|2[0-4][0-9]\.|1[0-9]{2}\.|[0-9]{1,2}\.))((25[0-5]|2[0-4][0-9]|1[0-9]{2}|[0-9]{1,2})\.){2}(25[0-5]|2[0-4][0-9]|1[0-9]{2}|[0-9]{1,2})\]?$)/i); return pattern.test(emailAddress); } </text>

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  • How far should one take e-mail address validation?

    - by Mike Tomasello
    I'm wondering how far people should take the validation of e-mail address. My field is primarily web-development, but this applies anywhere. I've seen a few approaches: simply checking if there is an "@" present, which is dead simply but of course not that reliable. a more complex regex test for standard e-mail formats a full regex against RFC 2822 - the problem with this is that often an e-mail address might be valid but it is probably not what the user meant DNS validation SMTP validation As many people might know (but many don't), e-mail addresses can have a lot of strange variation that most people don't usually consider (see RFC 2822 3.4.1), but you have to think about the goals of your validation: are you simply trying to ensure that an e-mail address can be sent to an address, or that it is what the user probably meant to put in (which is unlikely in a lot of the more obscure cases of otherwise 'valid' addresses). An option I've considered is simply giving a warning with a more esoteric address but still allowing the request to go through, but this does add more complexity to a form and most users are likely to be confused. While DNS validation / SMTP validation seem like no-brainers, I foresee problems where the DNS server/SMTP server is temporarily down and a user is unable to register somewhere, or the user's SMTP server doesn't support the required features. How might some experienced developers out here handle this? Are there any other approaches than the ones I've listed? Edit: I completely forgot the most obvious of all, sending a confirmation e-mail! Thanks to answerers for pointing that one out. Yes, this one is pretty foolproof, but it does require extra hassle on the part of everyone involved. The user has to fetch some e-mail, and the developer needs to remember user data before they're even confirmed as valid.

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  • how to use a regex to search backwards effectively?

    - by Asaf
    hi, i'm searching forward in an array of strings with a regex, like this: for (int j = line; j < lines.length; j++) { if (lines[j] == null || lines[j].isEmpty()) { continue; } matcher = pattern.matcher(lines[j]); if (matcher.find(offset)) { offset = matcher.end(); line = j; System.out.println("found \""+matcher.group()+"\" at line "+line+" ["+matcher.start()+","+offset+"]"); return true; } offset = 0; } return false; note that in my implementation above i save the line and offset for continuous searches. anyway, now i want to search backwards from that [line,offset]. my question: is there a way to search backwards with a regex efficiently? if not, what could be an alternative? 10x, asaf :-) clarification: by backwards i mean finding the previous match. for example, say that i'm searching for "dana" in "dana nama? dana kama! lama dana kama?" and got to the 2nd match. if i do matcher.find() again, i'll search forward and get the 3rd match. but i want to seach backwards and get to the 1st match. the code above should then output something like: found "dana" at line 0 [0,3] // fwd found "dana" at line 0 [11,14] // fwd found "dana" at line 0 [0,3] // bwd

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  • Help write regex that will surround certain text with <strong> tags, only if the <strong> tag isn't

    - by sahil
    I have several posts on a website; all these posts are chat conversations of this type: AD: Hey! BC: What's up? AD: Nothing BC: Okay They're marked up as simple paragraphs surrounded by <p> tags. Using the javascript replace function, I want all instances of "AD" in the beginning of a conversation (ie, all instances of "AD" at the starting of a line followed by a ":") to be surrounded by <strong> tags, but only if the instance isn't already surrounded by a <strong> tag. What regex should I use to accomplish this? Am I trying to do what this advises against? The code I'm using is like this: var posts = document.getElementsByClassName('entry-content'); for (var i = 0; i < posts.length; i++) { posts[i].innerHTML = posts[i].innerHTML.replace(/some regex here/, 'replaced content here'); }

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  • C# Find and Replace a section of a string with wildcard type search using RegEx (while retaining som

    - by fraXis
    Hello, I am trying to replace some text in a string with different text while retaining some of the text and I am having trouble doing it. My code is: StreamReader reader = new StreamReader(fDialog.FileName.ToString()); string content = reader.ReadToEnd(); reader.Close(); /Replace M2 with M3 (this works fine) content = Regex.Replace(content, "M2", "M3"); I want to replace a string that contains this: Z0.1G0H1E1 and turn it into: G54G43Z.1H1M08 (Note the Z value and the H value contain the same numeric value before the text change) The trouble I am having is that when I replace the values, I need to retain the H value and the Z value from the first set of text. For example, Z0.5G0H5E1 I need to add the new text, but also add the H5 and Z0.5 back into the text such as: G54G43Z0.5H5M08 But the Z values and H values will be different every time, so I need to capture those values and reinsert them back into the string when add the new G54G43 values. Can someone please show me how to do this using Regex.Replace? Thanks so much, Shawn

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  • Why does this data YYYY-MM-DD regex fail in Java?

