Search Results

Search found 4848 results on 194 pages for 'expression blend 3'.

Page 29/194 | < Previous Page | 25 26 27 28 29 30 31 32 33 34 35 36  | Next Page >

  • Regular Expression to Match Specific "Values" in Isolated Group

    - by Gandarez
    If have this regular expression to test (\&TRUNC)[\(]{1,}(.+)[\)]{1,} And I have this "tester" ((((&TRUNC((1800,000 / 510)) * 510) * 920) + (2 * (510 * 700)) + ((&TRUNC((1800,000 / 510)) - 1) * 2 * 510 * 80)) / 1000000) * 85,715 My expected value is (inside the personal command "&TRUNC(command)") (1800,000 / 510) I got this value 1800,000 / 510)) * 510) * 920) + (2 * (510 * 700)) + ((&TRUNC((1800,000 / 510)) - 1) * 2 * 510 * 80)) / 1000000 How can I get only expected value in a separated group? PS:. The expressions inside the command called for me as "&TRUNC(command)" is variable.

    Read the article

  • Regular expression (PCRE) for url matching

    - by zerkms
    The input: we get some plain text as input string and we have to highlighight all urls there with {url For some time i've used regex taken from http://flanders.co.nz/2009/11/08/a-good-url-regular-expression-repost/, which i modified several times, but it's built for another issue - to check whether the whole input string is an url or no. So, what regex do you use in such issues?

    Read the article

  • SQLite user_version() expression

    - by ralf.w.
    how can I get user_version into an expression (for SELECT or WHERE) ? I would also appreciate any clue to how to save ANY global variable in SQLite. Does anyone know of such an extension (dll) ? BTW: I know that PRAGMA user_version; gives me the right answer, but I cannot use that inside a WHERE-clause !

    Read the article

  • A regular expression question

    - by Hellnar
    Hello, I am in dire need of a such regular expression where my alphabet is made up of 0s and 1s. Now I need a language that accepts all words as long as it has three 0s. IE: 000 10001 0001 1000 10000101

    Read the article

  • Regular expression for a list of numbers without other characters

    - by JamesW
    I need a regular expression for a list of numbers in a text box with carriage returns and line feeds, but no other characters. e.g. 1234 5678 5874 3478 I have this: Regex(@"\d\r\n${0,}?"); ... but it is accepting commas when I paste them into my text box: e.g. 1234, 5678 5874, 3478 Can someone please tell me what I'm doing wrong? Thanks

    Read the article

  • How to match parameter names in an expression?

    - by burak ozdogan
    Hi, I have a set of expressions representing some formula with some parameters inside. Like: [parameter1] * [parameter2] * [multiplier] And many others like this. I want to use a regular expression so that I can get a list of strings (List<string>) which will have the following inside: [paramter1] [paramter2] [multiplier] I am not using regular expressions so often; if you have already used something like this I would appreciate if you can share. Thanks!

    Read the article

  • Regular Expression Fails

    - by Meander365
    Anyone help? When I run this I get " invalid quantifier ?<=href= " var aHrefMatch = new RegExp("(?<=href\=")[^]+?(?=")"); var matchedLink = mystring.match(aHrefMatch); But I know the regular expression is valid. Any ideas?

    Read the article

  • Regular Expression Routes in Rails

    - by Kevin Sylvestre
    I am looking to create a rails route that is capable of accepting requests using a regular expression. Specifically, I need optional paths. As an example: "(/first)?(/second)?" Would match: /first /second /first/second But not: /second/first Is this possible? Thanks.

    Read the article

  • Regular expression - starting and ending with a letter, accepting only letters, numbers and _

    - by jreid9001
    I'm trying to write a regular expression which specifies that text should start with a letter, every character should be a letter, number or underscore, there should not be 2 underscores in a row and it should end with a letter or number. At the moment, the only thing I have is ^[a-zA-Z]\w[a-zA-Z1-9_] but this doesn't seem to work properly since it only ever matches 3 characters, and allows repeated underscores. I also don't know how to specify requirements for the last character.

    Read the article

  • regular expression breaking on new line

    - by shyam
    I'm trying to use a regular expression as below: preg_match_all('|<table.*</table>|',$html,$matches, PREG_SET_ORDER); But this is not working, and I think the problem is the new line inside the string $html. Could someone tell me a work around?

