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  • UITableView Inactive Until Scroll iPhone

    - by dubbeat
    HI, I'm trying to find the cause of some annoying behaviour with my UITableView. In each cell of my table (10 of them ) I asynchronously load an image. The problem is, is that if I don't touch the app at all the imageviews will permanently show a loading icon. However as soon as I scroll a cell off the screen and back on again the image shows straight away. What could be causing this "stuck in the mud" behaviour? Is there a way to force whatever gets called when the list is scrolled to make the needed update happen? #define ASYNC_IMAGE_TAG 9999 #define LABEL_TAG 8888 // Customize the appearance of table view cells. - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { PromoListItem *promo = [promoList objectAtIndex:indexPath.row]; static NSString *CellIdentifier = @"Cell"; AsyncImageView *asyncImageView = nil; UILabel *label = nil; UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier]; if (cell == nil) { cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:CellIdentifier] autorelease]; CGRect frame; frame.origin.x = 0; frame.origin.y = 0; frame.size.width = 100; frame.size.height = 100; asyncImageView = [[[AsyncImageView alloc] initWithFrame:frame] autorelease]; asyncImageView.tag = ASYNC_IMAGE_TAG; [cell.contentView addSubview:asyncImageView]; frame.origin.x = 110; frame.size.width =100; label = [[[UILabel alloc] initWithFrame:frame] autorelease]; label.tag = LABEL_TAG; [cell.contentView addSubview:label]; cell.accessoryType = UITableViewCellAccessoryDisclosureIndicator; } else { asyncImageView = (AsyncImageView *) [cell.contentView viewWithTag:ASYNC_IMAGE_TAG]; label = (UILabel *) [cell.contentView viewWithTag:LABEL_TAG]; } NSURL *url = [NSURL URLWithString:promo.imgurl]; [asyncImageView loadImageFromURL:url]; label.text = promo.artistname; return cell; }

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  • how to handle out of memory error?

    - by UMMA
    dear friends, i am using following code to display bitmap in my imageview. when i try to load image of size for example bigger than 1.5MB it give me error any one suggest me solution? try { URL aURL = new URL(myRemoteImages[val]); URLConnection conn = aURL.openConnection(); conn.connect(); InputStream is = null; try { is= conn.getInputStream(); }catch(IOException e) { return 0; } int a= conn.getConnectTimeout(); BufferedInputStream bis = new BufferedInputStream(is); Bitmap bm; try { bm = BitmapFactory.decodeStream(bis); }catch(Exception ex) { bis.close(); is.close(); return 0; } bis.close(); is.close(); img.setImageBitmap(bm); } catch (IOException e) { return 0; } return 1; Log cat 06-14 12:03:11.701: ERROR/AndroidRuntime(443): Uncaught handler: thread main exiting due to uncaught exception 06-14 12:03:11.861: ERROR/AndroidRuntime(443): java.lang.OutOfMemoryError: bitmap size exceeds VM budget 06-14 12:03:11.861: ERROR/AndroidRuntime(443): at android.graphics.BitmapFactory.nativeDecodeStream(Native Method)

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  • Java webapp: how to implement a web bug (1x1 pixel)?

    - by NoozNooz42
    In the accepted answer in the following question, a SO regular with 13K+ rep suggests to use a "web bug" (non-cacheable 1x1 img) to be able to track requests in the logs: http://stackoverflow.com/questions/1784893 How can I do this in Java? Basically, I've got two issues: how to make sure the 1x1 image is not cacheable (how to set the header)? how to make sure the query for these 1x1 image will appear in the logs? I'm looking for exact piece of code because I know how to write a .jsp/servlet and I know how to serve an 1x1 image :) My question is really about the exact .jsp/servlet that I should write and how/what needs to be done so that Tomcat logs the request. For example I plan to use the following mapping: <servlet-mapping> <servlet-name>WebBugServlet</servlet-name> <url-pattern>/webbug*</url-pattern> </servlet-mapping> and then use an img tag referencing a "webbug.png" (or .gif), so how do I write the .jsp/servlet? What/where should I look for in the logs?

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  • Too many open files in one of my java routine.

