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  • How do you construct an array suitable for numpy sorting?

    - by Alex
    I need to sort two arrays simultaneously, or rather I need to sort one of the arrays and bring the corresponding element of its associated array with it as I sort. That is if the array is [(5, 33), (4, 44), (3, 55)] and I sort by the first axis (labeled below dtype='alpha') then I want: [(3.0, 55.0) (4.0, 44.0) (5.0, 33.0)]. These are really big data sets and I need to sort first ( for nlog(n) speed ) before I do some other operations. I don't know how to merge my two separate arrays though in the proper manner to get the sort algorithm working. I think my problem is rather simple. I tried three different methods: import numpy x=numpy.asarray([5,4,3]) y=numpy.asarray([33,44,55]) dtype=[('alpha',float), ('beta',float)] values=numpy.array([(x),(y)]) values=numpy.rollaxis(values,1) #values = numpy.array(values, dtype=dtype) #a=numpy.array(values,dtype=dtype) #q=numpy.sort(a,order='alpha') print "Try 1:\n", values values=numpy.empty((len(x),2)) for n in range (len(x)): values[n][0]=y[n] values[n][1]=x[n] print "Try 2:\n", values #values = numpy.array(values, dtype=dtype) #a=numpy.array(values,dtype=dtype) #q=numpy.sort(a,order='alpha') ### values = [(x[0], y[0]), (x[1],y[1]) , (x[2],y[2])] print "Try 3:\n", values values = numpy.array(values, dtype=dtype) a=numpy.array(values,dtype=dtype) q=numpy.sort(a,order='alpha') print "Result:\n",q I commented out the first and second trys because they create errors, I knew the third one would work because that was mirroring what I saw when I was RTFM. Given the arrays x and y (which are very large, just examples shown) how do I construct the array (called values) that can be called by numpy.sort properly? *** Zip works great, thanks. Bonus question: How can I later unzip the sorted data into two arrays again?

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  • Help with a compiler warning: Initialization from distinct Objective-C type when types match

    - by Alex Gosselin
    Here is the function where I get the compiler warning, I can't seem to figure out what is causing it. Any help is appreciated. -(void)displaySelector{ //warning on the following line: InstanceSelectorViewController *controller = [[InstanceSelectorViewController alloc] initWithCreator:self]; [self.navController pushViewController:controller animated:YES]; [controller release]; } Interface and implementation for the initWithCreator: method -(InstanceSelectorViewController*)initWithCreator:(InstanceCreator*)creator; -(InstanceSelectorViewController*)initWithCreator:(InstanceCreator*)crt{ if (self = [self initWithNibName:@"InstanceSelectorViewController" bundle:nil]) { creator = crt; } return self; }

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  • How do you pass user credentials from WebClient to a WCF REST service?

