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  • Sorting a list in OCaml

    - by darkie15
    Hi All, Here is the code on sorting any given list: let rec sort lst = match lst with [] -> [] | head :: tail -> insert head (sort tail) and insert elt lst = match lst with [] -> [elt] | head :: tail -> if elt <= head then elt :: lst else head :: insert elt tail;; [Source: Code However, I am getting an Unbound error: Unbound value tail # let rec sort lst = match lst with [] -> [] | head :: tail -> insert head (sort tail) and insert elt lst = match lst with [] -> [elt] | head :: tail -> if elt <= head then elt :: lst else head :: insert elt tail;; Characters 28-29: | head :: tail -> if elt <= head then elt :: lst else head :: insert elt tail;; ^ Error: Syntax error Can anyone please help me understand the issue here?? I did not find head or tail to be predefined anywhere nor in the code

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  • MySQL Full-Text Search Across Multiple Tables - Quick/Long Solution?

    - by Kerry
    Hello all, I have been doing a bit of research on full-text searches as we realized a series of LIKE statements are terrible. My first find was MySQL full-text searches. I tried to implement this and it worked on one table, failed when I was trying to join multiple tables, and so I consulted stackoverflow's articles (look at the end for a list of the ones I've been to) I didn't see anything that clearly answered my questions. I'm trying to get this done literally in an hour or two (quick solution) but I also want to do a better long term solution. Here is my query: SELECT a.`product_id`, a.`name`, a.`slug`, a.`description`, b.`list_price`, b.`price`, c.`image`, c.`swatch`, e.`name` AS industry FROM `products` AS a LEFT JOIN `website_products` AS b ON (a.`product_id` = b.`product_id`) LEFT JOIN ( SELECT `product_id`, `image`, `swatch` FROM `product_images` WHERE `sequence` = 0) AS c ON (a.`product_id` = c.`product_id`) LEFT JOIN `brands` AS d ON (a.`brand_id` = d.`brand_id`) INNER JOIN `industries` AS e ON (a.`industry_id` = e.`industry_id`) WHERE b.`website_id` = 96 AND b.`status` = 1 AND b.`active` = 1 AND MATCH( a.`name`, a.`sku`, a.`description`, d.`name` ) AGAINST ( 'ashley sofa' ) GROUP BY a.`product_id` ORDER BY b.`sequence` LIMIT 0, 9 The error I get is: Incorrect arguments to MATCH If I remove d.name from the MATCH statement it works. I have a full-text index on that column. I saw one of the articles say to use an OR MATCH for this table, but won't that lose the effectiveness of being able to rank them together or match them properly? Other places said to use UNIONs but I don't know how to do that properly. Any advice would be greatly appreciated. In the idea of a long term solution it seems that either Sphinx or Lucene is best. Now by no means and I a MySQL guru, and I heard that Lucene is a bit more complicated to setup, any recommendations or directions would be great. Articles: http://stackoverflow.com/questions/1117005/mysql-full-text-search-across-multiple-tables http://stackoverflow.com/questions/668371/mysql-fulltext-search-across-1-table http://stackoverflow.com/questions/2378366/mysql-how-to-make-multiple-table-fulltext-search http://stackoverflow.com/questions/737275/pros-cons-of-full-text-search-engine-lucene-sphinx-postgresql-full-text-searc http://stackoverflow.com/questions/1059253/searching-across-multiple-tables-best-practices

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  • Efficiently Combine MatchCollections in .Net Regex

    - by Laramie
    In the simplified example, there are 2 Regular Expressions, one case sensitive, the other not. The idea would be to efficiently create an IEnumerable collection (see "combined" below) combining the results. string test = "abcABC"; string regex = "(?<grpa>a)|(?<grpb>b)|(?<grpc>c)]"; Regex regNoCase = new Regex(regex, RegexOptions.IgnoreCase); Regex regCase = new Regex(regex); MatchCollection matchNoCase = regNoCase.Matches(test); MatchCollection matchCase = regCase.Matches(test); //Combine matchNoCase and matchCase into an IEnumerable IEnumerable<Match> combined= null; foreach (Match match in combined) { //Use the Index and (successful) Groups properties //of the match in another operation } In practice, the MatchCollections might contain thousands of results and be run frequently using long dynamically created REGEXes, so I'd like to shy away from copying the results to arrays, etc. I am still learning LINQ and am fuzzy on how to go about combining these or what the performance hits to an already sluggish process will be.

