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  • Ruby and RSS2 Feed not displaying image

    - by pcasa
    Trying to create a simple RSS2 Feed that I could later pass on to FeedBurner but can't get RSS feed to display images at all. Also, from what I have read having xml.instruct! on top might cause IE to complain it's not a valid feed. Is this true? My Code looks like xml.instruct! xml.rss "version" => "2.0", "xmlns:dc" => "http://purl.org/dc/elements/1.1/" do xml.channel do xml.title "Store" xml.link url_for :only_path => false, :controller => 'products' xml.description "Store" xml.pubDate @products.first.updated_at.rfc822 if @products.any? @products.each do |product| xml.item do xml.title product.name xml.pubDate (product.updated_at.rfc822) xml.image do xml.url domain_host + product.product_image.url(:small) xml.title "Store" xml.link url_for :only_path => false, :controller => 'products' end xml.link url_for :only_path => false, :controller => 'products', :action => 'show', :id => product.permalink xml.description product.fine_print xml.guid url_for :only_path => false, :controller => 'products', :action => 'show', :id => product.permalink end end end end

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  • How can I bind a simple Javascript array to an MVC3 controller action method?

    - by Sergio Tapia
    Here is the javascript code I use to create the array and send it on it's way: <script type="text/javascript" language="javascript"> $(document).ready(function () { $("#update-cart-btn").click(function() { var items = []; $(".item").each(function () { var productKey = $(this).find("input[name='item.ProductId']").val(); var productQuantity = $(this).find("input[type='text']").val(); items[productKey] = productQuantity; }); $.ajax({ type: "POST", url: "@Url.Action("UpdateCart", "Cart")", data: items, success: function () { alert("Successfully updated your cart!"); } }); }); }); </script> The items object is properly constructed with the values I need. What data type must my object be on the backend of my controller? I tried this but the variable remains null and is not bound. [Authorize] [HttpPost] public ActionResult UpdateCart(object[] items) // items remains null. { // Some magic here. return RedirectToAction("Index"); }

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  • Dynamic loading a class in java with a different package name

    - by C. Ross
    Is it possible to load a class in Java and 'fake' the package name/canonical name of a class? I tried doing this, the obvious way, but I get a "class name doesn't match" message in a ClassDefNotFoundException. The reason I'm doing this is I'm trying to load an API that was written in the default package so that I can use it directly without using reflection. The code will compile against the class in a folder structure representing the package and a package name import. ie: ./com/DefaultPackageClass.class // ... import com.DefaultPackageClass; import java.util.Vector; // ... My current code is as follows: public Class loadClass(String name) throws ClassNotFoundException { if(!CLASS_NAME.equals(name)) return super.loadClass(name); try { URL myUrl = new URL(fileUrl); URLConnection connection = myUrl.openConnection(); InputStream input = connection.getInputStream(); ByteArrayOutputStream buffer = new ByteArrayOutputStream(); int data = input.read(); while(data != -1){ buffer.write(data); data = input.read(); } input.close(); byte[] classData = buffer.toByteArray(); return defineClass(CLASS_NAME, classData, 0, classData.length); } catch (MalformedURLException e) { throw new UndeclaredThrowableException(e); } catch (IOException e) { throw new UndeclaredThrowableException(e); } }

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  • HTML href with css ie Problem

    - by Jordan Pagaduan
    <style type="text/css"> .web_westloh { background-image: url(images/web_westloh.png); background-repeat: no-repeat; height: 100px; width: 350px; } .web_westloh:hover { border-bottom-width: 2px; border-bottom-style: dashed; border-bottom-color: #999999; padding-bottom: 5px; } .web_money { background-image: url(images/web_money.png); background-repeat: no-repeat; height: 100px; width: 350px; } .web_money:hover { border-bottom-width: 2px; border-bottom-style: dashed; border-bottom-color: #999999; padding-bottom: 5px; } </style> <a href="http://www.westloh.com" title="Click to Visit http://www.westloh.com" target="_blank" class="web_westloh"> <div class="web_westloh"></div> </a> <a href="http://www.money-mind-set.com" title="Click to Visit http://www.money-mind-set.com" target="_blank"> <div class="web_money"></div> </a> The Problem is: In mozilla linking is ok. No problem. But in IE the link is a problem, it will not link in the target. See this page to see Problem: http://replytowest.com -- at the bottom. Thank You