    - by Irfan Zulfiqar
    I have a multithreaded code that has to generated a set of objects and write them to a file. When I run it I sometime get "Too many open files" message in Exception. I have checked the code to make sure that all the file streams are being closed properly. Here is the stack trace. When I do ulimit -a, open files allowed is set to 1024. We think increasing this number is not a viable option / solution. [java] java.io.FileNotFoundException: /export/event_1_0.dtd (Too many open files) [java] at java.io.FileInputStream.open(Native Method) [java] at java.io.FileInputStream.<init>(FileInputStream.java:106) [java] at java.io.FileInputStream.<init>(FileInputStream.java:66) [java] at sun.net.www.protocol.file.FileURLConnection.connect(FileURLConnection.java:70) [java] at sun.net.www.protocol.file.FileURLConnection.getInputStream(FileURLConnection.java:161) [java] at java.net.URL.openStream(URL.java:1010) Now what we have identified so far by looking closely at the list of open files is that the VM is opening same class file multiple times. /export/BaseEvent.class 236 /export/EventType1BaseEvent.class 60 /export/EventType2BaseEvent.class 48 /export/EventType2.class 30 /export/EventType1.class 14 Where BaseEvent is partent of all the classes and EventType1 ant EventType2 inherits EventType1BaseEvent and EventType2BaseEvent respectively. Why would a class loader load the same class file 200+ times. It seems it is opening up the base class as many time it create any child instance. Is this normal? Can it be handler any other way apart from increasing the number of open files?

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  • ASP.Net MVC 2 - How to set up a Cancel button with client side navigation

    - by arame3333
    Thanks to a previous question I found a useful link on multiple buttons. http://weblogs.asp.net/dfindley/archive/2009/05/31/asp-net-mvc-multiple-buttons-in-the-same-form.aspx What I want to do is have a cancel button on my page, similar to this; <button name="button" type="button" onclick="document.location.href=$('#cancelUrl').attr('href')">Cancel</button> <a id="cancelUrl" href="<%: Url.Action("Index", "Home") %>" style="display:none;"></a> However although this code works, I really want to go back to the previous page. For Web Forms I could use the javascript Back() or Go(-1) functions, but they relied on postbacks. I could of course hard code the previous page and controller as I have done above. However I am struggling to find links that explain to me how Url.Action works. Because if I do this, I also need to include an index parameter, and I am not clear how the syntax works for that. It seems odd the amount of coding to do this. Out of curiosity, I am also wondering how you TDD client side code like this.

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  • What techniques can be used to detect so called "black holes" (a spider trap) when creating a web crawler?

    - by Tom
    When creating a web crawler, you have to design somekind of system that gathers links and add them to a queue. Some, if not most, of these links will be dynamic, which appear to be different, but do not add any value as they are specifically created to fool crawlers. An example: We tell our crawler to crawl the domain evil.com by entering an initial lookup URL. Lets assume we let it crawl the front page initially, evil.com/index The returned HTML will contain several "unique" links: evil.com/somePageOne evil.com/somePageTwo evil.com/somePageThree The crawler will add these to the buffer of uncrawled URLs. When somePageOne is being crawled, the crawler receives more URLs: evil.com/someSubPageOne evil.com/someSubPageTwo These appear to be unique, and so they are. They are unique in the sense that the returned content is different from previous pages and that the URL is new to the crawler, however it appears that this is only because the developer has made a "loop trap" or "black hole". The crawler will add this new sub page, and the sub page will have another sub page, which will also be added. This process can go on infinitely. The content of each page is unique, but totally useless (it is randomly generated text, or text pulled from a random source). Our crawler will keep finding new pages, which we actually are not interested in. These loop traps are very difficult to find, and if your crawler does not have anything to prevent them in place, it will get stuck on a certain domain for infinity. My question is, what techniques can be used to detect so called black holes? One of the most common answers I have heard is the introduction of a limit on the amount of pages to be crawled. However, I cannot see how this can be a reliable technique when you do not know what kind of site is to be crawled. A legit site, like Wikipedia, can have hundreds of thousands of pages. Such limit could return a false positive for these kind of sites. Any feedback is appreciated. Thanks.