    - by ProfessionalAmateur
    Hello StackOverFlow, My first question and Im excited... I've lurked since go-live and love the site, however I apologize for any newbie errors, formatting, etc... I'm attempting to validate the format of a string field that contains a date in Java. We will receive the date in a string, I will validate its format before parsing it into a real Date object. The format being passed in is in the YYYY-MM-DD format. However I'm stuck on one of my tests, if I pass in "1999-12-33" the test will fail (as it should with a day number 33) with this incomplete pattern: ((19|20)\\d{2})-([1-9]|0[1-9]|1[0-2])-([12][0-9]|3[01]) However as soon as I add the characters in bold below it passes the test (but should not) ((19|20)\\d{2})-([1-9]|0[1-9]|1[0-2])-(0[1-9]|[1-9]|[12][0-9]|3[01]) *additional note, I know I can change the 0[1-9]|[1-9] into 0?[1-9] but I wanted to break everything down to its most simple format to try and find why this isn't working. Here is the scrap test I've put together to run through all the different date scenarios: import java.util.regex.Matcher; import java.util.regex.Pattern; public class scrapTest { public scrapTest() { } public static void main(String[] args) { scrapTest a = new scrapTest(); boolean flag = a.verfiyDateFormat("1999-12-33"); } private boolean verfiyDateFormat(String dateStr){ Pattern datePattern = Pattern.compile("((19|20)\\d{2})-([1-9]|0[1-9]|1[0-2])-(0[1-9]|[1-9]|[12][0-9]|3[01])"); Matcher dateMatcher = datePattern.matcher(dateStr); if(!dateMatcher.find()){ System.out.println("Invalid date format!!! -> " + dateStr); return false; } System.out.println("Valid date format."); return true; } } Ive been programming for ~10 years but extremely new to Java, so please feel free to explain anything as elementary as you see fit.

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  • PHP Regex to match lines with all-caps with occaisional hyphens.

    - by Yaaqov
    I'm trying to to convert an existing PHP Regular Expression match case to apply to a slightly different style of document. Here's the original style of the document: **FOODS - TYPE A** ___________________________________ **PRODUCT** 1) Mi Pueblito Queso Fresco Authentic Mexican Style Fresh Cheese; 2) La Fe String Cheese **CODE** Sell by date going back to February 1, 2009 And the successfully-running PHP Regex match code that only returns "true" if the line is surrounded by asterisks, and stores each side of the "-" as $m[1] and $m[2], respectively. if ( preg_match('#^\*\*([^-]+)(?:-(.*))?\*\*$#', $line, $m) ) { // only for **header - subheader** $m[2] is set. if ( isset($m[2]) ) { return array(TYPE_HEADER, array(trim($m[1]), trim($m[2]))); } else { return array(TYPE_KEY, array($m[1])); } } So, for line 1: $m[1] = "FOODS" AND $m[2] = "TYPE A"; Line 2 would be skipped; Line 3: $m[1] = "PRODUCT", etc. The question: How would I re-write the above regex match if the headers did not have the asterisks, but still was all-caps, and was at least 4 characters long? For example: FOODS - TYPE A ___________________________________ PRODUCT 1) Mi Pueblito Queso Fresco Authentic Mexican Style Fresh Cheese; 2) La Fe String Cheese CODE Sell by date going back to February 1, 2009 Thank you.

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  • A regex I have working in Ruby doesn't in PHP; what could the cause be?

    - by Alex R
    I do not know ruby. I am trying to use the following regex that was generated by ruby (namely by http://www.a-k-r.org/abnf/ running on the grammar given rfc1738) in php. It is failing to match in php, but it is successfully matching in ruby. Does anyone see what differences between php's and ruby's handling of regexes that might explain this discrepancy? http:\/\/(?:(?:(?:(?:[0-9a-z]|[0-9a-z](?:[\x2d0-9a-z]?)*[0-9a-z])\x2e)?)*(?:[a-z]|[a-z](?:[\x2d0-9a-z]?)*[0-9a-z])|\d+\x2e\d+\x2e\d+\x2e\d+)(?::\d+)?(?:\/(?:(?:[!\x24'-\x2e0-9_a-z]|%[0-9a-f][0-9a-f]|[&:;=@])?)*(?:(?:\/(?:(?:[!\x24'-\x2e0-9_a-z]|%[0-9a-f][0-9a-f]|[&:;=@])?)*)?)*(?:\x3f(?:(?:[!\x24'-\x2e0-9_a-z]|%[0-9a-f][0-9a-f]|[&:;=@])?)*)?)?/i Since you all love regexes so much, how about an alternate solution. Given the ABNF in an rfc, I want a way (in php) to check if an arbitrary string is in the grammar. APG fails to compile on a 64-bit system, VTC is not Free, and I have not found any other such tools. I would also prefer not to use a regex, but it's the closest I've come to success.

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  • How do I get rid of this "(" using regex?

