Search Results

Search found 118 results on 5 pages for 'matcher'.

Page 3/5 | < Previous Page | 1 2 3 4 5  | Next Page >

  • Why does this data YYYY-MM-DD regex fail in Java?

    - by ProfessionalAmateur
    Hello StackOverFlow, My first question and Im excited... I've lurked since go-live and love the site, however I apologize for any newbie errors, formatting, etc... I'm attempting to validate the format of a string field that contains a date in Java. We will receive the date in a string, I will validate its format before parsing it into a real Date object. The format being passed in is in the YYYY-MM-DD format. However I'm stuck on one of my tests, if I pass in "1999-12-33" the test will fail (as it should with a day number 33) with this incomplete pattern: ((19|20)\\d{2})-([1-9]|0[1-9]|1[0-2])-([12][0-9]|3[01]) However as soon as I add the characters in bold below it passes the test (but should not) ((19|20)\\d{2})-([1-9]|0[1-9]|1[0-2])-(0[1-9]|[1-9]|[12][0-9]|3[01]) *additional note, I know I can change the 0[1-9]|[1-9] into 0?[1-9] but I wanted to break everything down to its most simple format to try and find why this isn't working. Here is the scrap test I've put together to run through all the different date scenarios: import java.util.regex.Matcher; import java.util.regex.Pattern; public class scrapTest { public scrapTest() { } public static void main(String[] args) { scrapTest a = new scrapTest(); boolean flag = a.verfiyDateFormat("1999-12-33"); } private boolean verfiyDateFormat(String dateStr){ Pattern datePattern = Pattern.compile("((19|20)\\d{2})-([1-9]|0[1-9]|1[0-2])-(0[1-9]|[1-9]|[12][0-9]|3[01])"); Matcher dateMatcher = datePattern.matcher(dateStr); if(!dateMatcher.find()){ System.out.println("Invalid date format!!! -> " + dateStr); return false; } System.out.println("Valid date format."); return true; } } Ive been programming for ~10 years but extremely new to Java, so please feel free to explain anything as elementary as you see fit.

    Read the article

  • Java Scanner newline parsing with regex (Bug?)

    - by SEK
    I'm developing a syntax analyzer by hand in Java, and I'd like to use regex's to parse the various token types. The problem is that I'd also like to be able to accurately report the current line number, if the input doesn't conform to the syntax. Long story short, I've run into a problem when I try to actually match a newline with the Scanner class. To be specific, when I try to match a newline with a pattern using the Scanner class, it fails. Almost always. But when I perform the same matching using a Matcher and the same source string, it retrieves the newline exactly as you'd expect it too. Is there a reason for this, that I can't seem to discover, or is this a bug, as I suspect? FYI: I was unable to find a bug in the Sun database that describes this issue, so if it is a bug, it hasn't been reported. Example Code: Pattern newLinePattern = Pattern.compile("(\\r\\n?|\\n)", Pattern.MULTILINE); String sourceString = "\r\n\n\r\r\n\n"; Scanner scan = new Scanner(sourceString); scan.useDelimiter(""); int count = 0; while (scan.hasNext(newLinePattern)) { scan.next(newLinePattern); count++; } System.out.println("found "+count+" newlines"); // finds 7 newlines Matcher match = newLinePattern.matcher(sourceString); count = 0; while (match.find()) { count++; } System.out.println("found "+count+" newlines"); // finds 5 newlines

    Read the article

  • how to read string part in java

    - by Gandalf StormCrow
    Hello everyone, I have this string : <meis xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" uri="localhost/naro-nei" onded="flpSW531213" identi="lemenia" id="75" lastStop="bendi" xsi:noNamespaceSchemaLocation="http://localhost/xsd/postat.xsd xsd/postat.xsd"> How can I get lastStop property value in JAVA? This regex worked when tested on http://www.myregexp.com/ But when I try it in java I don't see the matched text, here is how I tried : import java.util.regex.Pattern; import java.util.regex.Matcher; public class SimpleRegexTest { public static void main(String[] args) { String sampleText = <meis xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" uri=\"localhost/naro-nei\" onded=\"flpSW531213\" identi=\"lemenia\" id=\"75\" lastStop=\"bendi\" xsi:noNamespaceSchemaLocation=\"http://localhost/xsd/postat.xsd xsd/postat.xsd\">"; String sampleRegex = "(?<=lastStop=[\"']?)[^\"']*"; Pattern p = Pattern.compile(sampleRegex); Matcher m = p.matcher(sampleText); if (m.find()) { String matchedText = m.group(); System.out.println("matched [" + matchedText + "]"); } else { System.out.println("didn’t match"); } } }