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  • ListView Parsing Persian xml

    - by Namikaze Minato
    I used a tutorial about listview parsing xm from internet and using LasyAdapter shows the items in listview. When I add persian characters in xml (into one of childnodes) the result is some boxes in listview (after showing the text in listview). The format of xml is UTF-8, too. I used typeface too (but didn't work). Besides when I type Pwesian into the application it shows alright but it can't show Persiann content parsed from xml. Thanks in advance. I updated the post with original XMLparser (which was the problem). public String getXmlFromUrl(String url) { String xml = null; try { // defaultHttpClient DefaultHttpClient httpClient = new DefaultHttpClient(); HttpPost httpPost = new HttpPost(url); HttpResponse httpResponse = httpClient.execute(httpPost); HttpEntity httpEntity = httpResponse.getEntity(); xml = EntityUtils.toString(httpEntity); } catch (UnsupportedEncodingException e) { e.printStackTrace(); } catch (ClientProtocolException e) { e.printStackTrace(); } catch (IOException e) { e.printStackTrace(); } // return XML return xml; } public Document getDomElement(String xml) { Document doc = null; DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance(); try { DocumentBuilder db = dbf.newDocumentBuilder(); InputSource is = new InputSource(); is.setCharacterStream(new StringReader(xml)); doc = db.parse(is); } catch (ParserConfigurationException e) { Log.e("Error: ", e.getMessage()); return null; } catch (SAXException e) { Log.e("Error: ", e.getMessage()); return null; } catch (IOException e) { Log.e("Error: ", e.getMessage()); return null; } return doc; }

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  • unable to get data from iphone application

    - by user317192
    Hi, I have made a php web services that takes the username and password from iPhone application and saves the data in the users table. I call that web service from my button touchup event like this: NSLog(userName.text); NSLog(password.text); NSString * dataTOB=[userName.text stringByAppendingString:password.text]; NSLog(dataTOB); NSData * postData=[dataTOB dataUsingEncoding:NSUTF8StringEncoding allowLossyConversion:YES]; NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]]; NSLog(postLength); NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease]; NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"http://localhost:8888/write.php"]]; [request setURL:url]; [request setHTTPMethod:@"POST"]; [request setValue:postLength forHTTPHeaderField:@"Content-Length"]; [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"]; [request setHTTPBody:postData]; NSURLResponse *response; NSError *error; [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error]; if(error==nil) NSLog(@"Error is nil"); else NSLog(@"Error is not nil"); NSLog(@"success!"); I am getting nothing from Iphone side into the web service and I am unable to understand why this is happening. Although i debugged and found that everything is fine at the iphone application level but the call to the web service doesn't works as expected...... Please suggest..... Thanks Ashish

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  • Problem with my Jquery loading in my Codeigniter views

    - by sico87
    Hello, I am working on a one page website that allows the users to add and remove pages from there navigation as and when they would like too, the way it works is that if the click 'Blog' on the main nav a 'Blog' section should appear on the page, if they then click 'News' the 'News' section should also be visible, however the way I have started to implement this it seems I can only have one section at a time, can my code be adpated to allow multiple sections to shown on the main page. Here is my code for the page that has the main menu and the users selections on it. <!DOCTYPE html> <head> <title>Development Site</title> <link rel="stylesheet" href="/media/css/reset.css" media="screen"/> <link rel="stylesheet" href="/media/css/generic.css" media="screen"/> <script type="text/javascript" src="/media/javascript/jquery-ui/js/jquery.js"></script> <script type="text/javascript" src="/media/javascript/jquery-ui/development-bundle/ui/ui.core.js"></script> <script type="text/javascript" src="/media/javascript/jquery-ui/development-bundle/ui/ui.accordion.js"></script> <script type="text/javascript"> $(document).ready(function() { $('a.menuitem').click(function() { var link = $(this), url = link.attr("href"); $("#content_pane").load(url); return false; // prevent default link-behavior }); }); </script> </head> <body> <li><a class="menuitem" href="inspiration">Inspiration</a></li> <li><a class="menuitem" href="blog">Blog</a></li> <div id="content_pane"> </div> </body> </html>

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  • PHP: Best solution for links breaking in a mod_rewrite app