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  • Kohana ORM Aliasing and "Trying to get property of non-object"

    - by Toto
    I have the following tables in the database: teams: id name matches: id team1_id team2_id I've defined the following ORM models in my Kohana application: class Match_Model extends ORM { protected $belongs_to = array('team1_id' => 'team', 'team2_id' => 'team'); } class Team_Model extends ORM { protected $has_many = array('matches'); } The following code in a controller: $match = ORM::factory('match',1); echo $match->team1_id->name; /* <-- */ Is throwing the following error on the linke marked with /* <--- */: Trying to get property of non-object The framework is yielding the value of the foreign key instead of a reference to a Match_Model instance as it should (giving the has_many and belongs_to properties stated). Am I missing something? Note: Just in case, I've added the irregular plural 'match' => 'matches' in application/config/inflector.php

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  • Mysql query problem....

    - by Avinash
    I have below values in my database. been Lorem Ipsum and scrambled ever scrambledtexttextofandtooktooktypetexthastheunknownspecimenstandardsincetypesett Here is my query: SELECT nBusinessAdID, MATCH (`sHeadline`) AGAINST ("text" IN BOOLEAN MODE) AS score FROM wiki_businessads WHERE MATCH (`sHeadline`) AGAINST ("text" IN BOOLEAN MODE) AND bDeleted ="0" AND nAdStatus ="1" ORDER BY score DESC, bPrimeListing DESC, dDateCreated DESC It's not fetching first result, why? It should fetch first result because its contain text word in it. I have disabled the stopword filtering. This one is also not working SELECT nBusinessAdID, MATCH (`sHeadline`) AGAINST ('"text"' IN BOOLEAN MODE) AS score FROM wiki_businessads WHERE MATCH (`sHeadline`) AGAINST ('"text"' IN BOOLEAN MODE) AND bDeleted ="0" AND nAdStatus ="1" ORDER BY score DESC, bPrimeListing DESC, dDateCreated DESC Thanks Avinash

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  • Can I write this regex in one step?

    - by Marin Doric
    This is the input string "23x +y-34 x + y+21x - 3y2-3x-y+2". I want to surround every '+' and '-' character with whitespaces but only if they are not allready sourrounded from left or right side. So my input string would look like this "23x + y - 34 x + y + 21x - 3y2 - 3x - y + 2". I wrote this code that does the job: Regex reg1 = new Regex(@"\+(?! )|\-(?! )"); input = reg1.Replace(input, delegate(Match m) { return m.Value + " "; }); Regex reg2 = new Regex(@"(?<! )\+|(?<! )\-"); input = reg2.Replace(input, delegate(Match m) { return " " + m.Value; }); explanation: reg1 // Match '+' followed by any character not ' ' (whitespace) or same thing for '-' reg2 // Same thing only that I match '+' or '-' not preceding by ' '(whitespace) delegate 1 and 2 just insert " " before and after m.Value ( match value ) Question is, is there a way to create just one regex and just one delegate? i.e. do this job in one step? I am a new to regex and I want to learn efficient way.

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  • How does MatchEvaluator works? ( C# regex replace)

    - by Marin Doric
    This is the input string 23x * y34x2. I want to insert " * " (star surrounded by whitespaces) after every number followed by letter, and after every letter followed by number. So my input string would look like this: 23 * x * y * 34 * x * 2. This is the regex that does the job: @"\d(?=[a-z])|a-z". This is the function that I wrote that inserts the " * ". Regex reg = new Regex(@"\d(?=[a-z])|[a-z](?=\d)"); MatchCollection matchC; matchC = reg.Matches(input); int ii = 1; foreach (Match element in matchC)//foreach match I will find the index of that match { input = input.Insert(element.Index + ii, " * ");//since I' am inserting " * " ( 3 characters ) ii += 3; //I must increment index by 3 } return input; //return modified input My question how to do same job using .net MatchEvaluator? I'am new to regex and don't understand good replacing with MatchEvaluator. This is the code that I tried to wrote: Regex reg = new Regex(@"\d(?=[a-z])|[a-z](?=\d)"); MatchEvaluator matchEval = new MatchEvaluator(ReplaceStar); input = reg.Replace(input, matchEval); return input; } public string ReplaceStar( Match match ) { //return What?? }

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  • How to find a word within text using XSLT 2.0 and REGEX (which doesn't have \b word boundary)?