    - by Alex
    I am trying to expose a WCT REST service and only users with valid username and password would be able to access it. The username and password are stored in a SQL database. Here is the service contract: public interface IDataService { [OperationContract] [WebGet(ResponseFormat = WebMessageFormat.Json)] byte[] GetData(double startTime, double endTime); } Here is the WCF configuration: <bindings> <webHttpBinding> <binding name="SecureBinding"> <security mode="Transport"> <transport clientCredentialType="Basic"/> </security> </binding> </webHttpBinding> </bindings> <behaviors> <serviceBehaviors> <behavior name="DataServiceBehavior"> <serviceMetadata httpGetEnabled="true"/> <serviceCredentials> <userNameAuthentication userNamePasswordValidationMode="Custom" customUserNamePasswordValidatorType= "CustomValidator, WCFHost" /> </serviceCredentials> </behavior> </serviceBehaviors> </behaviors> <services> <service behaviorConfiguration="DataServiceBehavior" name="DataService"> <endpoint address="" binding="webHttpBinding" bindingConfiguration="SecureBinding" contract="IDataService" /> </service> </services> I am accessing the service via the WebClient class within a Silverlight application. However, I have not been able to figure out how to pass the user credentials to the service. I tried various values for client.Credentials but none of them seems to trigger the code in my custom validator. I am getting the following error: The underlying connection was closed: An unexpected error occurred on a send. Here is some sample code I have tried: WebClient client = new WebClient(); client.Credentials = new NetworkCredential("name", "password", "domain"); client.OpenReadCompleted += new OpenReadCompletedEventHandler(GetData); client.OpenReadAsync(new Uri(uriString)); If I set the security mode to None, the whole thing works. I also tried other clientCredentialType values and none of them worked. I also self-hosted the WCF service to eliminate the issues related to IIS trying to authenticate a user before the service gets a chance. Any comment on what the underlying issues may be would be much appreciated. Thanks. Update: Thanks to Mehmet's excellent suggestions. Here is the tracing configuration I had: <system.diagnostics> <sources> <source name="System.ServiceModel" switchValue="Information, ActivityTracing" propagateActivity="true"> <listeners> <add name="xml" /> </listeners> </source> <source name="System.IdentityModel" switchValue="Information, ActivityTracing" propagateActivity="true"> <listeners> <add name="xml" /> </listeners> </source> </sources> <sharedListeners> <add name="xml" type="System.Diagnostics.XmlWriterTraceListener" initializeData="c:\Traces.svclog" /> </sharedListeners> </system.diagnostics> But I did not see any message coming from my Silverlight client. As for https vs http, I used https as follows: string baseAddress = "https://localhost:6600/"; _webServiceHost = new WebServiceHost(typeof(DataServices), new Uri(baseAddress)); _webServiceHost.Open(); However, I did not configure any SSL certificate. Is this the problem?

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  • [Ruby] [gem] A GEM error shown during run the commend: gem update --system

    - by Alex
    I’m a freshman on Ruby and now trying to install ruby on my machine according to the Tutorial on http://wiki.openqa.org/display/WTR/Tutorial However, after I installed the ruby186-26, and run the command “gem update --system”, the following error occurred: C:\Documents and Settings\e482090\Desktopgem update --system c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:in initialize': Inval id argument - <Not Set>/.gemrc (Errno::EINVAL) from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:inopen' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:in initi alize' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:36:innew' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:36:in do_con figuration' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:25:inrun' from c:/ruby/bin/gem:23 C:\Documents and Settings\e482090\Desktopgem install watir c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:in initialize': Inval id argument - <Not Set>/.gemrc (Errno::EINVAL) from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:inopen' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:in initi alize' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:36:innew' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:36:in do_con figuration' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:25:inrun' from c:/ruby/bin/gem:23 Meanwhile, we have tried this on other machines and the result turned out ok. Thus, my question is why the error happened on my pc? Have you met this kind of error before?

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  • Why is it so difficult to get a working IDE for Scala?

    - by Alex R
    I recently gave up trying to use Scala in Eclipse (basic stuff like completion doesn't work). So now I'm trying IntelliJ. I'm not getting very far. This was the original error. See below for update: Scala signature Predef has wrong version Expected 5.0 found: 4.1 in .... scala-library.jar I tried both versions 2.7.6 and 2.8 RC1 of scala-*.jar, the result was the same. JDK is 1.6.u20. UPDATE Today I uninstalled IntelliJ 9.0.1, and installed 9.0.2 Early Availability, with the 4/14 stable version of the Scala plug-in. Then I setup a project from scratch through the wizards: new project from scratch JDK is 1.6.u20 accept the default (project) instead of global / module accept the download of Scala 2.8.0beta1 into project's lib folder Created a new class: object hello { def main(args: Array[String]) { println("hello: " + args); } } For my efforts, I now have a brand-new error :) Here it is: Scalac internal error: class java.lang.ClassNotFoundException [java.net.URLClassLoader$1.run(URLClassLoader.java:202), java.security.AccessController.doPrivileged(Native Method), java.net.URLClassLoader.findClass(URLClassLoader.java:190), java.lang.ClassLoader.loadClass(ClassLoader.java:307), sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301), java.lang.ClassLoader.loadClass(ClassLoader.java:248), java.lang.Class.forName0(Native Method), java.lang.Class.forName(Class.java:169), org.jetbrains.plugins.scala.compiler.rt.ScalacRunner.main(ScalacRunner.java:72)] Thanks

