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  • Regular Expression for username

    - by neobie
    I need help on regular expression on the condition (4) below: Begin with a-z End with a-z0-9 allow 3 special characters like ._- The characters in (3) must be followed by alphanumeric characters, and it cannot be followed by any characters in (3) themselves. Not sure how to do this. Any help is appreciated, with the sample and some explanations.

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  • How can I match a match a null byte (0x00) in the Visual Studio binary editor with a find using a re

    - by Paul K
    Open a file in the Visual Studio binary editor that contains a null byte (0x00), then use the Quick Find feature (Ctrl +F) to find null bytes. I would have thought I could use a regular expression such as \x00 to match null bytes but it doesn't work. Searching for any other hex value using this method works fine. Is this a VS bug, 'feature', or am I just missing something? Is there a work around?

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  • Simple regular expression for decimal numbers?

    - by finch
    I know this may be the simplest question ever asked on Stack Overflow, but what is the regular expression for a decimal with a precision of 2? Valid examples: 123.12 2 56754 92929292929292.12 0.21 3.1 Invalid examples: 12.1232 2.23332 e666.76 Sorry for the lame question, but for the life of me I haven't been able to find anyone that can help! The decimal place may be option, and that integers may also be included.

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  • Need help parsing HTML with a regex in python

    - by laspal
    Hi, My string is mystring = "<tr><td><span class='para'><b>Total Amount : </b>INR (Indian Rupees) 100.00</span></td></tr>" My problem here is I have to search and get the total amount test = re.search("(Indian Rupees)(\d{2})(?:\D|$)", mystring) but my test give me None. How can I get the values and values can be 10.00, 100.00, 1000.00 Thanks

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  • How Do I grep For non-ASCII Characters in UNIX

    - by Peter Conrey
    I have several very large XML files and I'm trying to find the lines that contain non-ASCII characters. I've tried the following: grep -e "[\x{00FF}-\x{FFFF}]" file.xml But this returns every line in the file, regardless of whether the line contains a character in the range specified. Do I have the syntax wrong or am I doing something else wrong? I've also tried: egrep "[\x{00FF}-\x{FFFF}]" file.xml (with both single and double quotes surrounding the pattern).

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  • String pattern matching in Javascript

    - by kwokwai
    Hi all, I am doing some self learning about Patern Matching in Javascript. I got a simple input text field in a HTML web page, and I have done some Javascript to capture the string and check if there are any strange characters other than numbers and characters in the string. But I am not sure if it is correct. Only numbers, characters or a mixture of numbers and characters are allowed. var pattern = /^[a-z]+|[A-Z]+|[0-9]+$/; And I have another question about Pattern Matching in Javascript, what does the percentage symbol mean in Pattern matching. For example: var pattern = '/[A-Z0-9._%-]+@[A-Z0-9.-]+\.[A-Z]{2,4}/';

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  • 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)?

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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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  • How to with extract url from tweet using Regular Expressions

    - by neutreno
    Ok so i'm executing the following line of code in javascript RegExp('(http:\/\/t.co\/)[a-zA-Z0-9\-\.]{8}').exec(tcont); where tcont is equal to some string like 'Test tweet to http://t.co/GXmaUyNL' (the content of a tweet obtained by jquery). However it is returning, in the case above for example, 'http://t.co/GXmaUyNL,http://t.co/'. This is frustracting because I want the url without the bit on the end - after and including the comma. Any ideas why this is appearing? Thanks

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  • parse string with regular exression

    - by llamerr
    I trying to parse this string: $right = '34601)S(1,6)[2] - 34601)(11)[2] + 34601)(3)[2,4]'; with following regexp: const word = '(\d{3}\d{2}\)S{0,1}\([^\)]*\)S{0,1}\[[^\]]*\])'; preg_match('/'.word.'{1}(?:\s{1}([+-]{1})\s{1}'.word.'){0,}/', $right, $matches); print_r($matches); i want to return array like this: Array ( [0] => 34601)S(1,6)[2] - 34601)(11)[2] + 34601)(3)[2,4] [1] => 34601)S(1,6)[2] [2] => - [3] => 34601)(11)[2] [4] => + [5] => 34601)(3)[2,4] ) but i return only following: Array ( [0] => 34601)S(1,6)[2] - 34601)(11)[2] + 34601)(3)[2,4] [1] => 34601)S(1,6)[2] [2] => + [3] => 34601)(3)[2,4] ) i think, its becouse of [^)]* or [^]]* in the word, but how i should correct regexp for matching this in another way? i tryied to specify it: \d+(?:[,#]\d+){0,} so word become const word = '(\d{3}\d{2}\)S{0,1}\(\d+(?:[,#]\d+){0,}\)S{0,1}\[\d+(?:[,#]\d+){0,}\])'; but it gives nothing

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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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  • regular expression: extract last 2 characters

    - by dotnet-practitioner
    what is the best way to extract last 2 characters of a string using regular expression. For example, I want to extract state code from the following "A_IL" I want to extract IL as string.. please provide me C# code on how to get it.. string fullexpression = "A_IL"; string StateCode = some regular expression code.... thanks

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  • Filter list of phone numbers using php

    - by LiveEn
    I have a list of phone numbers that start with the below numbers and in different formats...i need to grab the numbers that start only with the below numbers/format using php...... 020 8 07974 +44 (0) 20 +44 0 440203 any help will be appreciated..