    Read the article

  • Trying to parse links in an HTML directory listing using Java regex

    - by DiskCrasher
    Ok I know everyone is going to tell me not to use RegEx for parsing HTML, but I'm programming on Android and don't have ready access to an HTML parser (that I'm aware of). Besides, this is server generated HTML which should be more consistent than user-generated HTML. The regex looks like this: Pattern patternMP3 = Pattern.compile( "<A HREF=\"[^\"]+.+\\.mp3</A>", Pattern.CASE_INSENSITIVE | Pattern.UNICODE_CASE); Matcher matcherMP3 = patternMP3.matcher(HTML); while (matcherMP3.find()) { ... } The input HTML is all on one line, which is causing the problem. When the HTML is on separate lines this pattern works. Any suggestions?

    Read the article

  • Attempting to extract a pattern within a string

    - by Brian
    I'm attempting to extract a given pattern within a text file, however, the results are not 100% what I want. Here's my code: import java.util.regex.Matcher; import java.util.regex.Pattern; public class ParseText1 { public static void main(String[] args) { String content = "<p>Yada yada yada <code> foo ddd</code>yada yada ...\n" + "more here <2004-08-24> bar<Bob Joe> etc etc\n" + "more here again <2004-09-24> bar<Bob Joe> <Fred Kej> etc etc\n" + "more here again <2004-08-24> bar<Bob Joe><Fred Kej> etc etc\n" + "and still more <2004-08-21><2004-08-21> baz <John Doe> and now <code>the end</code> </p>\n"; Pattern p = Pattern .compile("<[1234567890]{4}-[1234567890]{2}-[1234567890]{2}>.*?<[^%0-9/]*>", Pattern.MULTILINE); Matcher m = p.matcher(content); // print all the matches that we find while (m.find()) { System.out.println(m.group()); } } } The output I'm getting is: <2004-08-24> bar<Bob Joe> <2004-09-24> bar<Bob Joe> <Fred Kej> <2004-08-24> bar<Bob Joe><Fred Kej> <2004-08-21><2004-08-21> baz <John Doe> and now <code> The output I want is: <2004-08-24> bar<Bob Joe> <2004-08-24> bar<Bob Joe> <2004-08-24> bar<Bob Joe> <2004-08-21> baz <John Doe> In short, the sequence of "date", "text (or blank)", and "name" must be extracted. Everything else should be avoided. For example the tag "Fred Kej" did not have any "date" tag before it, therefore, it should be flagged as invalid. Also, as a side question, is there a way to store or track the text snippets that were skipped/rejected as were the valid texts. Thanks, Brian

    Read the article

  • java regex illegal escape character error not occurring from command line arguments

    - by Shades88
    This simple regex program import java.util.regex.*; class Regex { public static void main(String [] args) { System.out.println(args[0]); // #1 Pattern p = Pattern.compile(args[0]); // #2 Matcher m = p.matcher(args[1]); boolean b = false; while(b = m.find()) { System.out.println(m.start()+" "+m.group()); } } } invoked by java regex "\d" "sfdd1" compiles and runs fine. But if #1 is replaced by Pattern p = Pattern.compile("\d");, it gives compiler error saying illegal escape character. In #1 I also tried printing the pattern specified in the command line arguments. It prints \d, which means it is just getting replaced by \d in #2. So then why won't it throw any exception? At the end it's string argument that Pattern.compile() is taking, doesn't it detect illegal escape character then? Can someone please explain why is this behaviour?

    Read the article

  • split a string based on pattern in java - capital letters and numbers

    - by rookie
    Hi all I have the following string "3/4Ton". I want to split it as -- word[1] = 3/4 and word[2] = Ton. Right now my piece of code looks like this:- Pattern p = Pattern.compile("[A-Z]{1}[a-z]+"); Matcher m = p.matcher(line); while(m.find()){ System.out.println("The word --> "+m.group()); } It carries out the needed task of splitting the string based on capital letters like:- String = MachineryInput word[1] = Machinery , word[2] = Input The only problem is it does not preserve, numbers or abbreviations or sequences of capital letters which are not meant to be separate words. Could some one help me out with my regular expression coding problem. Thanks in advance...