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  • Browser window popups - risks and special features

    - by Sandeepan Nath
    1. What exactly is the security risk with popups? The new browsers provide settings to block window popups (on blocking, sites with active popups display a message to user). What exactly is the security risk with popups? If allowing popups can execute something dangerous, then the main window can too. Is it not the case. I think I don't know about some special powers of window popups. 2. Any special features of popup windows? Take for example the HDFC bank netbanking site. The entire netbanking session happens in a new window popup and a user neither manually edit the URL or paste the URL in the main browser window. it does not work. Is a popup window needed for this feature? Does it improve security? (Asking because everything that is there in this site revolves around security - so they must have done that for a reason too). Why otherwise they would implement the entire netbanking on a popup window? 3. Is it possible to override browser's popup blocking settings Lastly, the HDFC site succcessfully displays popup window even when in the browser settings popups are blocked. So, how do they do it? Is that a browser hack? To see this - go to http://hdfcbank.com/ Under the "Login to your account" section select "HDFC Bank NetBanking" and click the "Login" button. You can verify that even if popups are blocked/popup blocker is enabled in the browser settings, this site is able to display popups. The answers to this question say that it is not possible to display popup windows if it has been blocked in browser settings. Solved Concluded with Pointy's solution and comments under that. Here is a fiddle demonstrating the same.

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  • POSTmethod and PHP- login verification

    - by Neethusha
    I wrote a code for login verification..I got output with GET. But i need output with POST since it is more secure.pls let me know if there is any error in my code. javascript code: var xml; function verifyusernamepasswd(pass) { //pass is password that will be passed as parameter xml=new XMLHttpRequest(); var url="http://localhost/loginvalidate.php"; var para="q="+username+"&p="+pass;//username is global xml.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); xml.setRequestHeader("Content-length", para.length); xml.setRequestHeader("Connection", "close"); xml.open("POST",url,true); xml.onreadystatechange=statechanged1; xml.send(para); } function statechanged1() { if(xml.readyState==4) alert(xml.responseText); } php code: <?php $username=$_POST["q"]; $password=$_POST["p"]; $con=mysql_connect("localhost","root","blaze"); if(!$con) { die('Could not connect: '.mysql.error()); } mysql_select_db("BLAZE",$con) or die("No such Db"); $result=mysql_query("SELECT Passwword FROM USERTABLE WHERE Userhandle='$username'"); if($result==null) echo "false"; else if($result!=null) { $row=mysql_fetch_array($result); if((strcmp($row['Passwword'],$password)==0)) echo "true"; else echo "false"; } ?> the verification does not return anything, cos my alert is not displayed at all...pls tell me whats wrong....

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  • How to map (large) integer on (small in size( alphanumeric string with PHP? (Cantor?)

    - by Glooh
    Dear all, I can't figure out how to optimally do the following in PHP: In a database, I have messages with a unique ID, like 19041985. Now, I want to refer to these messages in a short-url service but not using generated hashes but by simply 'calculate' the original ID. In other words, for example: http://short.url/sYsn7 should let me calculate the message ID the visitor would like to request. To make it more obvious, I wrote the following in PHP to generate these 'alphanumeric ID versions' and of course, the other way around will let me calculate the original message ID. The question is: Is this the optimal way of doing this? I hardly think so, but can't think of anything else. $alphanumString = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ-_'; for($i=0;$i < strlen($alphanumString);$i++) { $alphanumArray[$i] = substr($alphanumString,$i,1); } $id = 19041985; $out = ''; for($i=0;$i < strlen($id);$i++) { if(isset($alphanumString["".substr($id,$i,2).""]) && strlen($alphanumString["".substr($id,$i,2).""]) 0) { $out.=$alphanumString["".substr($id,$i,2).""]; } else { $out.=$alphanumString["".substr($id,$i,1).""]; $out.=$alphanumString["".substr($id,($i+1),1).""]; } $i++; } print $out;

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  • Paging not working in my wordpress installation

    - by Bootcamp
    I recently started a blog site and wanted to give it a magazine look. I used Wordpress for my blog and used the Arthemia theme with it. I also changed the permalink structure to point to /%year%/%monthnum%/%day%/%postname%/ structure. Now the problem that i have is that the paging has stopped working on my home page. When i click on the next page link i get a 404 error. My /page/2 url does not show the next page. I check on google and found out that it was due to the redirection that is being performed due to the permalink change. The solution given was that i need to skip the url rewriting for the /page/* urls. This is the link to an article which said this http://www.yoursearchadvisor.com/blog/wordpress-next_posts_link-broken/ . I was not able to follow this article and solve my problem, as i could not find the permanent redirect manager under the settings section as said in this article. Can somebody please guide me how to solve this problem. I am using the latest Wordpress version and Arthemia theme with it. Thanks.

