Search Results

Search found 9970 results on 399 pages for 'regular john'.

Page 27/399 | < Previous Page | 23 24 25 26 27 28 29 30 31 32 33 34  | Next Page >

• 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

• PHP & Regular expression: keyword just occurs once

  - by lauthiamkok

  Hi, how can I make sure a certain keyword just occurs once in the input with regular expression? I think there is some mistakes in the expression below as I can repeat the same keywords, if (!preg_match('/\b(.php?){1}\b/', $cfg_path)) { $error = true; echo '<error elementid="cfg_path" message="PATH - make sure you have a \'.php?\' in the path."/>'; } I just want this to be true, form.php?category=something or form.php? but not this, form.php?.php?category=something or form.php?.php? please let me know how to fix it. thanks.

  Read the article

• 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?

  Read the article

• Which is more efficient regular expression?

  - by Vagnerr

  I'm parsing some big log files and have some very simple string matches for example if(m/Some String Pattern/o){ #Do something } It seems simple enough but in fact most of the matches I have could be against the start of the line, but the match would be "longer" for example if(m/^Initial static string that matches Some String Pattern/o){ #Do something } Obviously this is a longer regular expression and so more work to match. However I can use the start of line anchor which would allow an expression to be discarded as a failed match sooner. It is my hunch that the latter would be more efficient. Can any one back me up/shoot me down :-)

  Read the article

• Dealing with regular expressions, Python

  - by Gusto

  I want to remove some symbols from a string using a regular expression, for example: == (that occur both at the beginning and at the end of a line), * (at the beginning of a line ONLY). def some_func(): clean = re.sub(r'= {2,}', '', clean) #Removes 2 or more occurrences of = at the beg and at the end of a line. clean = re.sub(r'^\* {1,}', '', clean) #Removes 1 or more occurrences of * at the beginning of a line. What's wrong with my code? It seems like expressions are wrong. How do I remove a character/symbol if it's at the beginning or at the end of the line (with one or more occurrences)?

  Read the article

• Regular expression in Ruby

  - by Sainath Mallidi

  Hi, Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles. The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string. I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.

  Read the article

• Regular non-low energy bluetooth for IOS

  - by user3712524

  Ok, so I can't get my app to connect with a peripheral using bluetooth low-energy(my app is installed on an iPhone 5s and I have LightBlue installed on an iPhone 4S, simulating a peripheral. I also have light blue installed on the 5s, simulating a central, to verify that the two can connect and they do.) I have verified that the centralManager is instantiating and the State is powered on. So, my question is, is regular bluetooth still a viable option? Can I just use the GameKit API and go that route? Thanks in advance!

  Read the article

• Regular Expression - Password Validation is not working

  - by Kesavan

  Hi, I have to validate the password using regex. The password rule is like at least 1 uppercase and at least 2 numeric. It works fine except if the character comes at the end of the string. The regular expression which i am using is "^(?=.*\d.{2})(?=.*[A-Z].{1})(?=.*[@#$%^&+=].{2}).{8,12}$" Rules: minimum length = 8 minimum uppercase = 1 minimum numeric = 2 minimum special character = 1 It works for Test123$$, Test$123, TEST123$s, Test123$1, Test12$3 but it fails if the character specified comes at the end of the string like Test123$, Test$a12, Test12aa@, 123aa@@T. Please let me know if there is any fix for this.

  Read the article

• Jointure in linq with a regular expression

  - by Graveen

  I'm actually using a join in linqtosql (via dblinq). I'm trying to include a regular expression in the join part of the linq query. from i in collectiona join j in collectionb on Regex.IsMatch(i.name, j.jokered_name) equals true (...) I agree i can push the RegExp check in the where part of the linq query, but i was wondering if it is possible in the join part ? The above code wants an "i equals j" code structure. One thing i think to perform is overriding Equals() which 'll contains the RegEx.IsMatch() stuff and put a simple i equals j in the join part. Any suggestions about my problem ?

  Read the article

• how to write regular expression to validate a string using regex in C#

  - by Charu

  I need to validate a user input based on condition. i wrote a regular expression to do so, but it's failing not sure why. Can somebody point where i am making mistake? Regex AccuracyCodeHexRegex = new Regex(@"^[PTQA]((0|8)[01234567]){2}$"); This is what i am trying to validate(If the string is a subset of these strings then it is valid): Phh, Thh, Qhh, Ahh where 'h' is a hex digit in the set {00, 80, 01, 81, 02, 82, 03, 83, 04, 84, 05, 85, 06, 86, 07, 87} Ex: P00 is valid P20 is not valid

  Read the article

• Mutually exclusive regular expressions

  - by CaptnCraig

  If I have a list of regular expressions, is there an easy way to determine that no two of them will both return a match for the same string? That is, the list is valid if and only if for all strings a maximum of one item in the list will match the entire string. It seems like this will be very hard (maybe impossible?) to prove definitively, but I can't seem to find any work on the subject. The reason I ask is that I am working on a tokenizer that accepts regexes, and I would like to ensure only one token at a time can match the head of the input.