    Read the article

  • understanding this regex

    - by DarthVader
    I m trying to understand what the following does. ^([^=]+)(?:(?:\\=)(.+))?$ Any ideas? This is being used here. Obviously it s command line parser but i m trying to understand the syntax so i can actually run the program. This is from commandline-jmxclient , they have no documents on setting JMX properties but in their source code, there is such an option, so i just want to understand how i can invoke that method. Matcher m = Client.CMD_LINE_ARGS_PATTERN.matcher(command); if ((m == null) || (!m.matches())) { throw new ParseException("Failed parse of " + command, 0); } this.cmd = m.group(1); if ((m.group(2) != null) && (m.group(2).length() > 0)) this.args = m.group(2).split(","); else this.args = null;

    Read the article

  • Need help with using regular expression in Java

    - by richard
    Hi, I am trying to match pattern like '@(a-zA-Z0-9)+ " but not like 'abc@test'. So this is what I tried: Pattern MY_PATTERN = Pattern.compile("\\s@(\\w)+\\s?"); String data = "[email protected] #gogasig @jytaz @tibuage"; Matcher m = MY_PATTERN.matcher(data); StringBuffer sb = new StringBuffer(); boolean result = m.find(); while(result) { System.out.println (" group " + m.group()); result = m.find(); } But I can only see '@jytaz', but not @tibuage. How can I fix my problem? Thank you.

    Read the article

  • Java regex, need help with escape characters

    - by Blankman
    My HTML looks like: <td class="price" valign="top"><font color= "blue">&nbsp;&nbsp;$&nbsp; 5.93&nbsp;</font></td> I tried: String result = ""; Pattern p = Pattern.compile("\"blue\">&nbsp;&nbsp;$&nbsp;(.*)&nbsp;</font></td>"); Matcher m = p.matcher(text); if(m.find()) result = m.group(1).trim(); Doesn't seem to be matching. Am I missing an escape character?

    Read the article

  • How can I fill in the blanks to display just my image?

    - by Ignacio
    <h:inputText id="email" value="#{user.user.email}" title="Email" onchange="this.form.submit()" required="true" requiredMessage="_____"> <f:validator validatorId="checkvalidemail"/> </h:inputText> <h:message for="email" styleClass="error"/> validation: String enteredEmail = (String)object; Pattern p = Pattern.compile(".+@.+\\.[a-z]+"); Matcher m = p.matcher(enteredEmail); boolean matchFound = m.matches(); if (!matchFound) { FacesMessage message = new FacesMessage(); message.setSummary("____"); throw new ValidatorException(message); and css .error { background-image: url('includes/style/delete2.png'); text-align: left; font-size: 36px; } Thank you very much Best Regards Ignacio

    Read the article

  • 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()); } }

    Read the article

  • Loop and ListView

    - by monomi
    I find a match with Regexp. How to correctly organize a loop, if more than one match and write it in the ListView? listAlbums = (ListView)findViewById(R.id.listAlbums); ... Pattern patternNameToId = Pattern.compile(kRegexp); String nameTo = "" Matcher matcherName = patternNameToId.matcher(response); if (matcherName.find()) { nameTo = matcherTopicTitle.group(2); albumsList = new ArrayList<String>(); albumsList.add(nameTo); ArrayAdapter<String> adapter = new ArrayAdapter<String>(context, android.R.layout.simple_list_item_1, albumsList); listAlbums.setAdapter(adapter); } else Toast.makeText(context, "?? ???????", Toast.LENGTH_LONG).show();

    Read the article

  • Java regex skipping matches

    - by Mihail Burduja
    I have some text; I want to extract pairs of words that are not separated by punctuation. Thi is the code: //n-grams Pattern p = Pattern.compile("[a-z]+"); if (n == 2) { p = Pattern.compile("[a-z]+ [a-z]+"); } if (n == 3) { p = Pattern.compile("[a-z]+ [a-z]+ [a-z]+"); } Matcher m = p.matcher(text.toLowerCase()); ArrayList<String> result = new ArrayList<String>(); while (m.find()) { String temporary = m.group(); System.out.println(temporary); result.add(temporary); } The problem is that it skips some matches. For example "My name is James", for n = 3, must match "my name is" and "name is james", but instead it matches just the first. Is there a way to solve this?

    Read the article

  • My Java regex isn't capturing the group

    - by Geo
    I'm trying to match the username with a regex. Please don't suggest a split. USERNAME=geo Here's my code: String input = "USERNAME=geo"; Pattern pat = Pattern.compile("USERNAME=(\\w+)"); Matcher mat = pat.matcher(input); if(mat.find()) { System.out.println(mat.group()); } why doesn't it find geo in the group? I noticed that if I use the .group(1), it finds the username. However the group method contains USERNAME=geo. Why?