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  • GnuPG PHP gnupg Folder & Files Permission

    - by Michael Robinson
    Situation: we plan on using PHP's GnuPG extension to encrypt/decrypt files. Currently we've setup some test cases, using keys generated with GPG. The generated files reside in: /Users/username/.gnupg/ I am able to get keyinfo for the key I want to use to encrypt/decrypt, but when I attempt to use addencryptkey, I get: (E_WARNING: 2): gnupg::addencryptkey() [gnupg.addencryptkey]: get_key failed I think this is due to the permissions on the ~/.gnupg folder & enclosed files. The files are owned by me - username, but apache runs as www. A few days ago I did have this working, but it seems each time I use GPG Keychain Access to import / export a key, the folder's permissions are changed. Question: What are the exact permissions required to allow PHP's GnuPG to add encrypt & decrypt keys?

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  • "Launch Failed. Binary Not Found." Snow Leopard and Eclipse C/C++ IDE issue.

    - by Alex
    Not a question, I've just scoured the internet in search of a solution for this problem and thought I'd share it with the good folks of SO. I'll put it in plain terms so that it's accessible to newbs. :) (Apologies if this is the wrong place -- just trying to be helpful.) This issue occurs with almost any user OS X Snow Leopard who tries to use the Eclipse C/C++ IDE, but is particularly annoying for the people (like me) who were using the Eclipse C/C++ IDE in Leopard, and were unable to work with Eclipse anymore when they upgraded. The issue occurs When users go to build/compile/link their software. They get the following error: Launch Failed. Binary Not Found. Further, the "binaries" branch in the project window on the left is simply nonexistent. THE PROBLEM: is that GCC 4.2 (the GNU Compiler Collection) that comes with Snow Leopard compiles binaries in 64-bit by default. Unfortunately, the linker that Eclipse uses does not understand 64-bit binaries; it reads 32-bit binaries. There may be other issues here, but in short, they culminate in no binary being generated, at least not one that Eclipse can read, which translates into Eclipse not finding the binaries. Hence the error. One solution is to add an -arch i686 flag when making the file, but manually making the file every time is annoying. Luckily for us, Snow Leopard also comes with GCC 4.0, which compiles in 32 bits by default. So one solution is merely to link this as the default compiler. This is the way I did it. THE SOLUTION: The GCCs are in /usr/bin, which is normally a hidden folder, so you can't see it in the Finder unless you explicitly tell the system that you want to see hidden folders. Anyway, what you want to do is go to the /usr/bin folder and delete the path that links the GCC command with GCC 4.2 and add a path that links the GCC command with GCC 4.0. In other words, when you or Eclipse try to access GCC, we want the command to go to the compiler that builds in 32 bits by default, so that the linker can read the files; we do not want it to go to the compiler that compiles in 64 bits. The best way to do this is to go to Applications/Utilities, and select the app called Terminal. A text prompt should come up. It should say something like "(Computer Name):~ (Username)$ " (with a space for you user input at the end). The way to accomplish the tasks above is to enter the following commands, entering each one in sequence VERBATIM, and pressing enter after each individual line. cd /usr/bin rm cc gcc c++ g++ ln -s gcc-4.0 cc ln -s gcc-4.0 gcc ln -s c++-4.0 c++ ln -s g++-4.0 g++ Like me, you will probably get an error that tells you you don't have permission to access these files. If so, try the following commands instead: cd /usr/bin sudo rm cc gcc c++ g++ sudo ln -s gcc-4.0 cc sudo ln -s gcc-4.0 gcc sudo ln -s c++-4.0 c++ sudo ln -s g++-4.0 g++ Sudo may prompt you for a password. If you've never used sudo before, try just pressing enter. If that doesn't work, try the password for your main admin account. OTHER POSSIBLE SOLUTIONS You may be able to enter build variables into Eclipse. I tried this, but I don't know enough about it. If you want to feel it out, the flag you will probably need is -arch i686. In earnest, GCC-4.0 worked for me all this time, and I don't see any reason to switch now. There may be a way to alter the default for the compiler itself, but once again, I don't know enough about it. Hope this has been helpful and informative. Good coding!