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  • Parsing the Youtube API with DOM

    - by Kirk
    I'm using the Youtube API and I can retrieve the date information without a problem, but don't know how to retrieve the description information. My Code: <?php $v = "dQw4w9WgXcQ"; $url = "http://gdata.youtube.com/feeds/api/videos/". $v; $doc = new DOMDocument; $doc->load($url); $pub = $doc->getElementsByTagName("published")->item(0)->nodeValue; $desc = $doc->getElementsByTagName("media:description")->item(0)->nodeValue; echo "<b>Video Uploaded:</b> "; echo date( "F jS, Y", strtotime( $pub ) ); echo '<br>'; if (isset ($desc)) { echo "<b>Description:</b> "; echo $desc; echo '<br>'; } ?> Here's a link to the feed: http://gdata.youtube.com/feeds/api/videos/dQw4w9WgXcQ?prettyprint=true And the excerpt of code I don't know how to retrieve data from: <media:group> <media:description type='plain'>Music video by Rick Astley performing Never Gonna Give You Up. (C) 1987 PWL</media:description> </media:group> Thanks in advance.

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  • Rails - any fancy ways to handle 404s?

    - by jyoseph
    I have a rails app I built for an old site I converted from another cms (in a non-rails language, hehe). Most of the old pages are mapped to the new pages using routes.rb. But there are still a few 404s. I am a rails newb so I'm asking if there are any advanced ways to handle 404s. For example, if I was programming in my old language I'd do this: Get the URL (script_name) that was being accessed and parse it. Do a lookup in the database for any keywords, ids, etc found in the new URL. If found, redirect to the page (or if multiple records are found, show them all on a results page and let user choose). With rails I'd probably want to do :status = :moved_permanently I'm guessing? If not found, show a 404. Are there any gems/plugins or tutorials you know of that would handle such a thing, if it's even possible. Or can you explain on a high level how that can be done? I don't need a full code sample, just a push in the right direction. PS. It's a simple rails 3 app that uses a single Content model.

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  • Jquery Sorting by Letter

    - by Batfan
    I am using jquery to sort through a group of paragraph tags (kudos to Aaron Harun). It pulls the value "letter" (a letter) from the url string and displays only paragraphs that start with that letter. It hides all others and also consolidates the list so that there are no duplicates showing. See the code: var letter = '<?php echo(strlen($_GET['letter']) == 1) ? $_GET['letter'] : ''; ?>' function finish(){ var found_first = []; jQuery('p').each(function(){ if(jQuery(this).text().substr(0,1).toUpperCase() == letter){ if(found_first[jQuery(this).text()] != true){ jQuery(this).addClass('current-series'); found_first[jQuery(this).text()] = true; }else{ jQuery(this).hide(); } } else{ jQuery(this).hide();} }) } Been working with this all day and I have 2 Questions on this: Is there a way to get it to ignore the word 'The', if it's first? For example, if a paragraph starts with 'The Amazing', I would like it to show up on the 'A' page, not the 'T' page, like it currently is. Is there a way to have a single page for (all) numbers? For example, the url to the page would be something similar to domain.com/index.php?letter=0 and this would show only the paragraph tags that start with a number, any number. I can currently do this with single numbers but, I would like 1 page for all numbers.

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  • Django: Serving a Download in a Generic View

    - by TheLizardKing
    So I want to serve a couple of mp3s from a folder in /home/username/music. I didn't think this would be such a big deal but I am a bit confused on how to do it using generic views and my own url. urls.py url(r'^song/(?P<song_id>\d+)/download/$', song_download, name='song_download'), The example I am following is found in the generic view section of the Django documentations: http://docs.djangoproject.com/en/dev/topics/generic-views/ (It's all the way at the bottom) I am not 100% sure on how to tailor this to my needs. Here is my views.py def song_download(request, song_id): song = Song.objects.get(id=song_id) response = object_detail( request, object_id = song_id, mimetype = "audio/mpeg", ) response['Content-Disposition'= "attachment; filename=%s - %s.mp3" % (song.artist, song.title) return response I am actually at a loss of how to convey that I want it to spit out my mp3 instead of what it does now which is to output a .mp3 with all of the current pages html contained. Should my template be my mp3? Do I need to setup apache to serve the files or is Django able to retrieve the mp3 from the filesystem(proper permissions of course) and serve that? If it do need to configure Apache how do I tell Django that? Thanks in advanced. These files are all on the HD so I don't need to "generate" anything on the spot and I'd like to prevent revealing the location of these files if at all possible. A simple /song/1234/download would be fantastic.