    - by Mads Hansen
    I am attempting to scan a string of words and look for the presence of a particular word(case insensitive) in an XSLT 2.0 stylesheet using REGEX. I have a list of words that I wish to iterate over and determine whether or not they exist within a given string. I want to match on a word anywhere within the given text, but I do not want to match within a word (i.e. A search for foo should not match on "food" and a search for bar should not match on "rebar"). XSLT 2.0 REGEX does not have a word boundary(\b), so I need to replicate it as best I can.

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  • Regex to check that a character in range doesn't repeat

    - by Aly
    Hi, I want to match against Strings such as AhKs & AdKs (i.e. two cards Ah = Ace of Hearts). I want to match two off-suit cards with a regex, what I currently have is "^[AKQJT2-9][hscd]{2}$", but this could match hands such as AhKh (suited) and AhAh. Is there a way to possibly use backreferences to say the second [hscd] cannot be the same as the firs (similarly for [AKQJT2-9])

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  • ruby syntactic sugar: dealing with nils..

    - by luca
    probably asked already but I couldn't find it.. here are 2 common situation (for me while programming rails..) that are frustrating to write in ruby: "a string".match(/abc(.+)abc/)[1] in this case I get an error because the string doesn't match, therefore the [] operator is called upon nil. What I'd like to find is a nicer alternative to the following: temp="a string".match(/abc(.+)abc/); temp.nil? ? nil : temp[1] in brief, if it didn't match simply return nil without the error The second situation is this one: var = something.very.long.and.tedious.to.write var = something.other if var.nil? In this case I want to assign something to var only if it's not nil, in case it's nil I'll assign something.other.. Any suggestion? Thanks!

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  • Binary Search Help

    - by aloh
    Hi, for a project I need to implement a binary search. This binary search allows duplicates. I have to get all the index values that match my target. I've thought about doing it this way if a duplicate is found to be in the middle: Target = G Say there is this following sorted array: B, D, E, F, G, G, G, G, G, G, Q, R S, S, Z I get the mid which is 7. Since there are target matches on both sides, and I need all the target matches, I thought a good way to get all would be to check mid + 1 if it is the same value. If it is, keep moving mid to the right until it isn't. So, it would turn out like this: B, D, E, F, G, G, G, G, G, G (MID), Q, R S, S, Z Then I would count from 0 to mid to count up the target matches and store their indexes into an array and return it. That was how I was thinking of doing it if the mid was a match and the duplicate happened to be in the mid the first time and on both sides of the array. Now, what if it isn't a match the first time? For example: B, D, E, F, G, G, J, K, L, O, Q, R, S, S, Z Then as normal, it would grab the mid, then call binary search from first to mid-1. B, D, E, F, G, G, J Since G is greater than F, call binary search from mid+1 to last. G, G, J. The mid is a match. Since it is a match, search from mid+1 to last through a for loop and count up the number of matches and store the match indexes into an array and return. Is this a good way for the binary search to grab all duplicates? Please let me know if you see problems in my algorithm and hints/suggestions if any. The only problem I see is that if all the matches were my target, I would basically be searching the whole array but then again, if that were the case I still would need to get all the duplicates. Thank you BTW, my instructor said we cannot use Vectors, Hash or anything else. He wants us to stay on the array level and get used to using them and manipulating them.