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  • Connection Reset on MySQL query

    - by sunwukung
    OK, I'm flummoxed. I'm trying to execute a query on a database (locally) and I keep getting a connection reset error. I've been using the method below in a generic DAO class to build a query string and pass to Zend_Db API. public function insert($params) { $loop = false; $keys = $values = ''; foreach($params as $k => $v){ if($loop == true){ $keys .= ','; $values .= ','; } $keys .= $this->db->quoteIdentifier($k); $values .= $this->db->quote($v); $loop = true; } $sql = "INSERT INTO " . $this->table_name . " ($keys) VALUES ($values)"; //formatResult returns an array of info regarding the status and any result sets of the query //I've commented that method call out anyway, so I don't think it's that try { $this->db->query($sql); return $this->formatResult(array( true, 'New record inserted into: '.$this->table_name )); }catch(PDOException $e) { return $this->formatResult($e); } } So far, this has worked fine - the errors have been occurring since we generated new tables to record user input. The insert string looks like this: INSERT INTO tablename(`id`,`title`,`summary`,`description`,`keywords`,`type_id`,`categories`) VALUES ('5539','Sample Title','Sample content',' \'Lorem ipsum dolor sit amet, consectetur adipiscing elit. In et pellentesque mauris. Curabitur hendrerit, leo id ultrices pellentesque, est purus mattis ligula, vitae imperdiet neque ligula bibendum sapien. Curabitur aliquet nisi et odio pharetra tincidunt. Phasellus sed iaculis nisl. Fusce commodo mauris et purus vehicula dictum. Nulla feugiat molestie accumsan. Donec fermentum libero in risus tempus elementum aliquam et magna. Fusce vitae sem metus. Aenean commodo pharetra risus, nec pellentesque augue ullamcorper nec. Class aptent taciti sociosqu ad litora torquent per conubia nostra, per inceptos himenaeos. Nullam vel elit libero. Vestibulum in turpis nunc.\'','this,is,a,sample,array',1,'category title') You'll probably notice the big chunk of whitespace before the Lorem Ipsum string. The description field is being populated from a TinyMCE textarea - I'm guessing it's chucking in some line returns, so I've tried stripping those out. However, even if I disable the TinyMCE field, the reset error still occurs. The next port of call was checking the limits on the table, since it seems to insert if the length of "description" is around the 300 mark (it varies between 310 - 330). The field limit is set to VARCHAR(1500) and the validation on this field won't allow anything past bigger than 1200 with HTML, 800 without. The real kicker is that if I take this sql string and execute it via the command line, it works fine - so I can't for the life of me figure out what's wrong. So, in a nutshell, I'm stumped. Any ideas?

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  • Is this so bad when using MySQL queries in PHP?

    - by alex
    I need to update a lot of rows, per a user request. It is a site with products. I could... Delete all old rows for that product, then loop through string building a new INSERT query. This however will lose all data if the INSERT fails. Perform an UPDATE through each loop. This loop currently iterates over 8 items, but in the future it may get up to 15. This many UPDATEs doesn't sound like too good an idea. Change DB Schema, and add an auto_increment Id to the rows. Then first do a SELECT, get all old rows ids in a variable, perform one INSERT, and then a DELETE WHERE IN SET. What is the usual practice here? Thanks

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  • matplotlib: how to refresh figure.canvas