    Read the article

  • 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...

    Read the article

  • Updating table from async task android

    - by CantChooseUsernames
    I'm following this tutorial: http://huuah.com/android-progress-bar-and-thread-updating/ to learn how to make progress bars. I'm trying to show the progress bar on top of my activity and have it update the activity's table view in the background. So I created an async task for the dialog that takes a callback: package com.lib.bookworm; import android.app.ProgressDialog; import android.content.Context; import android.os.AsyncTask; public class UIThreadProgress extends AsyncTask<Void, Void, Void> { private UIThreadCallback callback = null; private ProgressDialog dialog = null; private int maxValue = 100, incAmount = 1; private Context context = null; public UIThreadProgress(Context context, UIThreadCallback callback) { this.context = context; this.callback = callback; } @Override protected Void doInBackground(Void... args) { while(this.callback.condition()) { this.callback.run(); this.publishProgress(); } return null; } @Override protected void onProgressUpdate(Void... values) { super.onProgressUpdate(values); dialog.incrementProgressBy(incAmount); }; @Override protected void onPreExecute() { super.onPreExecute(); dialog = new ProgressDialog(context); dialog.setCancelable(true); dialog.setMessage("Loading..."); dialog.setProgress(0); dialog.setProgressStyle(ProgressDialog.STYLE_HORIZONTAL); dialog.setMax(maxValue); dialog.show(); } @Override protected void onPostExecute(Void result) { super.onPostExecute(result); if (this.dialog.isShowing()) { this.dialog.dismiss(); } this.callback.onThreadFinish(); } } And in my activity, I do: final String page = htmlPage.substring(start, end).trim(); //Create new instance of the AsyncTask.. new UIThreadProgress(this, new UIThreadCallback() { @Override public void run() { row_id = makeTableRow(row_id, layout, params, matcher); //ADD a row to the table layout. } @Override public void onThreadFinish() { System.out.println("FINISHED!!"); } @Override public boolean condition() { return matcher.find(); } }).execute(); So the above creates an async task to run to update a table layout activity while showing the progress bar that displays how much work has been done.. However, I get an error saying that only the thread that started the activity can update its views. I tried doing: MainActivity.this.runOnUiThread(new Runnable() { @Override public void run() { row_id = makeTableRow(row_id, layout, params, matcher); //ADD a row to the table layout. } } But this gives me synchronization errors.. Any ideas how I can display progress and at the same time update my table in the background? Currently my UI looks like:

    Read the article

  • match word '90%' using regular expression

    - by amadhu
    Hi All, I want word '90%' to be matched with my String "I have 90% shares of this company". how can I write regular expression for same? I tried something like this: Pattern p = Pattern.compile("\\b90\\%\\b", Pattern.CASE_INSENSITIVE | Pattern.MULTILINE); Matcher m = p.matcher("I have 90% shares of this company"); while (m.find()){ System.out.println(m.group()); } but no luck. Can any one thow some lights on this? Many thanks, Archi

    Read the article

  • Infinite loop in regex in java

    - by carpediem
    Hello, My purpose is to match this kind of different urls: url.com my.url.com my.extended.url.com a.super.extended.url.com and so on... So, I decided to build the regex to have a letter or a number at start and end of the url, and to have a infinite number of "subdomains" with alphanumeric characters and a dot. For example, in "my.extended.url.com", "m" from "my" is the first class of the regex, "m" from "com" is the last class of the regex, and "y.", "extended." and "url." are the second class of the regex. Using the pattern and subject in the code below, I want the find method to return me a false because this url must not match, but it uses 100% of CPU and seems to stay in an infinite loop. String subject = "www.association-belgo-palestinienne-be"; Pattern pattern = Pattern.compile("^[A-Za-z0-9]\\.?([A-Za-z0-9_-]+\\.?)*[A-Za-z0-9]\\.[A-Za-z]{2,6}"); Matcher m = pattern.matcher(subject); System.out.println(" Start"); boolean hasFind = m.find(); System.out.println(" Finish : " + hasFind); Which only prints: Start I can't reproduce the problem using regex testers. Is it normal ? Is the problem coming from my regex ? Could it be due to my Java version (1.6.0_22-b04 / JVM 64 bit 17.1-b03) ? Thanks in advance for helping.