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  • Render only a portion of a PDF on iPhone/iPad

    - by Infinity
    Hello guys! I am rendering my pdf in an UIView's drawRect method. Here's my code: - (void)drawRect:(CGRect)rect2 { NSString *filename = @"lol.pdf"; CFStringRef path = CFStringCreateWithCString (NULL, [filename UTF8String], kCFStringEncodingUTF8); CFURLRef url = CFURLCreateWithFileSystemPath (NULL, path, kCFURLPOSIXPathStyle, 0); CGPDFDocumentRef pdf = CGPDFDocumentCreateWithURL(url); CGPDFPageRef page = CGPDFDocumentGetPage (pdf, 1); CGAffineTransform m; CGContextRef context = UIGraphicsGetCurrentContext(); CGRect aRect = CGRectMake(0, 0, 768, 1024); CGContextTranslateCTM(context, 0.0, aRect.size.height); CGContextScaleCTM(context, 1.0, -1.0); m = CGPDFPageGetDrawingTransform (page, kCGPDFMediaBox, aRect, 0, YES); CGContextSaveGState (context); CGContextConcatCTM (context, m); CGContextClipToRect (context,CGPDFPageGetBoxRect (page, kCGPDFCropBox)); CGContextDrawPDFPage (context, page); CGContextRestoreGState (context); } It renders the whole pdf. How can I render only a part from it? Can you help me with it?

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  • tomcat resource missing, servlet not running

    - by user2837260
    import javax.servlet.*; import java.io.*; public class MyServlet implements Servlet { public void init(ServletConfig con) {} public void service(ServletRequest req, ServletResponse res) throws IOException,ServletException { res.setContentType("text/html"); PrintWriter out=res.getWriter(); String s="blah"; String s1="blah"; out.println("<html><body>"); if((s.equals(req.getParameter("firstname")))&&(s1.equals(req.getParameter("pwd")))) out.println("passwords match"); else out.println("password and name combo does not match"); out.println("</body></html>"); } public void destroy() {} public ServletConfig getServletConfig() { return null;} public String getServletInfo() { return null;} } this is my java file with the servlet class.its saved with the name MyServlet.java and so is the class file. and here is the xml file: <web-app> <servlet> <servlet-name>demoo</servlet-name> <servlet-class>MyServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>demoo</servlet-name> <url-pattern>/demo</url-pattern> </servlet-mapping> </web-app> i have made the folder as WEB-INF and then classes... WEB-INF also contains the .xml file but when i try to run the servlet , it says resource not found ps- i am already looking for the servlet with the name :demo localhost:8081/s1/demo* s1 is the war file * a html file in the war file seems to run fine on the server though. *

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  • MVC 2 Beta DefaultControllerFactory with Areas

    - by stoto
    Why default factory WON'T return full name of the controllers (with namespaces)? I'm using Service Locator and autofac. using System.Web.Mvc; using Microsoft.Practices.ServiceLocation; namespace Application.Core.MVC { public override IController CreateController(System.Web.Routing.RequestContext requestContext, string **controllerName**) { return ServiceLocator.Current.GetInstance<IController>(controllerName); } } I had two home controllers (one under area Blog) http://localhost/Home http://localhost/Blog/Home controllerName return only "Home" without full qualified name for both in above code. This creates a problem when I try to regiser controllers' names for dependency injection. Here is how I register controllers right now according to this situation. Even this brings up the pages without exception. But When I access http://localhost/Home, both controllers invoked regardlessly. foreach (var tp in currentAssemblyControllersTypes) builder.Register(tp).FactoryScoped().Named(tp.Name.Replace("Controller", "")); Anyone can help?Thanks.