    - by Alex
    Hello, I can't understand how to refresh FigureCanvasWxAgg instance. Here is the example: import wx import matplotlib from matplotlib.backends.backend_wxagg import FigureCanvasWxAgg as FigureCanvas from matplotlib.figure import Figure class MainFrame(wx.Frame): def __init__(self): wx.Frame.__init__(self, None, wx.NewId(), "Main") self.sizer = wx.BoxSizer(wx.VERTICAL) self.figure = Figure(figsize=(1,2)) self.axe = self.figure.add_subplot(111) self.figurecanvas = FigureCanvas(self, -1, self.figure) self.buttonPlot = wx.Button(self, wx.NewId(), "Plot") self.buttonClear = wx.Button(self, wx.NewId(), "Clear") self.sizer.Add(self.figurecanvas, proportion=1, border=5, flag=wx.ALL | wx.EXPAND) self.sizer.Add(self.buttonPlot, proportion=0, border=2, flag=wx.ALL) self.sizer.Add(self.buttonClear, proportion=0, border=2, flag=wx.ALL) self.SetSizer(self.sizer) self.figurecanvas.Bind(wx.EVT_LEFT_DCLICK, self.on_dclick) self.buttonPlot.Bind(wx.EVT_BUTTON, self.on_button_plot) self.buttonClear.Bind(wx.EVT_BUTTON, self.on_button_clear) self.subframe_opened = False def on_dclick(self, evt): self.subframe = SubFrame(self, self.figure) self.subframe.Show(True) self.subframe_opened = True def on_button_plot(self, evt): self.axe.plot(range(10), color='green') self.figurecanvas.draw() def on_button_clear(self, evt): if self.subframe_opened: self.subframe.Close() self.figure.set_canvas(self.figurecanvas) self.axe.clear() self.figurecanvas.draw() class SubFrame(wx.Frame): def __init__(self, parent, figure): wx.Frame.__init__(self, parent, wx.NewId(), "Sub") self.sizer = wx.BoxSizer(wx.VERTICAL) self.figurecanvas = FigureCanvas(self, -1, figure) self.sizer.Add(self.figurecanvas, proportion=1, border=5, flag=wx.ALL | wx.EXPAND) self.SetSizer(self.sizer) self.Bind(wx.EVT_CLOSE, self.on_close) def on_close(self, evt): self.GetParent().subframe_opened = False evt.Skip() class MyApp(wx.App): def OnInit(self): frame = MainFrame() frame.Show(True) self.SetTopWindow(frame) return True app = MyApp(0) app.MainLoop() I'm interested in the following sequence of operations: run a script resize the main frame press Plot button double click on plot press Clear button Now I get a mess on main frame plot. If I resize the frame it redraws properly. My question is what should I add to my code to do that without resizing? Thanks in advance.

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  • How does C# lambda work?

    - by Alex
    I'm trying to implement method Find that searches the database. I want it to be like that: var user = User.Find(a => a.LastName == "Brown"); Like it's done in List class. But when I go to List's source code (thanks, Reflector), I see this: public T Find(Predicate<T> match) { if (match == null) { ThrowHelper.ThrowArgumentNullException(ExceptionArgument.match); } for (int i = 0; i < this._size; i++) { if (match(this._items[i])) { return this._items[i]; } } return default(T); } How can I implement this thing? I need to get those parameters to make the search.

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  • Recommended Post-SP1 Visual Studio 2008 Hotfixes

    - by Alex Angas
    Today I had to reinstall. I used to have some hotfixes installed for VS2008 but no longer have them and can't remember why they were necessary. I'm expecting any security-related hotfixes to come through Microsoft Update, but I'm interested in VS bug fixes. Does anyone have a list of hotfixes that they recommend installing for Visual Studio 2008 SP1?

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  • DNSCurve vs DNSSEC

    - by Bill Gray
    Can someone informed, please give a lengthy reply about the differences and advantages/disadvantages of both approaches? I am not a DNS expert, not a programmer. I have a decent basic understanding of DNS, and enough knowledge to understand how things like the kaminsky bug work. From what I understand, DNSCurve has stronger encryption, is far simpler to setup, and an altogether better solution. DNSSEC is needlessly complicated and uses breakable encryption, however it provides end to end security, something DNSCurve does not. However, many of the articles I have read have seemed to indicate that end to end security is of little use or makes no difference. So which is true? Which is the better solution, or what are the disadvantages/advantages of each? edit: I would appreciate if someone could explain what is gained by encrypting the message contents, when the goal is authentication rather than confidentiality. The proof that keys are 1024bit RSA keys is here.

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  • How can I set fancyBox's contents dynamically?