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  • Word VBA - Find text between delimiters and convert to lower case

    - by jJack
    I would like to find text which is between the < and characters, and then turn any found text into "normal" case, where first letter of word is capitalized. Here is what I have thus far: Function findTextBetweenCarots() As String Dim strText As String With Selection .Find.Text = "<" ' what about <[^0-9]+> ? .Find.Forward = True .Find.Wrap = wdFindContinue End With Selection.Find.Execute ' Application.Selection. ' how do I get the text between the other ">"? findCarotSymb = Application.Selection.Text End Function Or, is there a better way of doing this? I also approached the problem using the VBScript Regex 5.5 library, which worked on simple documents, but not on certain documents with complex tables. For example, trying to just bold the text (for simplicity): Sub BoldUpperCaseWords() Dim regEx, Match, Matches Dim rngRange As Range Set regEx = New RegExp regEx.Pattern = "<[^0-9]+>" regEx.IgnoreCase = False regEx.Global = True Set Matches = regEx.Execute(ActiveDocument.Range.Text) For Each Match In Matches ActiveDocument.Range(Match.FirstIndex, Match.FirstIndex + Len(Match.Value)).Bold = True Next End Sub would not work in a document with tables. In fact, it would not even bold the correct text (the text between the <. This leads me to believe I have a broader issue here that I am missing. Here is what a sample doc looks like. Notice the wrong text is bold:

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  • RegEx to ignore / skip everything in html tags

    - by Scott Sumpter
    Looking for a way to combine two Regular Expressions. One to catch the urls and the other to ensure is skips text within html tags. See sample text below functions. Need to pass a block of news text and format text by wrapping urls and email addresses in html tags so users don't have to. The below code works great until there are already html tags within the text. In that case it doubles the html tags. There are plenty of examples to strip html, but I want to just ignore it since the url is already linkified. Also - if there is an easier was to accomplish this, with or without Regex, please let me know. none of my attempts to combine Regexs have worked. coding in ASP.NET VB but will take any workable example/direction. Thanks! ===== Functions ============= Public Shared Function InsertHyperlinks(ByVal inText As String) As String Dim strBuf As String Dim objMatches As Object Dim iStart, iEnd As Integer strBuf = "" iStart = 1 iEnd = 1 Dim strRegUrlEmail As String = "\b(www|http|\S+@)\S+\b" 'RegEx to find urls and email addresses Dim objRegExp As New Regex(strRegUrlEmail, RegexOptions.IgnoreCase) 'Match URLs and emails Dim MatchList As MatchCollection = objRegExp.Matches(inText) If MatchList.Count <> 0 Then objMatches = objRegExp.Matches(inText) For Each Match In MatchList iEnd = Match.Index strBuf = strBuf & Mid(inText, iStart, iEnd - iStart + 1) If InStr(1, Match.Value, "@") Then strBuf = strBuf & HrefGet(Match.Value, "EMAIL", "_BLANK") Else strBuf = strBuf & HrefGet(Match.Value, "WEB", "_BLANK") End If iStart = iEnd + Match.Length + 1 Next strBuf = strBuf & Mid(inText, iStart) InsertHyperlinks = strBuf Else 'No hyperlinks to replace InsertHyperlinks = inText End If End Function Shared Function HrefGet(ByVal url As String, ByVal urlType As String, ByVal Target As String) As String Dim strBuf As String strBuf = "<a href=""" If UCase(urlType) = "WEB" Then If LCase(Left(url, 3)) = "www" Then strBuf = "<a href=""http://" & url & """ Target=""" & _ Target & """>" & url & "</a>" Else strBuf = "<a href=""" & url & """ Target=""" & _ Target & """>" & url & "</a>" End If ElseIf UCase(urlType) = "EMAIL" Then strBuf = "<a href=""mailto:" & url & """ Target=""" & _ Target & """>" & url & "</a>" End If HrefGet = strBuf End Function ===== Sample Text ============= This would be the inText parameter. Midway through the ride, we see a Skip this too. But sometimes we go here [insert normal www dot link dot com]. If you'd like to join us contact Bill Smith at [email protected]. Thanks! sorry stack overflow won't allow multiple hyperlinks to be added. ===== End Sample Text =============

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  • Regular Expression Longes Possible Matching

    - by syker
    So I have an input string which is a directory addres: Example: ProgramFiles/Micro/Telephone And I want to match it against a list of words very strictly: Example: Tel|Tele|Telephone I want to match against Telephone and not Tel. Right now my reg looks like this: my( $output ) = ( $input =~ m/($list)/o ); The regex above will match against Tel. What can I do to fix it?

