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  • Haskell: how to get through 'no instance for' ?

    - by artemave
    I am learning Haskell. I am on the 8th chapter of this book. The main thing I've learned so far is that Haskell is very unfriendly to me and it bites my ass where possible. Moreover... Heck! Enough mourning, to business. Here is the code: module GlobRegex ( globToRegex, matchesGlob ) where import Text.Regex.Posix import Text.Regex.Posix.String import Text.Regex.Base.RegexLike data CaseOpt = Case | NoCase deriving (Eq) matchesGlob :: String -> String -> CaseOpt -> Bool matchesGlob name pat caseopt = match regex name where regex = case caseopt of NoCase -> makeRegexOpts (defaultCompOpt + compIgnoreCase) defaultExecOpt (globToRegex pat) Case -> makeRegex (globToRegex pat) globToRegex :: String -> String ... And here is how it fails to compile: Prelude Text.Regex.Posix Text.Regex.Base.RegexLike> :load globtoregex\GlobRegex. hs [1 of 1] Compiling GlobRegex ( globtoregex\GlobRegex.hs, interpreted ) globtoregex\GlobRegex.hs:14:31: No instance for (RegexLike regex [Char]) arising from a use of `match' at globtoregex\GlobRegex.hs:14:31-46 Possible fix: add an instance declaration for (RegexLike regex [Char]) In the expression: match regex name In the definition of `matchesGlob': matchesGlob name pat caseopt = match regex name where regex = case caseopt of { NoCase -> makeRegexOpts (defaultCompOpt + compIgnoreCase) defaultExecOpt (globToRegex pat) Case -> makeRegex (globToRegex pat) } globtoregex\GlobRegex.hs:17:23: No instance for (RegexMaker regex CompOption execOpt String) arising from a use of `makeRegex' at globtoregex\GlobRegex.hs:17:23-49 Possible fix: add an instance declaration for (RegexMaker regex CompOption execOpt String) In the expression: makeRegex (globToRegex pat) In a case alternative: Case -> makeRegex (globToRegex pat) In the expression: case caseopt of { NoCase -> makeRegexOpts (defaultCompOpt + compIgnoreCase) defaultExecOpt (globToRegex p at) Case -> makeRegex (globToRegex pat) } Failed, modules loaded: none. To my best understanding, Text.Regex.Posix.String provides instances for RegexLike Regex String and RegexMaker Regex CompOption ExecOption String, so it should work. On the other hand, I can see that regex in the error message is type variable, not a concrete type, so, perhaps not... Anyway, this is where I am stuck. May be there is a common pattern for resolving no instance for type of problems? Or, in Haskell terms, instance of SmartGuess typeclass for no instance for?

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  • Java Regular Expressions

    - by david robers
    Hi All, Im struggling to understand the regex documentation. How would I find the strings that contain exactly one C in the following text: ABCCAMNL YOOBABCCA XNABCCA ZDXUABCCA TAQABCC ISABCCA REABCCA CABCAMONPT Edit: private void matchIt(String regex, ArrayList<String> d) { Pattern p = Pattern.compile("[\\w^C]"); Matcher m = p.matcher(regex); for (int i = 0; i < d.size(); i++) { p.matcher(d.get(i)); if(m.find()){ out.println(d.get(i)); } } } i have the above function and it only outputs: ABCCAMNL YOOBABCCA Why is that?

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  • Word mergefield wildcard not correctly matching

    - by aZn137
    Hello, Below is my mergefield code: { IF { MERGEFIELD Subs_State } = "GA" "blah blah" "{ IF { MERGEFIELD CEOrgStates } = "GA" "blah blah" ""} "} I'm pulling records from a MS Access db. My goal is to check whether a record has Subs_State field matching "GA", or the CEOrgStates has the word "GA" (some records have stuff like "|FL|CA|GA|CT|KY|" (no quotes)). When I merged the docs, Word doesnt seem to be able to match with the wildcards: If I use and compare "*GA" (fields ending with GA), it works; however, the double wildcards "*GA*" dont seem to work at all. Here are the things I’ve tried: Have data in lowercase, then compare with lowercase Have data in lowercase, convert to and then compare with uppercase Do the opposite of the above 2 with uppercase data Use “*GA*” and “*ga*” (no pipe) Use different delimiters Nothing seems to work with the double wildcard matching. What am I doing wrong? Thanks!