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  • Load images into separate movie clips from a XML, Flash, Actionscript 3.0

    - by James Dunay
    I have an xml image bank, pretty standard, and I have a loader, along with movie clips that I want the images loaded into, the problem that I am running into is I want the images to load into separate movie clips, so I’m using a case statement to specify where they go. However, I can only get them to load into a single movie clip, I assume they are loading ontop of each other and I don’t know how to get them to separate out. I’ll post my code. It doesn’t make any sense to me but if you have any suggestions that would be real great. I can make separate loaders and then just do 1 image per loader, but that just doesn’t sound right to me. var counterNumber:Number = 0; function callThumbs():void{ for (var i:Number = 0; i <3; i++){ thumbLoaded(); counterNumber++; } } function thumbLoaded(){ var photoLoader = new Loader(); switch (counterNumber){ case 1: photoLoader.load(new URLRequest(MovieClip(this.parent).xml.photos.imageOne.image.@url[0])); whole.boxOne.pictureLoader.addChild(photoLoader); trace("1Done"); break; case 2: photoLoader.load(new URLRequest(MovieClip(this.parent).xml.photos.imageTwo.image.@url[0])); whole.boxTwo.pictureLoader.addChild(photoLoader); trace("2Done"); break; } }

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  • Add class active when clicking the menu link with Jquery

    - by Adrian
    I have HTML <div id="top" class="shadow"> <ul class="gprc"> <li><a href="http://www.domain.com/">Home</a></li> <li><a href="http://www.domain.com/link1/">Text1</a></li> <li><a href="http://www.domain.com/link2/">Text2</a></li> <li><a href="http://www.domain.com/link3/">Text3</a></li> <li><a href="http://www.domain.com/link4">Text4</a></li> </ul> </div> and JQUERY $(function () { var url = window.location.pathname, urlRegExp = new RegExp(url.replace(/\/$/, '') + "$"); $('#top a').each(function () { if (urlRegExp.test(this.href.replace(/\/$/, ''))) { $(this).addClass('active'); } }); }); The problem is that when i click on the Home link all tabs are getting active class and don't understand why. I need it for the first link to not get any active class.

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  • OpenMeetings + Python + Suds

    - by user366774
    Trying to integrate openmeetings with django website, but can't understand how properly configure ImportDoctor: (here :// replaced with __ 'cause spam protection) print url http://sovershenstvo.com.ua:5080/openmeetings/services/UserService?wsdl imp = Import('http__schemas.xmlsoap.org/soap/encoding/') imp.filter.add('http__services.axis.openmeetings.org') imp.filter.add('http__basic.beans.hibernate.app.openmeetings.org/xsd') imp.filter.add('http__basic.beans.data.app.openmeetings.org/xsd') imp.filter.add('http__services.axis.openmeetings.org') d = ImportDoctor(imp) client = Client(url, doctor = d) client.service.getSession() Traceback (most recent call last): File "", line 1, in File "/usr/lib/python2.6/site-packages/suds/client.py", line 539, in call return client.invoke(args, kwargs) File "/usr/lib/python2.6/site-packages/suds/client.py", line 598, in invoke result = self.send(msg) File "/usr/lib/python2.6/site-packages/suds/client.py", line 627, in send result = self.succeeded(binding, reply.message) File "/usr/lib/python2.6/site-packages/suds/client.py", line 659, in succeeded r, p = binding.get_reply(self.method, reply) File "/usr/lib/python2.6/site-packages/suds/bindings/binding.py", line 159, in get_reply resolved = rtypes[0].resolve(nobuiltin=True) File "/usr/lib/python2.6/site-packages/suds/xsd/sxbasic.py", line 63, in resolve raise TypeNotFound(qref) suds.TypeNotFound: Type not found: '(Sessiondata, http__basic.beans.hibernate.app.openmeetings.org/xsd, )' what i'm doing wrong? please help and sorry for my english, but you are my last chance to save position :( need webinars at morning (2.26 am now)

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  • android view web pictures in gallery