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  • Doubt about django model API

    - by Clash
    Hello guys! So, here is what I want to do. I have a model Staff, that has a foreign key to the User model. I also have a model Match that has a foreign key to the User model. I want to select how much Matches every Staff has. I don't know how to do that, so far I only got it working for the User model. From Staff, it will not allow to annonate Match. This is what is working right now User.objects.annotate(ammount=Count("match")).filter(Q(ammount__gt=0)).order_by("ammount") And this is what I wanted to do Staff.objects.annotate(ammount=Count("match")).filter(Q(ammount__gt=0)).order_by("ammount") And by the way, is there any way to filter the matches? I want to filter the matches by a certain column. Thanks a lot in advance!

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  • How to find the insertion point in an array using binary search?

    - by ????
    The basic idea of binary search in an array is simple, but it might return an "approximate" index if the search fails to find the exact item. (we might sometimes get back an index for which the value is larger or smaller than the searched value). For looking for the exact insertion point, it seems that after we got the approximate location, we might need to "scan" to left or right for the exact insertion location, so that, say, in Ruby, we can do arr.insert(exact_index, value) I have the following solution, but the handling for the part when begin_index >= end_index is a bit messy. I wonder if a more elegant solution can be used? (this solution doesn't care to scan for multiple matches if an exact match is found, so the index returned for an exact match may point to any index that correspond to the value... but I think if they are all integers, we can always search for a - 1 after we know an exact match is found, to find the left boundary, or search for a + 1 for the right boundary.) My solution: DEBUGGING = true def binary_search_helper(arr, a, begin_index, end_index) middle_index = (begin_index + end_index) / 2 puts "a = #{a}, arr[middle_index] = #{arr[middle_index]}, " + "begin_index = #{begin_index}, end_index = #{end_index}, " + "middle_index = #{middle_index}" if DEBUGGING if arr[middle_index] == a return middle_index elsif begin_index >= end_index index = [begin_index, end_index].min return index if a < arr[index] && index >= 0 #careful because -1 means end of array index = [begin_index, end_index].max return index if a < arr[index] && index >= 0 return index + 1 elsif a > arr[middle_index] return binary_search_helper(arr, a, middle_index + 1, end_index) else return binary_search_helper(arr, a, begin_index, middle_index - 1) end end # for [1,3,5,7,9], searching for 6 will return index for 7 for insertion # if exact match is found, then return that index def binary_search(arr, a) puts "\nSearching for #{a} in #{arr}" if DEBUGGING return 0 if arr.empty? result = binary_search_helper(arr, a, 0, arr.length - 1) puts "the result is #{result}, the index for value #{arr[result].inspect}" if DEBUGGING return result end arr = [1,3,5,7,9] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [] b = 60 arr.insert(binary_search(arr, b), b) p arr and result: Searching for 6 in [1, 3, 5, 7, 9] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 6, arr[middle_index] = 5, begin_index = 3, end_index = 2, middle_index = 2 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9] Searching for 6 in [1, 3, 5, 7, 9, 11] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 6, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 3, middle_index = 3 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9, 11] Searching for 60 in [1, 3, 5, 7, 9] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 60, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 60, arr[middle_index] = 9, begin_index = 4, end_index = 4, middle_index = 4 the result is 5, the index for value nil [1, 3, 5, 7, 9, 60] Searching for 60 in [1, 3, 5, 7, 9, 11] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 60, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 60, arr[middle_index] = 11, begin_index = 5, end_index = 5, middle_index = 5 the result is 6, the index for value nil [1, 3, 5, 7, 9, 11, 60] Searching for -60 in [1, 3, 5, 7, 9] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 9, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9] Searching for -60 in [1, 3, 5, 7, 9, 11] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 11, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9, 11] Searching for -60 in [1] a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 0, the index for value 1 [-60, 1] Searching for 60 in [1] a = 60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 1, the index for value nil [1, 60] Searching for 60 in [] [60]

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  • Regex to use each letter only once?