    - by alex
    I have tried $('.selector').fancybox({ scrolling : 'no', titleShow : false, type : 'inline', content : '<p>hello</p>' But it did not work I have also tried adding the option onComplete: function() { $('#fancybox-inner').html(contents); } Except then when fancyBox is closed, it will not launch again. How can I dynamically set the contents of fancyBox, and retain all of it's functionality? Thanks

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  • Java Anagram Solver

    - by Alex
    I can work out how to create anagrams of a string but I don't know how I can compare them to a dictionary of real words to check if the anagram is a real word. Is there a class in the Java API that contains the entire English dictionary?

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  • Tips on implementing a custom UITextView interface on the iPhone?

    - by Alex
    I am trying to implement a control to edit text that will display the text in multiple colors. None of the solutions I have attempted yet have been good enough. UITextView cannot accomplish this. All of the text must be the same color. Using CoreGraphics to draw the text does not allow the text to be selected. Using a UIWebView, DIV and PRE tags cannot be set to contentEditable on Mobile Safari. Currently playing with using an off-screen TEXTAREA and an on-screen DIV to show the rendered text. This works pretty well, except supporting all of these at the same time seems impossible: click-to-type, click-to-move-cursor, click-and-hold-select/copy/paste. Anyone have any tips on this predicament? I've been trying to find any preexisting library out there that will accomplish this in a good way, to no luck. I'm open to any ideas!

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  • Tools for conceptual website design

    - by Alex Tang
    What tools (apart from Visio) will generate visually pleasing website site maps or diagrams of a conceptual website? We want to present nice diagrams to our client, but we're unsure about where to get started—we're all coders, not designers. Visio shapes or stencils are quite old. What tools are others in the industry using?

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  • Facebook Graph API - authorization types?

    - by Alex Cook
    I'm struggling with the new Facebook Graph API, perhaps someone here can help. Here is what I want to do: provide a ‘login w/ FB’ button, throw to /authorize, get a code, throw to /access_token, get an access_token, and be able to hit https://graph.facebook.com/me for info about the user. When I try to use type=client_cred in the /authorize call, I get an access_token that lets me hit URLs with userIDs or names, but not /me. I receive an error stating I need a valid token. If I can't hit /me, how do I figure out who the current user is? What exactly should I use in the type param if I want a website to access a users data? I've seen posts with type=web_server, etc, but I can't seem to find a sure fire way to do, what I think, is pretty simple... Thanks ahead of time for any help thats provided...

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  • Using a Generic Repository pattern with fluent nHibernate

    - by alex
    I'm currently developing a medium sized application, which will access 2 or more SQL databases, on different sites etc... I am considering using something similar to this: http://mikehadlow.blogspot.com/2008/03/using-irepository-pattern-with-linq-to.html However, I want to use fluent nHibernate, in place of Linq-to-SQL (and of course nHibernate.Linq) Is this viable? How would I go about configuring this? Where would my mapping definitions go etc...? This application will eventually have many facets - from a WebUI, WCF Library and Windows applications / services. Also, for example on a "product" table, would I create a "ProductManager" class, that has methods like: GetProduct, GetAllProducts etc... Any pointers are greatly received.

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  • Configuring RealVNC to accept copy/paste on a remote server

    - by AP257
    Hi So, I'm using RealVNC Viewer locally (Mac OSX 10.6) and connecting to a VNC Server on a remote machine (Debian - I'm no Unix expert). I want to be able to copy and paste at the command line, but Ctrl+V, Shift+V and Command+V all do nothing on the remote command line. First question: should I be trying a different combination of keys? Secondly, if it's not that I'm using the wrong combination of keys, how can I configure VNC Server to accept copy and paste? I have 'Share clipboard with VNC Server' checked locally, so I figure it must be a problem on the remote machine. I only have command-line access on the remote machine (though I am root) so I need to configure the option somehow via the command line.

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  • Minimal way to render Wiki Syntax (for example MoinMoin Wiki)

    - by Alex
    I enjoy using to Wikis to document all kind of stuff (recently I used MoinMoin, so I am used to that syntax). No I am looking for a more light weight solution, for documents were setting up a moinmoin server is to much hassle. What is the "easiest" way to render a .txt file in Wiki syntax. (for example by displaying it, or converting it to HTML). it should work on Linux, but the more platform in-dependent, the better. Maybe there is even a Javascript based solution?

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