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  • How do you match only valid roman numerals with a regular expression?

    - by Daniel Magliola
    Thinking about my other problem, i decided I can't even create a regular expression that will match roman numerals (let alone a context-free grammar that will generate them) The problem is matching only valid roman numerals. Eg, 990 is NOT "XM", it's "CMXC" My problem in making the regex for this is that in order to allow or not allow certain characters, I need to look back. Let's take thousands and hundreds, for example. I can allow M{0,2}C?M (to allow for 900, 1000, 1900, 2000, 2900 and 3000). However, If the match is on CM, I can't allow following characters to be C or D (because I'm already at 900). How can I express this in a regex? If it's simply not expressible in a regex, is it expressible in a context-free grammar? Thanks for any pointers!

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  • A regular expression that will allow a string with only one Capital Letter

    - by Phoenix
    The string should be 6 - 20 characters in length. And it should contain 1 Capital letter. I can do this in code using C# string st = "SomeString" Regex rg = new Regex("[A-Z]"); MatchCollection mc = rg.Matches(st); Console.WriteLine("Total Capital Letters: " + mc.Count); if (mc.Count > 1) { return false; } But what i really want is a Regular expression that will match my string if it only contains one capital. The string can start with a common letter and should have only letters. Thanks In advance. (I did look at some of the other RegEx questions but they did not help).

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  • NSPredicate error/behaving differently on 10.5 vs 10.6

    - by Tristan
    I am using a NSPredicate to determine if an entered email address is valid. On 10.6 it works perfectly as expected. I recently decided to get my app going on 10.5 and this is the only thing that doesn't work. The error i get is as follows: "Can't do regex matching, reason: Can't open pattern U_MALFORMED_SET (string [email protected], pattern ([\w-+]+(?:\.[\w-+]+)*@(?:[\w-]+\.)+[a-zA-Z]{2,7}), case 0, canon 0)" The code im using is as follows: NSString *regex = @"([\\w-+]+(?:\\.[\\w-+]+)*@(?:[\\w-]+\\.)+[a-zA-Z]{2,7})"; NSPredicate *regextest = [NSPredicate predicateWithFormat:@"SELF MATCHES %@", regex]; if ([regextest evaluateWithObject:[userEmail objectValue]] == YES) Does anyone know why this isn't working on 10.5? And how I might get it working or be able to do this test in a way compatible for both 10.5 and 10.6?

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  • Regular Expression Help in .NET

    - by Matt H.
    I have a simple pattern I am trying to match, any characters captured between parenthesis at the end of an HTML paragraph. I am running into trouble any time there is additional parentheticals in that paragraph: i.e. If the input string is "..... (321)</p" i want to get the value (321) However, if the paragraph has this text: "... (123) (321)</p" my regex is returning "(123) (321)" (everything between the opening "(" and closing ")" I am using the regex pattern "\s(.+)</p" How can I grab the correct value (using VB.NET) This is what I'm doing so far: Dim reg As New Regex("\s\(.+\)</P>", RegexOptions.IgnoreCase) Dim matchC As MatchCollection = reg.Matches(su.Question) If matchC.Count > 0 Then Dim lastMatch As Match = matchC(matchC.Count - 1) Dim DesiredValue As String = lastMatch.Value End If

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  • Matching tag in HTML keyboard shortcut

    - by kape123
    Is there a shortcut in Visual Studio (2008) that will allow me to jump to matching HTML tag... as CTRL+] does for matching braces when you are in code view? Example: <table> <tr> <td> </td> </tr> </table|> Cursor is on closing table tag and I would like to press something like CTRL+] to jump to opening table tag. Any ideas?

