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  • Regexp look-behind to match internet speeds

    - by Sandman
    So the user may search for "10 mbit" after which I want to capture the "10" so I can use it in a speed-search rather than a string-search. This isn't a problem, the below regexp does this fine: if (preg_match("/(\d+)\smbit/", $string)){ ... } But, the user may search for something like "10/10 mbit" or "10-100 mbit". I don't want to match those with the above regexp - they should be handled in another fashion. So I would like a regexp that matches "10 mbit" if the number is all-numeric as a whole word (i.e. contained by whitespace, newline or lineend/linestart) Using lookbehind, I did this: if (preg_match("#(?<!/)(\d+)\s+mbit#i", $string)){ Just to catch those that doesn't have "/" before them, but this matched true for this string: "10/10 mbit" so I'm obviously doing something wrong here, but what?

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  • Delete all characters in a multline string up to a given pattern

    - by biffabacon
    Using Python I need to delete all charaters in a multiline string up to the first occurrence of a given pattern. In Perl this can be done using regular expressions with something like: #remove all chars up to first occurrence of cat or dog or rat $pattern = 'cat|dog|rat' $pagetext =~ s/(.*?)($pattern)/$2/xms; What's the best way to do it in Python?

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  • How Do I Remove The First 4 Characters From A String If It Matches A Pattern In Ruby

    - by James
    I have the following string: "h3. My Title Goes Here" I basically want to remove the first 4 characters from the string so that I just get back: "My Title Goes Here". The thing is I am iterating over an array of strings and not all have the h3. part in front so I can't just ditch the first 4 characters blindly. I have checked the docs and the closest think I could find was chomp, but that only works for the end of a string. Right now I am doing this: "h3. My Title Goes Here".reverse.chomp(" .3h").reverse This gives me my desired output, but there has to be a better way right? I mean I don't want to reverse a string twice for no reason. I am new to programming so I might have missed something obvious, but I didn't see the opposite of chomp anywhere in the docs. Is there another method that will work? Thanks!

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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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  • Can a repeated piece of regular expression create multiple groups? Such as this example...

    - by Yousui
    Hi guys, I'm using RUBY 's regular expression to deal with text such as ${1:aaa|bbbb} ${233:aaa | bbbb | ccc ccccc } ${34: aaa | bbbb | cccccccc |d} ${343: aaa | bbbb | cccccccc |dddddd ddddddddd} ${3443:a aa|bbbb|cccccccc|d} ${353:aa a| b b b b | c c c c c c c c | dddddd} I want to get the trimed text between each pipe line. For example, for the first line of my upper example, I want to get the result aaa and bbbb, for the second line, I want aaa, bbbb and ccc ccccc. Now I have wrote a piece of regular expression and a piece of ruby code to test it: array = "${33:aaa|bbbb|cccccccc}".scan(/\$\{\s*(\d+)\s*:(\s*[^\|]+\s*)(?:\|(\s*[^\|]+\s*))+\}/) puts array Now my problem is the (?:\|(\s*[^\|]+\s*))+ part can't create multiple groups. I don't know how to solve this problem, because the number of text I need in each line is variable. Anyone can help? Great thanks.

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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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  • Regular Expression :match string containing only non repeating words

    - by nash
    I have this situation(Java code): 1) a string such as : "A wild adventure" should match. 2) a string with adjacent repeated words: "A wild wild adventure" shouldn't match. With this regular expression: .* \b(\w+)\b\s*\1\b.* i can match strings containing adjacent repeated words. How to reverse the situation i.e how to match strings which do not contain adjacent repeat words

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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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  • Extract a sentence out of sentences separated by delimitors

    - by Laura
    Below is a sample line I have extracted from a website: below a satisfactory level; &quot;an off year for tennis&quot;; &quot;his performance was off&quot; The output displays as: below a satisfactory level; "an off year for tennis"; "his performance was off" I want to get only the first sentence "below a satisfactory level"; Here is the code I have tried after exploring many stackoverflow posts: $data=explode('; ',$str); echo $data[0]; But somehow it is not working. Thanks in advance.