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  • Why does this regular expression for sed break inside Makefile?

    - by jcrocholl
    I'm using GNU Make 3.81, and I have the following rule in my Makefile: jslint : java org.mozilla.javascript.tools.shell.Main jslint.js mango.js \ | sed 's/Lint at line \([0-9]\+\) character \([0-9]\+\)/mango.js:\1:\2/' This works fine if I enter it directly on the command line, but the regular expression does not match if I run it with "make jslint". However, it works if I replace \+ with \{1,\} in the Makefile: jslint : java org.mozilla.javascript.tools.shell.Main jslint.js mango.js \ | sed 's/Lint at line \([0-9]\{1,\}\) character \([0-9]\{1,\}\)/mango.js:\1:\2/' Is there some special meaning to \+ in Makefiles, or is this a bug?

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  • Using awk to return only certain chunks of data

    - by Koriar
    I'm not 100% certain how to phrase my question simply, so I apologize if this has been answered somewhere and I was just unable to find it. What I have are debug logs with authentication packets in them along with a bunch of other output. I need to search through about 2 million lines of logs to find every packet that contains a certain mac address. The packets look something like this (slightly censored): -----------------[ header ]----------------- Event: Authd-Response (1900) Sequence: -54 Timestamp: 1969-12-31 19:30:00 (0) ---------------[ attributes ]--------------- Auth-Result = Auth-Accept Service-Profile-SID = 53 Service-Profile-SID = 49 RADIUS-Access-Accept-Attr/WiMAX-Capability = 0x(numbers) Session-Timeout = 3600 Service-Profile-SID = 4 Service-Profile-SID = 29 Chargeable-User-Identity = "(Numbers)" User-Password = "(the MAC address I'm looking for)" -------------------------------------------- However there are about 10 different possible types with different possible lengths. They all start with the header line and end with the all-dashes line. I've had success using awk to get the code blocks themselves using this: awk '/-----------------\[ header \]-----------------/,/--------------------------------------------/' filename.txt But I was hoping to be able to use it to return only the packets which contain the MAC address that I need. I've been trying to figure this out for a few days now and I'm pretty stuck. I could try and write a bash script, but I could swear that I've used awk to do something like this before...

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  • A more elegant way to parse a string with ruby regular expression using variable grouping?

    - by i0n
    At the moment I have a regular expression that looks like this: ^(cat|dog|bird){1}(cat|dog|bird)?(cat|dog|bird)?$ It matches at least 1, and at most 3 instances of a long list of words and makes the matching words for each group available via the corresponding variable. Is there a way to revise this so that I can return the result for each word in the string without specifying the number of groups beforehand? ^(cat|dog|bird)+$ works but only returns the last match separately , because there is only one group.

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  • re.sub emptying list

    - by jmau5
    def process_dialect_translation_rules(): # Read in lines from the text file specified in sys.argv[1], stripping away # excess whitespace and discarding comments (lines that start with '##'). f_lines = [line.strip() for line in open(sys.argv[1], 'r').readlines()] f_lines = filter(lambda line: not re.match(r'##', line), f_lines) # Remove any occurances of the pattern '\s*<=>\s*'. This leaves us with a # list of lists. Each 2nd level list has two elements: the value to be # translated from and the value to be translated to. Use the sub function # from the re module to get rid of those pesky asterisks. f_lines = [re.split(r'\s*<=>\s*', line) for line in f_lines] f_lines = [re.sub(r'"', '', elem) for elem in line for line in f_lines] This function should take the lines from a file and perform some operations on the lines, such as removing any lines that begin with ##. Another operation that I wish to perform is to remove the quotation marks around the words in the line. However, when the final line of this script runs, f_lines becomes an empty lines. What happened? Requested lines of original file: ## English-Geek Reversible Translation File #1 ## (Moderate Geek) ## Created by Todd WAreham, October 2009 "TV show" <=> "STAR TREK" "food" <=> "pizza" "drink" <=> "Red Bull" "computer" <=> "TRS 80" "girlfriend" <=> "significant other"

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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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  • Find and replace braced tags within a MySQL table

    - by Cy
    I have about 40000 records in that table that contains plain text and within the plain text, contains that kind of tags which its only characteristic is that they are braced between [ ] [caption id="attachment_2948" align="alignnone" width="480" caption="the caption goes here"] How could I remove those? (replace by nothing) I could also run a PHP program if necessary to do the cleanup.

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