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  • How to extract valid email from larger string in Scala

    - by luigi-prog
    My scala version 2.7.7 Im trying to extract an email adress from a larger string. the string itself follows no format. the code i've got: import scala.util.matching.Regex import scala.util.matching._ val Reg = """\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b""".r "yo my name is joe : [email protected]" match { case Reg(e) => println("match: " + e) case _ => println("fail") } the Regex passes in RegExBuilder but does not pass for scala. Also if there is another way to do this without regex that would be fine also. Thanks!

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  • Regular expression for parsing CSV in PHP

    - by Discodancer
    I already managed to split the CSV file using this regex: "/,(?=(?:[^\"]\"[^\"]\")(?![^\"]\"))/" But I ended up with an array of strings that contain the opening and ending double quotes. Now I need a regex that would strip those strings of the delimiter double quotes. As far as I know the CSV format can encapsulate strings in double quotes, and all the double quotes that are already a part of the string are doubled. For example: My "other" cat becomes "My ""other"" cat" What I basically need is a regex that will replace all sequences of N doublequotes with a sequence of (N/2 - rounded down) double quotes. Or is there a better way ? Thanks in advance.

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  • Oracle - Determine maximum supported size for regular expression

    - by FrustratedWithFormsDesigner
    I have a regular expression that throws ORA-12733, "regular expression is too long". How do I determine what the maximum supported size is? FYI: the offending regex is 892 characters. It's a generated regex, so I could change how I generate and execute it, but I would like to know what the limits to the max size are before I change how I am generating and executing. (running Oracle 10.2g) UPDATE: If it depends on the actual regex, here's the begining of it (the rest is just the same thing repeated, with different values between ^ and $): (^R_1A$|^R_2A$|^R_3A$|^R_4A$|^R_4B$|^R_5A$|^R_5B$...

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  • Need help with REGEXP_REPLACE

    - by Randy
    hey regex guru's I have a data column that contains a substring similar to this: 'This is a string with ID=123 contained inside' i need to replace the ID=123 with another fixed string say ID=1 i have a working REPLACE call that swaps out the values correctly, but this only works on a known original id number. I can extend to make it work when the number of digits are known by using some substr magic, however, the id number may have an arbitrary number of digits, so i'm thinking regex. any help with the regex magix to grab the entire ID=999 substring no matter how many digits would be very helpful. thanks

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  • Does lookaround affect which languages can be matched by regular expressions?

    - by sepp2k
    There are some features in modern regex engines which allow you to match languages that couldn't be matched without that feature. For example the following regex using back references matches the language of all strings that consist of a word that repeats itself: (.+)\1. This language is not regular and can't be matched by a regex, which does not use back references. My question: Does lookaround also affect which languages can be matched by a regular expression? I.e. are there any languages that can be matched using lookaround, which couldn't be matched otherwise? If so, is this true for all flavors of lookaround (negative or positive lookahead or lookbehind) or just for some of them?

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  • Get first character of each word and its position in a sentence/paragraph

    - by Radhika
    I am trying to create a map by taking the first character of each word and it's position in a sentence/paragraph. I am using regex pattern to achieve this. Regex is a costly operation. Are there are any ways to achieve this? Regex way: public static void getFirstChar(String paragraph) { Pattern pattern = Pattern.compile("(?<=\\b)[a-zA-Z]"); Map newMap = new HashMap(); Matcher fit = pattern.matcher(paragraph); while (fit.find()) { newMap.put((fit.group().toString().charAt(0)), fit.start()); } }

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  • PHP and MySQL - Printing rows matching a column value

    - by Michael
    Hello, I need to write a PHP script that will print out results from a MySQL database. For example, say I have 9 fields. Field 1 is an auto increasing number, field two is a three digit number. I need to be able to have a script read, find the matching number (it'll be from a POST), and then display all matching three digit results, and the 7 other fields as well. I am already logged in to the database in this script. I guess I'm really at a loss of where to begin. How would one start something like this? Thank you.