    Read the article

  • Regular Expression not disappearing

    - by user2439019
    I have 3 phone fields and any one is required. SO i had a custom validation class to make any one of them is required. And i am calling those class as follows [RegularExpression(@"^\(?([0-9]{3})\)?. ]?([0-9]{3})[-. ]?([0-9]{4})$", ErrorMessage = "Entered phone format is not valid. <br/> 10 digits are required.<br/> No spaces between digits. <br/> Numbers only.")] [AtLeastOneRequired("PhoneHome", "PhoneMobile", "PhoneOffice", ErrorMessage = "Please provide either of PhoneHome or PhoneMobile or phoneOffice. ")] public string PhoneHome { get; set; } [StringLength(11, MinimumLength = 10)] [RegularExpression(@"^\(?([0-9]{3})\)?[-. ]?([0-9]{3})[-. ]?([0-9]{4})$", ErrorMessage = "Entered phone format is not valid. <br/> 10 digits are required.<br/> No spaces between digits. <br/> Numbers only.")] [Display(Name = "Prompt_PhoneOffice", ResourceType = typeof(ResContactItems))] public string PhoneOffice { get; set; } [StringLength(11, MinimumLength = 10)] [RegularExpression(@"^\(?([0-9]{3})\)?[-. ]?([0-9]{3})[-. ]?([0-9]{4})$", ErrorMessage = "Entered phone format is not valid. <br/> 10 digits are required.<br/> No spaces between digits. <br/> Numbers only.")] [Display(Name = "Prompt_PhoneMobile", ResourceType = typeof(ResContactItems))] public string PhoneMobile { get; set; } The problem is with only "PhoneHome " field , if i enter incorrect format, it will display error message based ont he regular expression given .But the regular expression message is not disappearing wwhen we enter the correct format too.It will disappear only in f the field is empty.Other two fields are showing properly. Is this due to the custom class i am calling.? Please help me to sort out this issue Thanks, Vidya

    Read the article

  • .NET regular expression

    - by vert
    How would I write a regular expression (C#) which will check a given string to see if any of its characters are characters OTHER than the following: a-z A-Z Æ æ Å å Ø ø - '

    Read the article

  • Regular expression to process key value pairs

    - by user677680
    I am attempting to write a regular expression to process a string of key value(s) pairs formatted like so KEY/VALUE KEY/VALUE VALUE KEY/VALUE A key can have multiple values separated by a space. I want to match a keys values together, so the result on the above string would be VALUE VALUE VALUE VALUE I currently have the following as my regex [A-Z0-9]+/([A-Z0-9 ]+)(?:(?!^[A-Z0-9]+/)) but this returns VALUE KEY as the first result.