    Read the article

  • Regular expression either/or not matching everything

    - by dwatransit
    I'm trying to parse an HTTP GET request to determine if the url contains any of a number of file types. If it does, I want to capture the entire request. There is something I don't understand about ORing. The following regular expression only captures part of it, and only if .flv is the first int the list of ORd values. (I've obscured the urls with spaces because Stackoverflow limits hyperlinks) regex: GET.?(.flv)|(.mp4)|(.avi).? test text: GET http: // foo.server.com/download/0/37/3000016511/.flv?mt=video/xy match output: GET http: // foo.server.com/download/0/37/3000016511/.flv I don't understand why the .*? at the end of the regex isnt callowing it to capture the entire text. If I get rid of the ORing of file types, then it works. Here is the test code in case my explanation doesn't make sense: public static void main(String[] args) { // TODO Auto-generated method stub String sourcestring = "GET http: // foo.server.com/download/0/37/3000016511/.flv?mt=video/xy"; Pattern re = Pattern.compile("GET .?\.flv."); // this works //output: // [0][0] = GET http :// foo.server.com/download/0/37/3000016511/.flv?mt=video/xy // the match from the following ends with the ".flv", not the entire url. // also it only works if .flv is the first of the 3 ORd options //Pattern re = Pattern.compile("GET .?(\.flv)|(\.mp4)|(\.avi).?"); // output: //[0][0] = GET http: // foo.server.com/download/0/37/3000016511/.flv // [0][1] = .flv // [0][2] = null // [0][3] = null Matcher m = re.matcher(sourcestring); int mIdx = 0; while (m.find()){ for( int groupIdx = 0; groupIdx < m.groupCount()+1; groupIdx++ ){ System.out.println( "[" + mIdx + "][" + groupIdx + "] = " + m.group(groupIdx)); } mIdx++; } } }

    Read the article

  • Using java.util.regex in Android apps - are there issues with this?

    - by johnrock
    In an Android app I have a utility class that I use to parse strings for 2 regEx's. I compile the 2 patterns in a static initializer so they only get compiled once, then activities can use the parsing methods statically. This works fine except that the first time the class is accessed and loaded, and the static initializer compiles the pattern, the UI hangs for close to a MINUTE while it compiles the pattern! After the first time, it flies on all subsequent calls to parseString(). My regEx that I am using is rather large - 847 characters, but in a normal java webapp this is lightning fast. I am testing this so far only in the emulator with a 1.5 AVD. Could this just be an emulator issue or is there some other reason that this pattern is taking so long to compile? private static final String exp1 = "(insertratherlong---847character--regexhere)"; private static Pattern regex1 = null; private static final String newLineAndTagsExp = "[<>\\s]"; private static Pattern regexNewLineAndTags = null; static { regex1 = Pattern.compile(exp1, Pattern.CASE_INSENSITIVE); regexNewLineAndTags = Pattern.compile(newLineAndTagsExp); } public static String parseString(CharSequence inputStr) { String replacementStr = "replaceMentText"; String resultString = "none"; try { Matcher regexMatcher = regex1.matcher(inputStr); try { resultString = regexMatcher.replaceAll(replacementStr); } catch (IllegalArgumentException ex) { } catch (IndexOutOfBoundsException ex) { } } catch (PatternSyntaxException ex) { } return resultString; }

    Read the article

  • Getting dialogue snippets from text using regular expressions

    - by sheldon
    I'm trying to extract snippets of dialogue from a book text. For example, if I have the string "What's the matter with the flag?" inquired Captain MacWhirr. "Seems all right to me." Then I want to extract "What's the matter with the flag?" and "Seem's all right to me.". I found a regular expression to use here, which is "[^"\\]*(\\.[^"\\]*)*". This works great in Eclipse when I'm doing a Ctrl+F find regex on my book .txt file, but when I run the following code: String regex = "\"[^\"\\\\]*(\\\\.[^\"\\\\]*)*\""; String bookText = "\"What's the matter with the flag?\" inquired Captain MacWhirr. \"Seems all right to me.\""; Pattern p = Pattern.compile(regex); Matcher m = p.matcher(bookText); if(m.find()) System.out.println(m.group(1)); The only thing that prints is null. So am I not converting the regex into a Java string properly? Do I need to take into account the fact that Java Strings have a \" for the double quotes?