    - by bitma
    I am new to android, I just finished Hello gallery tutorial, and a web pciture tutorial. Now I want to know how can I show some web images in gallery? the hello gallery code is from andorid tutor this is Web gallery code, I want to load some pictures from web and then show them in gallery, how can I write it? public class WebGallery extends Activity { String imageUrl = "http://i.pbase.com/o6/92/229792/1/80199697.uAs58yHk.50pxCross_of_the_Knights_Templar_svg.png"; Bitmap bmImg; ImageView imView; public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); imView = (ImageView) findViewById(R.id.imview); imView.setImageBitmap(getRemoteImage(imageUrl)); } public Bitmap getRemoteImage(String imageUrl) { try { URL aURL = new URL(imageUrl); final URLConnection conn = aURL.openConnection(); conn.connect(); final BufferedInputStream bis = new BufferedInputStream(conn.getInputStream()); final Bitmap bm = BitmapFactory.decodeStream(bis); bis.close(); return bm; } catch (IOException e) { Log.d("DEBUGTAG", "Oh noooz an error..."); } return null; } }

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  • Address family not supported by protocol exception

    - by srg
    I'm trying to send a couple of values from an android application to a web service which I've setup. I'm using Http Post to send them but when I run the application I get the error- request time failed java.net.SocketException: Address family not supported by protocol. I get this while debugging with both the emulator as well as a device connected by wifi. I've already added the internet permission using: <uses-permission android:name="android.permission.INTERNET" /> This is the code i'm using to send the values void insertData(String name, String number) throws Exception { String url = "http://192.168.0.12:8000/testapp/default/call/run/insertdbdata/"; HttpClient client = new DefaultHttpClient(); HttpPost post = new HttpPost(url); try { List<NameValuePair> params = new ArrayList<NameValuePair>(2); params.add(new BasicNameValuePair("a", name)); params.add(new BasicNameValuePair("b", number)); post.setEntity(new UrlEncodedFormEntity(params)); HttpResponse response = client.execute(post); }catch(Exception e){ e.printStackTrace(); } Also I know that my web service work fine because when I send the values from an html page it works fine - <form name="form1" action="http://192.168.0.12:8000/testapp/default/call/run/insertdbdata/" method="post"> <input type="text" name="a"/> <input type="text" name="b"/> <input type="submit"/> I've seen questions of similar problems but haven't really found a solution. Thanks

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  • Django: Determining if a user has voted or not

    - by TheLizardKing
    I have a long list of links that I spit out using the below code, total votes, submitted by, the usual stuff but I am not 100% on how to determine if the currently logged in user has voted on a link or not. I know how to do this from within my view but do I need to alter my below view code or can I make use of the way templates work to determine it? I have read http://stackoverflow.com/questions/1528583/django-vote-up-down-method but I don't quite understand what's going on ( and don't need any ofjavascriptery). Models (snippet): class Link(models.Model): category = models.ForeignKey(Category, blank=False, default=1) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add=True) modified = models.DateTimeField(auto_now=True) url = models.URLField(max_length=1024, unique=True, verify_exists=True) name = models.CharField(max_length=512) def __unicode__(self): return u'%s (%s)' % (self.name, self.url) class Vote(models.Model): link = models.ForeignKey(Link) user = models.ForeignKey(User) created = models.DateTimeField(auto_now_add=True) def __unicode__(self): return u'%s vote for %s' % (self.user, self.link) Views (snippet): def hot(request): links = Link.objects.select_related().annotate(votes=Count('vote')).order_by('-created') for link in links: delta_in_hours = (int(datetime.now().strftime("%s")) - int(link.created.strftime("%s"))) / 3600 link.popularity = ((link.votes - 1) / (delta_in_hours + 2)**1.5) if request.user.is_authenticated(): try: link.voted = Vote.objects.get(link=link, user=request.user) except Vote.DoesNotExist: link.voted = None links = sorted(links, key=lambda x: x.popularity, reverse=True) links = paginate(request, links, 15) return direct_to_template( request, template = 'links/link_list.html', extra_context = { 'links': links, }) The above view actually accomplishes what I need but in what I believe to be a horribly inefficient way. This causes the dreaded n+1 queries, as it stands that's 33 queries for a page containing just 29 links while originally I got away with just 4 queries. I would really prefer to do this using Django's ORM or at least .extra(). Any advice?

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  • cURL + HTTP_POST, keep getting 500 error. Has no idea?