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  • g++ no matching function call error

    - by gufftan
    I've got a compiler error but I can't figure out why. the .hpp: #ifndef _CGERADE_HPP #define _CGERADE_HPP #include "CVektor.hpp" #include <string> class CGerade { protected: CVektor o, rv; public: CGerade(CVektor n_o, CVektor n_rv); CVektor getPoint(float t); string toString(); }; the .cpp: #include "CGerade.hpp" CGerade::CGerade(CVektor n_o, CVektor n_rv) { o = n_o; rv = n_rv.getUnitVector(); } the error message: CGerade.cpp:10: error: no matching function for call to ‘CVektor::CVektor()’ CVektor.hpp:28: note: candidates are: CVektor::CVektor(float, float, float) CVektor.hpp:26: note: CVektor::CVektor(bool, float, float, float) CVektor.hpp:16: note: CVektor::CVektor(const CVektor&) CGerade.cpp:10: error: no matching function for call to ‘CVektor::CVektor()’ CVektor.hpp:28: note: candidates are: CVektor::CVektor(float, float, float) CVektor.hpp:26: note: CVektor::CVektor(bool, float, float, float) CVektor.hpp:16: note: CVektor::CVektor(const CVektor&)

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  • Regular Expression Program

    - by david robers
    Hi I have the following text: SMWABCCA ABCCAEZZRHM NABCCAYJG XABCCA ABCCADK ABCCASKIYRH ABCCAKY PQABCCAK ABCCAKQ This method takes a regex in out by the user and SHOULD print out the Strings it applies to but seems to print out something completely different: private void matchIt(String regex) { Pattern p = Pattern.compile(regex); Matcher m = null; boolean found = false; for(int i = 0; i < data.length; i++){ m = p.matcher(data[i]); if(m.find()){ out.println(data[i]); found = true; } } if(!found){ out.println("Pattern Not Found"); } } When inputting "[C]" It outputs: SMWABCCA ABCCAEZZRHM NABCCAYJG XABCCA ABCCADK ABCCASKIYRH ABCCAKY PQABCCAK ABCCAKQ Any ideas why? I think I'm using m.find() improperly...

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  • RegExp to match fraction

    - by user3627265
    I'm trying to perform regex to match a fraction. The user will input a fraction eg., 1/4, 1 1/2 10/2 so on. I have tested this regex and it works, but the problem is when I type in 10, 20, 30, 40 so on It does not recognized these values. This is my regex As you can see, it first sorted out the integer and then the slash and lastly the integer after the slash. var check_zero_value = str1.match(/[1-9]\/[1-9]/g); if(!check_zero_value1) { return false; } Any idea on this?

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  • Get backreferences values and modificate these values