    - by Solignis
    Hi there, I was moving along on a regex expression and I have hit a road block I can't seem to get around. I am trying to get rid of "(" in the middle of a line of text using regex, there were 2 but I figured out how to get the one on the end of the line. its the one in the middle I can hack out. Here is the snippet I am searching for in the config file. I put 2 examples. guestOSAltName = "Ubuntu Linux (64-bit)" guestOSAltName = "Microsoft Windows 2000 Professional" Here is the snippet I am working on. if ($vmx_file =~ m/^\bguestOSAltName\b\s+\S\s+\W(?<GUEST_OS> .+[^")])\W/xm) { $virtual_machines{$vm}{"OS"} = "$+{GUEST_OS}"; } else { $virtual_machines{$vm}{"OS"} = "N/A"; } I am thinking the problem is I cannot make a match to "(" because the expression before that is to ".+" so that it matches everything in the line of text, be it alphanumeric or whitespace or even symbols like hypens. Any ideas how I can get this to work? This is what I am getting for an output from a hash dump. $VAR1 = { 'NS02' => { 'ID' => '144', 'Version' => '7', 'OS' => 'Ubuntu Linux (64-bit', 'VMX' => '/vmfs/volumes/datastore2/NS02/NS02.vmx', 'Architecture' => '64-bit' },

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  • Regular exp to validate email in C

    - by Liju Mathew
    Hi, We need to write a email validation program in C. We are planning to use GNU Cregex.h) regular expression. The regular expression we prepared is [a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])? But the below code is failing while compiling the regex. #include <stdio.h> #include <regex.h> int main(const char *argv, int argc) { const char *reg_exp = "[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?"; int status = 1; char email[71]; regex_t preg; int rc; printf("The regex = %s\n", reg_exp); rc = regcomp(&preg, reg_exp, REG_EXTENDED|REG_NOSUB); if (rc != 0) { if (rc == REG_BADPAT || rc == REG_ECOLLATE) fprintf(stderr, "Bad Regex/Collate\n"); if (rc == REG_ECTYPE) fprintf(stderr, "Invalid Char\n"); if (rc == REG_EESCAPE) fprintf(stderr, "Trailing \\\n"); if (rc == REG_ESUBREG || rc == REG_EBRACK) fprintf(stderr, "Invalid number/[] error\n"); if (rc == REG_EPAREN || rc == REG_EBRACE) fprintf(stderr, "Paren/Bracket error\n"); if (rc == REG_BADBR || rc == REG_ERANGE) fprintf(stderr, "{} content invalid/Invalid endpoint\n"); if (rc == REG_ESPACE) fprintf(stderr, "Memory error\n"); if (rc == REG_BADRPT) fprintf(stderr, "Invalid regex\n"); fprintf(stderr, "%s: Failed to compile the regular expression:%d\n", __func__, rc); return 1; } while (status) { fgets(email, sizeof(email), stdin); status = email[0]-48; rc = regexec(&preg, email, (size_t)0, NULL, 0); if (rc == 0) { fprintf(stderr, "%s: The regular expression is a match\n", __func__); } else { fprintf(stderr, "%s: The regular expression is not a match: %d\n", __func__, rc); } } regfree(&preg); return 0; } The regex compilation is failing with the below error. The regex = [a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])? Invalid regex main: Failed to compile the regular expression:13 What is the cause of this error? Whether the regex need to be modified? Thanks, Mathew Liju

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  • Minify CSS using preg_replace

    - by x3ro
    Hey folks, I'm trying to minify multiple CSS files using preg_replace. Actually, I'm only trying to remove any linebreaks/tabs and comments from the file. the following works for me: $regex = array('{\t|\r|\n}', '{(/\*(.*?)\*/)}'); echo preg_replace($regex, '', file_get_contents($file)); But I'd like to do it in a single multiline regex, like this: $regex = <<<EOF {( \t | \r | \n | /\*(.*?)\*/ )}x EOF; echo preg_replace($regex, '', file_get_contents($file)); However, this does not do anything at all. Is there any way to do this? Regards, x3ro Edit: Ok, so I'll take a look at existing minifiers, but it still leaves me with the question how I would do a multiline regex like this, because with the x-modifier multiline regexs should work fine even in php, shouldn't they?

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  • How to extract digit from the string using regex?

    - by Harikrishna
    I will have a different type of string(string will not have fixed format,they will be different every time) from them I want to remove some specific substring.Like the string can be OPTIDX 26FEB2009 NIFTY CE 2500 OPTIDX NIFTY 30 Jul 2009 4600.00 PE OPTSTK ICICIBANK 30 Jul 2009 700.00 PA I want to extract Rs.(digit) from those string and store it into one variable and then in those string there should not be Rs.(digit). What should be the regex for that to match Rupees ?

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  • Single Regex for filtering roman numerals from the text files.

    - by Nains
    I am stuck in between of a problem where only one pass of regular expression is allowed( some old hard code). I need the regex for roman numerals. I have tried the standard one i.e. ^(?i)M*(D?C{0,3}|C[DM])(L?X{0,3}|X[LC])(V?I{0,3}|I[VX])$, but the problem is it allows null( '') values also. Is there any way around to check is problem?

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