    Read the article

  • How to use Jquery UI in my Custom Function? (Autocomplete)

    - by bakazero
    I want to create a function to simplify configuration of jQuery UI AutoComplete. Here is my function code: (function($) { $.fn.myAutocomplete = function() { var cache = {}; var dataUrl = args.dataUrl; var dataSend = args.dataItem; $.autocomplete({ source: function(request, response) { if (cache.term == request.term && cache.content) { response(cache.content); } if (new RegExp(cache.term).test(request.term) && cache.content && cache.content.length < 13) { var matcher = new RegExp($.ui.autocomplete.escapeRegex(request.term), "i"); response($.grep(cache.content, function(value) { return matcher.test(value.value) })); } $.ajax({ url: dataUrl, dataType: "json", type: "POST", data: dataSend, success: function(data) { cache.term = request.term; cache.content = data; response(data); } }); }, minLength: 2, }); } }) (jQuery); but when I'm using this function like: $("input#tag").myAutocomplete({ dataUrl: "/auto_complete/tag", dataSend: { term: request.term, category: $("input#category").val() } }); It's give me an error: Uncaught ReferenceError: request is not defined

    Read the article

  • MacVim, Command-T: SEGV

    - by Ramon Tayag
    Details: OSX 10.7.4 I installed the latest MacVim via Homebrew: $ command-t brew install macvim ==> Downloading https://github.com/b4winckler/macvim/tarball/snapshot-64 Already downloaded: /Library/Caches/Homebrew/macvim-7.3-64.tgz ==> ./configure --with-features=huge --with-tlib=ncurses --enable-multibyte --with-macarchs=x86_64 --enable-perlinterp --enable-pythoninterp --enable-rubyinterp --enable-t ==> make getenvy ==> make ==> Caveats MacVim.app installed to: /usr/local/Cellar/macvim/7.3-64 To link the application to a normal Mac OS X location: brew linkapps or: ln -s /usr/local/Cellar/macvim/7.3-64/MacVim.app /Applications ==> Summary /usr/local/Cellar/macvim/7.3-64: 1733 files, 27M, built in 53 seconds $ command-t brew linkapps Linking /usr/local/Cellar/macvim/7.3-64/MacVim.app Finished linking. Find the links under ~/Applications. $ command-t ruby -v ruby 1.8.7 (2011-12-28 patchlevel 357) [universal-darwin11.0] $ command-t rvm list rvm rubies ree-1.8.7-2012.02 [ i686 ] ruby-1.8.7-p358 [ i686 ] ruby-1.9.2-p290 [ x86_64 ] ruby-1.9.2-p320 [ x86_64 ] ruby-1.9.3-p194 [ x86_64 ] # Default ruby not set. Try 'rvm alias create default <ruby>'. # => - current # =* - current && default # * - default $ command-t cd ~/.vim/bundle/vim-command-t/ruby/command-t ruby extconf.rb $ command-t ruby extconf.rb checking for ruby.h... yes creating Makefile $ command-t make cc -arch i386 -arch x86_64 -pipe -bundle -undefined dynamic_lookup -o ext.bundle ext.o match.o matcher.o -L. -L/System/Library/Frameworks/Ruby.framework/Versions/1.8/usr/lib -L. -arch i386 -arch x86_64 -lruby -lpthread -ldl -lobjc ld: warning: ignoring file ext.o, file was built for unsupported file format which is not the architecture being linked (i386) ld: warning: ignoring file match.o, file was built for unsupported file format which is not the architecture being linked (i386) ld: warning: ignoring file matcher.o, file was built for unsupported file format which is not the architecture being linked (i386) $ command-t mvim MacVim then opens here. But when I open Command-T, MacVim crashes and I see this in the command line: $ command-t dyld: lazy symbol binding failed: Symbol not found: _rb_intern2 Referenced from: /Users/ramon/.vim/bundle/vim-command-t/ruby/command-t/ext.bundle Expected in: flat namespace dyld: Symbol not found: _rb_intern2 Referenced from: /Users/ramon/.vim/bundle/vim-command-t/ruby/command-t/ext.bundle Expected in: flat namespace Vim: Caught deadly signal TRAP Vim: Finished. The problem I have is very similar to this, except that I switched to the system Ruby and still got the error.

    Read the article

  • How match 'other' applications to a tag in awesome-wm?

    - by Mnementh
    I use version 3.3.4 of awesome and it is fine. But I miss one thing I could do with an older version of awesome (without configuration via Lua): I could add a matcher with the regexp .* to add all windows without another tag to a specific tag: rule { name = ".*" tags = "9" } With that all applications I didn't made another rule for were added to tag 9. How can I do something similar with configuration in rc.lua?

    Read the article

< Previous Page | 1 2 3 4 5  | Next Page >