    - by roasted
    Could you please explain why im not able to get values of backreferences from a matched regex result and apply it some modification before effective replacement? The expected result is replacing for example string ".coord('X','Y')" by "X * Y". But if X to some value, divide this value by 2 and then use this new value in replacement. Here the code im currently testing: See /*>>1<<*/ & /*>>2<<*/ & /*>>3<<*/, this is where im stuck! I would like to be able to apply modification on backrefrences before replacement depending of backreferences values. Difference between /*>>2<<*/ & /*>>3<<*/ is just the self call anonymous function param The method /*>>2<<*/ is the expected working solution as i can understand it. But strangely, the replacement is not working correctly, replacing by alias $1 * $2 and not by value...? You can test the jsfiddle //string to test ".coord('125','255')" //array of regex pattern and replacement //just one for the example //for this example, pattern matching alphanumerics is not necessary (only decimal in coord) but keep it as it var regexes = [ //FORMAT is array of [PATTERN,REPLACEMENT] /*.coord("X","Y")*/ [/\.coord\(['"]([\w]+)['"],['"]?([\w:\.\\]+)['"]?\)/g, '$1 * $2'] ]; function testReg(inputText, $output) { //using regex for (var i = 0; i < regexes.length; i++) { /*==>**1**/ //this one works as usual but dont let me get backreferences values $output.val(inputText.replace(regexes[i][0], regexes[i][2])); /*==>**2**/ //this one should works as i understand it $output.val(inputText.replace(regexes[i][0], function(match, $1, $2, $3, $4) { $1 = checkReplace(match, $1, $2, $3, $4); //here want using $1 modified value in replacement return regexes[i][3]; })); /*==>**3**/ //this one is just a test by self call anonymous function $output.val(inputText.replace(regexes[i][0], function(match, $1, $2, $3, $4) { $1 = checkReplace(match, $1, $2, $3, $4); //here want using $1 modified value in replacement return regexes[i][4]; }())); inputText = $output.val(); } } function checkReplace(match, $1, $2, $3, $4) { console.log(match + ':::' + $1 + ':::' + $2 + ':::' + $3 + ':::' + $4); //HERE i should be able if lets say $1 > 200 divide it by 2 //then returning $1 value if($1 > 200) $1 = parseInt($1 / 2); return $1; }? Sure I'm missing something, but cannot get it! Thanks for your help, regards. EDIT WORKING METHOD: Finally get it, as mentionned by Eric: The key thing is that the function returns the literal text to substitute, not a string which is parsed for backreferences.?? JSFIDDLE So complete working code: (please note as pattern replacement will change for each matched pattern and optimisation of speed code is not an issue here, i will keep it like that) $('#btn').click(function() { testReg($('#input').val(), $('#output')); }); //array of regex pattern and replacement //just one for the example var regexes = [ //FORMAT is array of [PATTERN,REPLACEMENT] /*.coord("X","Y")*/ [/\.coord\(['"]([\w]+)['"],['"]?([\w:\.\\]+)['"]?\)/g, '$1 * $2'] ]; function testReg(inputText, $output) { //using regex for (var i = 0; i < regexes.length; i++) { $output.val(inputText.replace(regexes[i][0], function(match, $1, $2, $3, $4) { var checkedValues = checkReplace(match, $1, $2, $3, $4); $1 = checkedValues[0]; $2 = checkedValues[1]; regexes[i][1] = regexes[i][1].replace('$1', $1).replace('$2', $2); return regexes[i][1]; })); inputText = $output.val(); } } function checkReplace(match, $1, $2, $3, $4) { console.log(match + ':::' + $1 + ':::' + $2 + ':::' + $3 + ':::' + $4); if ($1 > 200) $1 = parseInt($1 / 2); if ($2 > 200) $2 = parseInt($2 / 2); return [$1,$2]; }

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  • Bash script to create mass series of directories

    - by Volomike
    I need to create a Bash script to go into every user's home folder, seek out a wp-content folder, create a directory uploads under it, and then chmod 0756 uploads. How do I achieve this? I imagine I need to use find with a regexp/regex, and then tell it to run another bash script on the results.

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  • Text-Editing program : muti-search-replace/multi-regex?

    - by rlb.usa
    I have a long and arduous text file, and I need to do lots and lots of the same search-replaces on it inside of selections. Is there a text editing program where I can do multiple find/replace (or regex) at one time? That is, I want to : (select text) - (do-find-replace-set-A) - (do other stuff) - (repeat) Instead of : (select text) - (f&r #1, f&r #2, f&r #3 ... ) - (do other stuff) - (repeat) I have textpad, but it's macro's won't handle find/replace.

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  • Deleting Duplicated Lines In TEXT File?

    - by echolab
    I am trying to cleanup a text and for some reason every line duplicated 3 times am i able to get ride of duplicates with regex or tricks or do you know a software which could do that , text file is like this Party Started 10:17 (89/1/2) Party Started 10:17 (89/1/2) Party Started 10:17 (89/1/2) Jessica At Dinner 17:54 (89/1/2) Jessica At Dinner 17:54 (89/1/2) Jessica At Dinner 17:54 (89/1/2) How can i clean it up , and get ride of duplicated lines , it's about 69,587 lines

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