  Read the article

• Complex regular expression

  - by Jose3d

  Hello, i will like to capture a substring part of a text choosing the number of characters but if any word is cut then get until de last blank. As example if this is the text: "This is an example of text lorem ipsum, etc..." and i would like to get for instance 12 characters that are: "This is an e". In this case example is cutted, then i would like to get "This is an". Its possible do this with Regular Expressions? Thanks in advance. Jose

  Read the article

• Regular Expression to match unlimited number of options

  - by Pekka

  I want to be able to parse file paths like this one: /var/www/index.(htm|html|php|shtml) into an ordered array: array("htm", "html", "php", "shtml") and then produce a list of alternatives: /var/www/index.htm /var/www/index.html /var/www/index.php /var/www/index.shtml Right now, I have a preg_match statement that can split two alternatives: preg_match_all ("/\(([^)]*)\|([^)]*)\)/", $path_resource, $matches); Could somebody give me a pointer how to extend this to accept an unlimited number of alternatives (at least two)? Just regarding the regular expression, the rest I can deal with. The rule is: The list needs to start with a ( and close with a ) There must be one | in the list (i.e. at least two alternatives) Any other occurrence(s) of ( or ) are to remain untouched.

  Read the article

• Regular Expression: Changes HTML Attributes Value to some pattern

  - by brain90

  Dear Engineers, I'm a newbie in RegEx I have thousands html tags, have wrote like this: <input type="text" name="CustomerName" /> <input type="text" name="SalesOrder"/> I need to match every name attribute values and convert them all to be like this: CustomerName -> cust[customer_name] SalesOrder -> cust[sales_order] So the results will be : <input type="text" name="cust[customer_name]" /> <input type="text" name="cust[sales_order]" /> My best try have stuck in this pattern: name=\"[a-zA-Z0-9]*\" - just found name="CustomerName" Please guide me wrote some Regular Expression magics to done this, I'm using Netbeans PDT. Thanks in advance for any pointers!.

  Read the article

• Constructing a regular expression to wrap images with <a>

  - by bobo

  A web page contains lots of image elements: <img src="myImage.gif" width="180" height="18" /> But they may not be very well-formed, for example, the width or height attribute may be missing. And it also may not be properly closed with /. The src attribute is always there. I need a regular expression that wraps these with a hyperlink having href set to the src of the img. <a href="myImage.gif" target="_blank"><img src="myImage.gif" width="180" height="18" /></a> I can successfully locate the images using this regexp in this editor: http://gskinner.com/RegExr/: <img src="([^<]*)"[^<]*> But what is the next step?

  Read the article

• Regular expression to truncate a String

  - by user470184

  To truncate a String here is what I'm using : String test1 = "this is test truncation 1.pdf"; String p1 = test1.substring(0, 10) + "..."; System.out.println(p1); The output is 'this is te...' How can I access the file name extension so that output becomes : 'this is te... pdf' I could use substring method to access the last three characters but other file extensions could be 4 chars in length such as .aspx Is there a regular expression I can use so that "this is test truncation 1.pdf" becomes "this is te... pdf"

  Read the article

• Using Regular Expressions

  - by bebeTech

  I am having problems trying to use the regular expression that I used in JavaScript. On a web page, you may have: <b>Renewal Date:</b> 03 May 2010</td> I just want to be able to pull out the 03 May 2010, remembering that a webpage has more than just the above content. The way I currently perform this using JavaScript is: DateStr = /<b>Renewal Date:<\/b>(.+?)<\/td>/.exec(returnedHTMLPage); I tried to follow some tutorials on java.util.regex.Pattern and java.util.regex.Matcher with no luck. I can't seem to be able to translate (.+?) into something they can understand?? thanks, Noeneel

  Read the article

• Regular expressions in java

  - by rookie

  String s= "(See <a href=\"/wiki/Grass_fed_beef\" title=\"Grass fed beef\" " + "class=\"mw-redirect\">grass fed beef.) They have been used for " + "<a href=\"/wiki/Paper\" title=\"Paper\">paper-making since " + "2400 BC or before."; In the string above I have inter-mixed html with text. Well the requirement is that the output looks like:- They have been used for paper-making since 2400 BC or before. Could some one help me with a generic regular expression that would produce the desired output from the given input? Thanks in advance!