    - by mysqllearner
    Okay, I want to make a HTTP_POST using cURL to a SSL site. I already imported the certificate to my server. This is my code: $url = "https://www.xxx.xxx"; $post = "";# all data that going to send $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_POST, true); curl_setopt($ch, CURLOPT_POSTFIELDS, $post); curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, FALSE); curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, FALSE); curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/4.0 (compatible; MSIE 5.01; Windows NT 5.0'); $exe = curl_exec($ch); $getInfo = curl_getinfo($ch); if ($exe === false) { $output = "Error in sending"; if (curl_error($ch)){ $output .= "\n". curl_error($ch); } } else if($geInfo['http_code'] != 777){ $output = "No data returned. Error: " . $geInfo['http_code']; if (curl_error($ch)){ $output .= "\n". curl_error($ch); } } curl_close($c); echo $output; It keep returned "500". Based on w3schools, 500 means Internal Server Error. Is my server having problem? How to solve/troubleshoot this?

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  • rails not recognizing project

    - by tipu
    I can create a new project using rails and I can use stuff like rails migration ... and i (correctly) get a error because the sqlite gem is missing. but when i try using rails migration ... with a project i checked out from github, it doesn't recognize that it is a rails project i get: Usage: rails new APP_PATH [options] Options: -d, [--database=DATABASE] # Preconfigure for selected database (options: mysql/oracle/postgresql/sqlite3/frontbase/ibm_db) # Default: sqlite3 -O, [--skip-active-record] # Skip Active Record files [--dev] # Setup the application with Gemfile pointing to your Rails checkout -J, [--skip-prototype] # Skip Prototype files -T, [--skip-test-unit] # Skip Test::Unit files -G, [--skip-git] # Skip Git ignores and keeps -b, [--builder=BUILDER] # Path to an application builder (can be a filesystem path or URL) [--edge] # Setup the application with Gemfile pointing to Rails repository -m, [--template=TEMPLATE] # Path to an application template (can be a filesystem path or URL) -r, [--ruby=PATH] # Path to the Ruby binary of your choice # Default: /usr/bin/ruby1.8 [--skip-gemfile] # Don't create a Gemfile and it goes on. any ideas? edit: it's probably an important detail that earlier my rails wasn't working at all. i had to cp /usr/bin/ruby to /usr/bin/local/ruby

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  • webclient download problem!!!

    - by user472018
    Hello all, if this problem was discussed before,sorry for asking again.. I want to download an image from an url with using System.Net.WebClient class. When i try to download an image (ie. google logo).it does not occur any errors,but some images are occurring errors.I dont understand why this errors. how can i fix this problem? my Code is: WebClient client = new WebClient(); try { //Downloads the file from the given url to the given destination client.DownloadFile(urltxt.Text, filetxt.Text); return true; } catch (WebException w) { MessageBox.Show(w.ToString()); return false; } catch (System.Security.SecurityException) { MessageBox.Show("securityexeption"); return false; } catch (Exception) { MessageBox.Show("exception"); return false; } Errors are: System.Net.WebException:The underlying connection was closed:An unexpected error occurred on a recieve.--System.IO.IOException:Unable to read data from the transport connection:An existing connection was forcibly closed by the remote host.--System.Net.Sockets.SocketException:An existing connection was forcibly closed by the remote host...bla bla Thanks for your help.

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  • problem configure JBoss to work with JNDI(2)

    - by Spiderman
    in continuation to the question from last week: http://stackoverflow.com/questions/2828237/problem-configure-jboss-to-work-with-jndi I'm trying to bind datasource in JBoss and use it in my application. In my struggling, I already managed to avoid the javax.naming.NameNotFoundException by: 1. using in java new InitialContext().lookup(connection); instead of new JndiObjectFactoryBean().setJndiName(connection); 2. changing the connection name from: 'jndi-name' to 'java:jndi-name' Now the problem is that the datasouce that I get from the lookup is null. I created the datsource file: <datasources> <local-tx-datasource> <jndi-name>bilby</jndi-name> <connection-url>jdbc:oracle:myURL</connection-url> <driver-class>oracle.jdbc.OracleDriver </driver-class> <user-name>myUsername</user-name> <password>myPassword</password> <exception-sorter-class- name>org.jboss.resource.adapter.jdbc.vendor.OracleExceptionSorter</exception-sorter-class-name> <metadata> <type-mapping>Oracle9i</type-mapping> </metadata> </local-tx-datasource> </datasources> and put it under \server\default\deploy\oracle-ds.xml I get during runtime the line: 18:37:56,560 INFO [ConnectionFactoryBindingService] Bound ConnectionManager 'jb oss.jca:service=DataSourceBinding,name=bilby' to JNDI name 'java:bilby' So my question is - why do I get null as my datasource???

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