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  • Match subpatterns in any order

    - by Yaroslav
    I have long regexp with two complicated subpatters inside. How i can match that subpatterns in any order? Simplified example: /(apple)?\s?(banana)?\s?(orange)?\s?(kiwi)?/ and i want to match both of apple banana orange kiwi apple orange banana kiwi It is very simplified example. In my case banana and orange is long complicated subpatterns and i don't want to do something like /(apple)?\s?((banana)?\s?(orange)?|(orange)?\s?(banana)?)\s?(kiwi)?/ Is it possible to group subpatterns like chars in character class? UPD Real data as requested: 14:24 26,37 Mb 108.53 01:19:02 06.07 24.39 19:39 46:00 my strings much longer, but it is significant part. Here you can see two lines what i need to match. First has two values: length (14 min 24 sec) and size 26.37 Mb. Second one has three values but in different order: size 108.53 Mb, length 01 h 19 m 02 s and date June, 07 Third one has two size and length Fourth has only length There are couple more variations and i need to parse all values. I have a regexp that pretty close except i can't figure out how to match patterns in different order without writing it twice. (?<size>\d{1,3}\[.,]\d{1,2}\s+(?:Mb)?)?\s? (?<length>(?:(?:01:)?\d{1,2}:\d{2}))?\s* (?<date>\d{2}\.\d{2}))? NOTE: that is only part of big regexp that forks fine already.

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  • Regular expression to match any table tag

    - by keeg
    I'm trying to write a regular expression to see if a string contains any of the typical table tags: <table></table> <td></td> <th></th> <tr></tr> <thead></thead> <tfoot></tfoot> <tbody></tbody> Along with tags that may contain other attributes e.g: <table border="1"> I've come up with this so far, however, it matches <br /> tag and I'm not sure why: /<\/?[table|td|th|tr|tfoot|thead|tbody]{1,}>?/ http://www.rexfiddle.net/20Xtqka

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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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  • 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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  • Regular expression for pipe delimited and double quoted string

    - by Hiren Amin
    I have a string something like this: "2014-01-23 09:13:45|\"10002112|TR0859657|25-DEC-2013>0000000000000001\"|10002112" I would like to split by pipe apart from anything wrapped in double quotes so I have something like (similar to how csv is done): [0] => 2014-01-23 09:13:45 [1] => 10002112|TR0859657|25-DEC-2013>0000000000000001 [2] => 10002112 I would like to know if there is a regular expression that can do this?

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  • Find and Replace with Notepad++

    - by Levi
    I have a document that was converted from PDF to HTML for use on a company website to be referenced and indexed for search. I'm attempting to format the converted document to meet my needs and in doing so I am attempting to clean up some of the junk that was pulled over from when it was a PDF such as page numbers, headers, and footers. luckily all of these lines that need to be removed are in blocks of 4 lines unfortunately they are not exactly the same therefore cannot be removed with a simple literal replace. The lines contain numbers which are incremental as they correlate with the pages. How can I remove the following example from my html file. Title<br> 10<br> <hr> <A name=11></a>Footer<br> I've tried many different regular expression attempts but as my skill in that area is limited I can't find the proper syntax. I'm sure i'm missing something fairly easy as it would seem all I need is a wildcard replace for the two numbers in the code and the rest is literal. any help is apprciated

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  • Help with Regular Expression

    - by shivesh
    Hello I need help with Regular Expression, I want to match each section (number and it's text - 2 groups), the text can be multi line, each section ends when another section starts (another number) or when .END is reached or EOF. Demo Expression: \(\d{1,3}\) ([\s\S]*?)(\.END|\(\d{1,3}\)) Input text: (1) some text some text some text some text some text some text (2) some text some textsome text (3) some textsome text some textsome textsome text (4) some text .END first group should match number (with brackets) and second group should match corresponded text.

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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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