  Read the article

• Need to create regular expression in Javascript to check the valid conditional string

  - by user1796078

  I want to create the regular expression in javascript which will check the valid conditional string like -1 OR (1 AND 2) AND 1 -1 OR (1 AND 2) -1 OR 2 -1 OR 1 OR 1 -1 AND 1 AND 1 The string should not contain 'AND' and 'OR'. For example - 1 OR 2 AND 3 is invalid. -It should be (1 OR 2) AND 3 or 1 or (2 AND 3). I tried the following Regex. It works for most of the conditions but it is failing to check the above condition. /^(\s*\(\d+\s(AND|OR)\s\d+\)|\s*\d+)((\s*(AND|OR)\s*)(\(\d+\s(AND|OR)\s\d+\)|\s*\d+))*$/ Can anyone please help me to sort out the above problem.

  Read the article

• using regular expression / Remove special characters with linq to sql

  - by Prasad

  How can i use regular expressions with linq to sql in the asp.net mvc(C#) application? The columns in my MSSQL table (Products) has some special characters like (, - % ',.....). While searching for a product i need to search without that special chareters. For ex.: I have a product say (ABC-Camp's / One), when i search for "abccamp", it should pull the product. I am trying the query like: from p in _context.pu_Products where p.User_Id == userId && p.Is_Deleted == false && p.Product_Name.ToLower().Contains(text.ToLower()) select new Product { ProductId = p.Product_Id, ProductName = p.Product_Name.Replace("’", "").Replace("\"", ""), RetailPrice = p.Retail_Price ?? 0M, LotSize = p.Lot_Size > 0 ? p.Lot_Size ?? 1 : 1, QuantityInHand = p.Quantity_In_Hand ?? 0 } But i need to search without any special characters...

  Read the article

• Regular Expression to Match All Characters after another Regex Match

  - by Anthony Wood

  I know this may sound a little confusing, so I am open to suggestions on renaming the title. Basically I have string such as C:...\Downloads\Folder\SubFolder\SubSubFolder. and I want to return the SubFolder and SubSubFolder only. So far, my Regex looks like (?=\\Downloads\\.*?\\).* which matches Downloads\Folder\SubFolder\SubSubFolder. Does anybody have any Ideas what I am missing???? All the solutions below seem to work (except if you didn't know "Folder"). Potentially a bug with the tool I was using to test the regular expressions.

  Read the article

• How to write a regular expression for "everything between X and Y" for use with preg_replace

  - by pg

  I want to take a variable called $encoded_str and and remove cd1, CD1 and anything between the first 'l' and the last blank space. So for example "lp6 id4 STRINGcd1" would return "STRING". I'm using PHP 4 for now so I can't use str_ireplace, I have this: $encoded_str=str_replace('CD1','',$encoded_str); $encoded_str=str_replace('cd1','',$encoded_str); $encoded_str=preg_replace('X','',$encoded_str); I've RTFM for preg_replace but am a bit confused. What should I replace the X with and can you suggest a decent introductory primer for writing regular expressions?

  Read the article

• Python Regular Expression TypeError

  - by spaghettiwestern

  I am writing my first python program and I am running into a problem with regex. I am using regular expression to search for a specific value in a registry key. import _winreg import re key = _winreg.OpenKey(_winreg.HKEY_LOCAL_MACHINE,"Software\\Microsoft\\Windows\\CurrentVersion\\Uninstall\\{26A24AE4-039D-4CA4-87B4-2F83216020FF}") results=[] v = re.compile(r"(?i)Java") try: i = 0 while 1: name, value, type = _winreg.EnumValue(key, i) if v.search(value): results.append((name,value,type)) i += 1 except WindowsError: print for x in results: print "%-50s%-80s%-20s" % x I am getting the following error: exceptions.TypeError: expected string or buffer I can use the "name" variable and my regex works fine. For example if I make the following changes regex doesn't complain: v = re.compile(r"(?i)DisplayName") if v.search(name): Thanks for any help.

  Read the article

• Saving substrings using Regular Expressions

  - by user362971

  I'm new to regular expressions in Java (or any language, for that matter) and I'm wanting to do a find using them. The tricky part that I don't understand how to do is replace something inside the string that matches. For example, if the line I'm looking for is Person item6 [can {item thing [wrap]}] I'm able to write a regex that finds that line, but finding what the word "thing" is (as it may differ among different lines) is my problem. I may want to either replace that word with something else or save it in a variable for later. Is there any easy way to do this using Java's regex engine?

  Read the article

• Regular expression one or more times JAVA

  - by user1381564

  Hi i am trying to match a string against a pattern this is the possible string signal CS, NS, dl: stateType := writeOrRead0; signal CS, pS : stateType := writeOrRead0; signal dS : stateType := writeOrRead0; i am only concerned with the pattern as far as the first colon. but the number of signals define can be more than one it could be three or four even this is the regular expression i have ^signal\\s*(\\w+),*\\s*(\\w+)\\s*: it will pick up the second two signal but and for the second one it picks up CS and pS and but the d and S in the next signal when i use matcher.group() come up seperately Can anyone give me an expression that will pick up all signal names whether there is one two three or more?

  Read the article

< Previous Page | 23 24 25 26 27 28 29 30 31 32 33 34  | Next Page >