    - by gtcaz
    Is it possible to construct a PCRE-style regular expression that will only match each letter in a list only once? For example, if you have the letters "lrsa" and you try matching a word list against: ^[lrsa]*m[lrsa]*$ you're going to match "lams" (valid), but also "lamas" (invalid for our purposes because you only had one "a"). If your letter set was "lrsaa", you would want to match "lamas". Is this possible with regular expressions, or should I handle it programmatically?

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  • Shortening code

    - by Misiur
    Nah, looks like it was hosting fault. Who can make this code shorter? private function replaceFunc($subject) { foreach($this->func as $t) { preg_match_all('/\{'.$t.'\([a-zA-Z,\']+\)\}/i', $subject, $res); for($j = 0; $j < sizeof($res[0]); $j++) { preg_match('/\([a-zA-Z,\']+\)/i', $res[0][$j], $match); if($match > 0) { $prep = explode(", ", substr($match[0], 1, -1)); $args = array(); for($i = 0; $i < sizeof($prep); $i++) { $args[] = substr($prep[$i], 1, -1); } } else { $args = array(); } $subject = preg_replace('/\{'.$t.preg_quote($match[0]).'\}/i', call_user_func_array($t, $args), $subject); } } return $subject; }

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  • If-elseif-else Logic Question

    - by Changeling
    I have a set of three values, call them x, y, and z. If value A happens to match only one in the set x, y, and z, then that means we have a proper match and we stop searching for a match, even if it is at y. It can match any one in that set. These values x, y, and z are non-constant so I cannot use a switch-case statement. How do I do this with an if-elseif-else statements without having to use GOTO. I am using C++ (no boost or any of that other fancy stuff). Now, I am trying to do this in code and it is racking my brain this morning (not enough coffee?)

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  • Is Minus Zero some sort of JavaScript performance trick?

    - by James Wiseman
    Looking in the jQuery core I found the folloiwng code convention: nth: function(elem, i, match){ return match[3] - 0 === i; }, And I was really curious about the snippet match[3] - 0 Hunting around for '-0' on google isn't too productive, and a search for 'minus zero' brings back a reference to a Bob Dylan song. So, can anyone tell me. Is this some sort of performance trick, or is there a reason for doing this rather than a parseInt or parseFloat? Thanks

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  • javascript string exec strange behavior

    - by Michael
    have funciton in my object which is called regularly. parse : function(html) { var regexp = /...some pattern.../ var match = regexp.exec(html); while (match != null) { ... match = regexp.exec(html); } ... var r = /...pattern.../g; var m = r.exec(html); } with unchanged html the m returns null each other call. let's say parse(html);// ok parse(html);// m is null!!! parse(html);// ok parse(html);// m is null!!! // ...and so on... is there any index or somrthing that has to be reset on html ... I'm really confused. Why match always returns proper result?

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  • Python Regular Expressions: Capture lookahead value (capturing text without consuming it)

    - by Lattyware
    I wish to use regular expressions to split words into groups of (vowels, not_vowels, more_vowels), using a marker to ensure every word begins and ends with a vowel. import re MARKER = "~" VOWELS = {"a", "e", "i", "o", "u", MARKER} word = "dog" if word[0] not in VOWELS: word = MARKER+word if word[-1] not in VOWELS: word += MARKER re.findall("([%]+)([^%]+)([%]+)".replace("%", "".join(VOWELS)), word) In this example we get: [('~', 'd', 'o')] The issue is that I wish the matches to overlap - the last set of vowels should become the first set of the next match. This appears possible with lookaheads, if we replace the regex as follows: re.findall("([%]+)([^%]+)(?=[%]+)".replace("%", "".join(VOWELS)), word) We get: [('~', 'd'), ('o', 'g')] Which means we are matching what I want. However, it now doesn't return the last set of vowels. The output I want is: [('~', 'd', 'o'), ('o', 'g', '~')] I feel this should be possible (if the regex can check for the second set of vowels, I see no reason it can't return them), but I can't find any way of doing it beyond the brute force method, looping through the results after I have them and appending the first character of the next match to the last match, and the last character of the string to the last match. Is there a better way in which I can do this? The two things that would work would be capturing the lookahead value, or not consuming the text on a match, while capturing the value - I can't find any way of doing either.