    Read the article

  • Regular expression to remove one parameter from query string

    - by Kip
    I'm looking for a regular expression to remove a single parameter from a query string, and I want to do it in a single regular expression if possible. Say I want to remove the foo parameter. Right now I use this: /&?foo\=[^&]+/ That works as long as foo is not the first parameter in the query string. If it is, then my new query string starts with an ampersand. (For example, "foo=123&bar=456" gives a result of "&bar=456".) Right now, I'm just checking after the regex if the query string starts with ampersand, and chopping it off if it does. Example edge cases: Input | Output -------------------------+----------------- foo=123 | (empty string) foo=123&bar=456 | bar=456 bar=456&foo=123 | bar=456 abc=789&foo=123&bar=456 | abc=789&bar=456 Edit OK as pointed out in comments there are there are way more edge cases than I originally considered. I got the following regex to work with all of them: /&foo(\=[^&]*)?(?=&|$)|^foo(\=[^&]*)?(&|$)/ This is modified from Mark Byers's answer, which is why I'm accepting that one, but Roger Pate's input helped a lot too. Here is the full suite of test cases I'm using, and a Perl script which tests them. Input | Output -------------------------+------------------- foo | foo&bar=456 | bar=456 bar=456&foo | bar=456 abc=789&foo&bar=456 | abc=789&bar=456 foo= | foo=&bar=456 | bar=456 bar=456&foo= | bar=456 abc=789&foo=&bar=456 | abc=789&bar=456 foo=123 | foo=123&bar=456 | bar=456 bar=456&foo=123 | bar=456 abc=789&foo=123&bar=456 | abc=789&bar=456 xfoo | xfoo xfoo&bar=456 | xfoo&bar=456 bar=456&xfoo | bar=456&xfoo abc=789&xfoo&bar=456 | abc=789&xfoo&bar=456 xfoo= | xfoo= xfoo=&bar=456 | xfoo=&bar=456 bar=456&xfoo= | bar=456&xfoo= abc=789&xfoo=&bar=456 | abc=789&xfoo=&bar=456 xfoo=123 | xfoo=123 xfoo=123&bar=456 | xfoo=123&bar=456 bar=456&xfoo=123 | bar=456&xfoo=123 abc=789&xfoo=123&bar=456 | abc=789&xfoo=123&bar=456 foox | foox foox&bar=456 | foox&bar=456 bar=456&foox | bar=456&foox abc=789&foox&bar=456 | abc=789&foox&bar=456 foox= | foox= foox=&bar=456 | foox=&bar=456 bar=456&foox= | bar=456&foox= abc=789&foox=&bar=456 | abc=789&foox=&bar=456 foox=123 | foox=123 foox=123&bar=456 | foox=123&bar=456 bar=456&foox=123 | bar=456&foox=123 abc=789&foox=123&bar=456 | abc=789&foox=123&bar=456 Test script (Perl) @in = ('foo' , 'foo&bar=456' , 'bar=456&foo' , 'abc=789&foo&bar=456' ,'foo=' , 'foo=&bar=456' , 'bar=456&foo=' , 'abc=789&foo=&bar=456' ,'foo=123' , 'foo=123&bar=456' , 'bar=456&foo=123' , 'abc=789&foo=123&bar=456' ,'xfoo' , 'xfoo&bar=456' , 'bar=456&xfoo' , 'abc=789&xfoo&bar=456' ,'xfoo=' , 'xfoo=&bar=456' , 'bar=456&xfoo=' , 'abc=789&xfoo=&bar=456' ,'xfoo=123', 'xfoo=123&bar=456', 'bar=456&xfoo=123', 'abc=789&xfoo=123&bar=456' ,'foox' , 'foox&bar=456' , 'bar=456&foox' , 'abc=789&foox&bar=456' ,'foox=' , 'foox=&bar=456' , 'bar=456&foox=' , 'abc=789&foox=&bar=456' ,'foox=123', 'foox=123&bar=456', 'bar=456&foox=123', 'abc=789&foox=123&bar=456' ); @exp = ('' , 'bar=456' , 'bar=456' , 'abc=789&bar=456' ,'' , 'bar=456' , 'bar=456' , 'abc=789&bar=456' ,'' , 'bar=456' , 'bar=456' , 'abc=789&bar=456' ,'xfoo' , 'xfoo&bar=456' , 'bar=456&xfoo' , 'abc=789&xfoo&bar=456' ,'xfoo=' , 'xfoo=&bar=456' , 'bar=456&xfoo=' , 'abc=789&xfoo=&bar=456' ,'xfoo=123', 'xfoo=123&bar=456', 'bar=456&xfoo=123', 'abc=789&xfoo=123&bar=456' ,'foox' , 'foox&bar=456' , 'bar=456&foox' , 'abc=789&foox&bar=456' ,'foox=' , 'foox=&bar=456' , 'bar=456&foox=' , 'abc=789&foox=&bar=456' ,'foox=123', 'foox=123&bar=456', 'bar=456&foox=123', 'abc=789&foox=123&bar=456' ); print "Succ | Input | Output | Expected \n"; print "-----+--------------------------+--------------------------+-------------------------\n"; for($i=0; $i <= $#in; $i++) { $out = $in[$i]; $out =~ s/_PUT_REGEX_HERE_//; $succ = ($out eq $exp[$i] ? 'PASS' : 'FAIL'); #if($succ eq 'FAIL') #{ printf("%s | %- 24s | %- 24s | %- 24s\n", $succ, $in[$i], $out, $exp[$i]); #} }

    Read the article

  • Regular Expressions, Checking for a range of occurrences

    - by gmcalab
    I have a phone number I want to match against a regular expression. The format of the phone number must match this: (123) 123-4567 x12345 The extension is optional. Also the extension must contain 1-5 numbers. Below is a regular expression I wrote that works. ^\(\d{3}\) \d{3}-\d{4}( x\d\d?\d?\d?\d?)?$ I was wondering if there is a better way to check for the extension instead of x\d\d?\d?\d?\d? Can I say 1-5 occurrences of \d instead of the above some how ?

    Read the article

< Previous Page | 25 26 27 28 29 30 31 32 33 34 35 36  | Next Page >