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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 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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  • 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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  • 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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  • 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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  • 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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  • Help with this reg. exp. in PHP

    - by Jonathan
    Hi, i don't know about regular expressions, I asked here for one that: gets either anything up to the first parenthesis/colon or the first word inside the first parenthesis. This was the answer: preg_match('/(?:^[^(:]+|(?<=^\\()[^\\s)]+)/', $var, $match); I need an improvement, I need to get either anything up to the first parenthesis/colon/quotation marks or the first word inside the first parenthesis. So if I have something like: $var = 'story "The Town in Hell"s Backyard'; // I get this: $match = 'story'; $var = "screenplay (based on)"; // I get this: $match = 'screenplay'; $var = "(play)"; // I get this: $match = 'play'; $var = "original screen"; // I get this: $match = 'original screen'; Thanks!

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  • Regexs in Ruby getting filename

    - by user1290757
    i am extracting file names of html files using line: filename = File.basename(input_filename, ".*") which currently prints full file name excluding .html extension All files are stored in the form of http^x.x.edu^1^2 all file names begin with http^ and contain edu^ what i want is to extract 2 (which changes) but it is always the second element after .edu I have attempted destructive gsub! but i m weak with regular expressions.

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  • Regular expression: who's greedier?

    - by polygenelubricants
    My primary concern is with the Java flavor, but I'd also appreciate information regarding others. Let's say you have a subpattern like this: (.*)(.*) Not very useful as is, but let's say these two capture groups (say, \1 and \2) are part of a bigger pattern that matches with backreferences to these groups, etc. So both are greedy, in that they try to capture as much as possible, only taking less when they have to. My question is: who's greedier? Does \1 get first priority, giving \2 its share only if it has to? What about: (.*)(.*)(.*) Let's assume that \1 does get first priority. Let's say it got too greedy, and then spit out a character. Who gets it first? Is it always \2 or can it be \3? Let's assume it's \2 that gets \1's rejection. If this still doesn't work, who spits out now? Does \2 spit to \3, or does \1 spit out another to \2 first?

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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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  • Perl Regular expression remove double tabs, line breaks, white spaces

    - by Scoox
    Hi guys, I want to write a perl script that removes double tabs, line breaks and white spaces. What I have so far is: $txt=~s/\r//gs; $txt=~s/ +/ /gs; $txt=~s/\t+/\t/gs; $txt=~s/[\t\n]*\n/\n/gs; $txt=~s/\n+/\n/gs; But, 1. It's not beautiful. Should be possible to do that with far less regexps. 2. It just doesn't work and I really do not know why. It leaves some double tabs, white spaces and empty lines (i.e. lines with only a tab or whitespace) I could solve it with a while, but that is very slow and ugly. Any suggestions?

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  • Regular Expression Sanitize (PHP)

    - by atif089
    Hello, I would like to sanitize a string in to a URL so this is what I basically need. Everything must be removed except alphanumeric characters and spaces and dashed. Spaces should be converter into dashes. Eg. This, is the URL! must return this-is-the-url Thanks

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  • PHP: Regular Expression to get a URL from a string

    - by Matthew Iselin
    I'm working on some PHP code which takes input from various sources and needs to find the URLs and save them somewhere. The kind of input that needs to be handled is as follows: http://www.youtube.com/watch?v=IY2j_GPIqRA Try google: http://google.com! (note exclamation mark is not part of the URL) Is http://somesite.com/ down for anyone else? Output: http://www.youtube.com/watch?v=IY2j_GPIqRA http://google.com http://somesite.com/ I've already borrowed one regular expression from the internet which works, but unfortunately wipes the query string out - not good! Any help putting together a regular expression, or perhaps another solution to this problem, would be appreciated.

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  • How to replace only part of the match with python re.sub

    - by Arty
    I need to match two cases by one reg expression and do replacement 'long.file.name.jpg' - 'long.file.name_suff.jpg' 'long.file.name_a.jpg' - 'long.file.name_suff.jpg' I'm trying to do the following re.sub('(\_a)?\.[^\.]*$' , '_suff.',"long.file.name.jpg") But this is cut the extension '.jpg' and I'm getting long.file.name_suff. instead of long.file.name_suff.jpg I understand that this is because of [^.]*$ part, but I can't exclude it, because I have to find last occurance of '_a' to replace or last '.' Is there a way to replace only part of the match?

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  • Convert a complicated string into an array in php

    - by Patrick Beardmore
    I have a php variable that comes from a form that needs tidying up. I hope you can help. The variable contains a list of items (possibly two or three word items with a space in between words). I want to convert it to a comma separated list with no superfluous white space. I want the divisions to fall only at commas, semi-colons or new-lines. Blank cannot be an item. Here's a comprehensive example (with a deliberately messy input): Variable In: "dog, cat ,car,tea pot,, ,,, ;;(++NEW LINE++)fly, cake" Variable Out "dog,cat,car,tea pot,fly,cake" Can anyone help?

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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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  • Best way to correct garbled data caused by false encoding

    - by ercan
    Hi all, I have a set of data that contains garbled text fields because of encoding errors during many import/exports from one database to another. Most of the errors were caused by converting UTF-8 to ISO-8859-1. Strangely enough, the errors are not consistent: the word 'München' appears as 'München' in some place and as 'MÜnchen'. Is there a trick in SQL server to correct this kind of crap? The first thing that I can think of is to exploit the COLLATE clause, so that ü is interpreted as ü, but I don't exactly know how. If it isn't possible to make it in the DB level, do you know any tool that helps for a bulk correction? (no manual find/replace tool, but a tool that guesses the garbled text somehow and correct them)

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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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  • validation of special characters

    - by jpallavi
    I want to validate login name with special characters !@#S%^*()+_-?/<:"';. space using regular expression in ruby on rails. These special characters should not be acceptable. What is the code for that? Thanks, Pallavi

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