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  • ArithmeticException thrown during BigDecimal.divide

    - by polygenelubricants
    I thought java.math.BigDecimal is supposed to be The Answer™ to the need of performing infinite precision arithmetic with decimal numbers. Consider the following snippet: import java.math.BigDecimal; //... final BigDecimal one = BigDecimal.ONE; final BigDecimal three = BigDecimal.valueOf(3); final BigDecimal third = one.divide(three); assert third.multiply(three).equals(one); // this should pass, right? I expect the assert to pass, but in fact the execution doesn't even get there: one.divide(three) causes ArithmeticException to be thrown! Exception in thread "main" java.lang.ArithmeticException: Non-terminating decimal expansion; no exact representable decimal result. at java.math.BigDecimal.divide It turns out that this behavior is explicitly documented in the API: In the case of divide, the exact quotient could have an infinitely long decimal expansion; for example, 1 divided by 3. If the quotient has a non-terminating decimal expansion and the operation is specified to return an exact result, an ArithmeticException is thrown. Otherwise, the exact result of the division is returned, as done for other operations. Browsing around the API further, one finds that in fact there are various overloads of divide that performs inexact division, i.e.: final BigDecimal third = one.divide(three, 33, RoundingMode.DOWN); System.out.println(three.multiply(third)); // prints "0.999999999999999999999999999999999" Of course, the obvious question now is "What's the point???". I thought BigDecimal is the solution when we need exact arithmetic, e.g. for financial calculations. If we can't even divide exactly, then how useful can this be? Does it actually serve a general purpose, or is it only useful in a very niche application where you fortunately just don't need to divide at all? If this is not the right answer, what CAN we use for exact division in financial calculation? (I mean, I don't have a finance major, but they still use division, right???).

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  • Create fulltext index on a VIEW

    - by kylex
    Is it possible to create a full text index on a VIEW? If so, given two columns column1 and column2 on a VIEW, what is the SQL to get this done? The reason I'd like to do this is I have two very large tables, where I need to do a FULLTEXT search of a single column on each table and combine the results. The results need to be ordered as a single unit. Suggestions? EDIT: This was my attempt at creating a UNION and ordering by each statements scoring. (SELECT a_name AS name, MATCH(a_name) AGAINST('$keyword') as ascore FROM a WHERE MATCH a_name AGAINST('$keyword')) UNION (SELECT s_name AS name,MATCH(s_name) AGAINST('$keyword') as sscore FROM s WHERE MATCH s_name AGAINST('$keyword')) ORDER BY (ascore + sscore) ASC sscore was not recognized.

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  • How to apply or chain multiple matching templates in XSLT?

    - by Ignatius
    I am working on a stylesheet employing many templates with match attributes: <xsl:template match="//one" priority="0.7"> <xsl:param name="input" select="."/> <xsl:value-of select="util:uppercase($input)"/> <xsl:next-match /> </xsl:template> <xsl:template match="/stuff/one"> <xsl:param name="input" select="."/> <xsl:value-of select="util:add-period($input)"/> </xsl:template> <xsl:function name="util:uppercase"> <xsl:param name="input"/> <xsl:value-of select="upper-case($input)"/> </xsl:function> <xsl:function name="util:add-period"> <xsl:param name="input"/> <xsl:value-of select="concat($input,'.')"/> </xsl:function> What I would like to do is be able to 'chain' the two functions above, so that an input of 'string' would be rendered in the output as 'STRING.' (with the period.) I would like to do this in such a way that doesn't require knowledge of other templates in any other template. So, for instance, I would like to be able to add a "util:add-colon" method without having to open up the hood and monkey with the existing templates. I was playing around with the <xsl:next-match/> instruction to accomplish this. Adding it to the first template above does of course invoke both util:uppercase and util:add-period, but the output is an aggregation of each template output (i.e. 'STRINGstring.') It seems like there should be an elegant way to chain any number of templates together using something like <xsl:next-match/>, but have the output of each template feed the input of the next one in the chain. Am I overlooking something obvious?

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