    - by psil
    I'm using mod rewrite to redirect all requests targeting non-existent files/directories to index.php?url=* This is surely the most common thing you do with mod_rewrite yet I have a problem: Naturally, if the page url is "mydomain.com/blog/view/1", the browser will look for images, stylesheets and relative links in the "virtual" directory "mydomain.com/blog/view/". Problem 1: Is using the base tag the best solution? I see that none of the PHP frameworks out there use the base tag, though. I'm currently having a regex replace all the relative links to point to the right path before output. Is that "okay"? Problem 2: It is possible that the server doesn't support mod_rewrite. However, all public files like images, stylesheets and the requests collector index.php are located in the directory /myapp/public. Normally mod_rewrite points all request to /public so it seems as if public was actually the root directory too all users. However if there is no mod_rewrite, I then have to point the users to /public from the root directory with a header() call. That means, however that all links are broken again because suddenly all images, etc. have to be called via /public/myimage.jpg Additional info: When there is no mod_rewrite the above request would look like this: mydomain.com/public/index.php/blog/view/1 What would be the best solutions for both problems?

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  • JSF works only with the .xhtml ending

    - by ThreeFingerMark
    Hello, i start with the programming of a JSF Website. At the moment all files have the .xhtml ending. When i go to http://localhost:8080/myProject/start.jsf everything is all right. But when i rename the file from start.xhtml to start.jsf i became a NoClassDefFound Error. What is my mistake? <servlet-name>Faces Servlet</servlet-name> <servlet-class>javax.faces.webapp.FacesServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>Faces Servlet</servlet-name> <url-pattern>*.jsf</url-pattern> </servlet-mapping> <context-param> <param-name>javax.faces.PROJECT_STAGE</param-name> <param-value>Development</param-value> </context-param>

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  • Using AJAX to get a specific DOM Element (using Javascript, not jQuery)

    - by Matt Frost
    How do you use AJAX (in plain JavaScript, NOT jQuery) to get a page (same domain) and display just a specific DOM Element? (Such as the DOM element marked with the id of "bodyContent"). I'm working with MediaWiki 1.18, so my methods have to be a little less conventional (I know that a version of jQuery can be enabled for MediaWiki, but I don't have access to do that so I need to look at other options). I apologize if the code is messy, but there's a reason I need to build it this way. I'm mostly interested in seeing what can be done with the Javascript. Here's the HTML code: <div class="mediaWiki-AJAX"><a href="http://www.domain.com/whatever"></a></div> Here's the Javascript I have so far: var AJAXs = document.getElementsByClassName('mediaWiki-AJAX'); if (AJAXs.length > 0) { for (var i = AJAXs.length - 1; i >= 0; i--) { var URL = AJAXs[i].getElementsByTagName('a')[0].href; xmlhttp = new XMLHttpRequest(); xmlhttp.onreadystatechange = function() { if (xmlhttp.readyState == 4 && xmlhttp.status == 200) { AJAXs[i].innerHTML = xmlhttp.responseText; } } xmlhttp.open('GET',URL,true); xmlhttp.send(); } }

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  • Google Reader API - feed/[FEEDURL]/ is coming back as Not found

    - by JustinXXVII
    There is one feed I'm subscribed to which always turns up as NOT FOUND when I try to use the API. I return an array of Dictionaries, containing 3 objects. The first in the list represents the user himself, like so: { FeedID = "user/MY_UNIQUE_NUMBER/state/com.google/reading-list"; Timestamp = 1273448807271463; Unread = 59; } The Unread count is very important. My client depends on downloading 59 items from Google before it refreshes. If a feed doesn't download properly, the count is off and the client won't update. An example of a working Feed is here: { FeedID = "feed/http://arstechnica.com/index.rssx"; Timestamp = 1273447158484528; Unread = 13; } The FeedID value combines with a specially formatted URL string and gives back a list of articles. The above example works fine. However, the following feed always returns NOT FOUND on Google, and if I paste the URL verbatim into a browser, it never turns up. See here: { FeedID = "feed/http://www.peopleofwalmart.com/?feed=rss2"; Timestamp = 1273424138183529; Unread = 6; } http://www.google.com/reader/api/0/stream/contents/feed/http://www.peopleofwalmart.com/?feed=rss2?ot=1&r=n&xt=user/-/state/com.google/read&n=6&ck=1273449028&client=testClient If you are at all proficient with the API, can you please help me? Like I said, since Google always says NOT FOUND when I search for that feed, my download count is off by N articles and won't update. I would rather not hack around it, honestly. Thanks!

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  • Apply a Quartz filter while saving PDF under Mac OS X 10.6.3

    - by olpa
    Using Mac OS X API, I'm trying to save a PDF file with a Quartz filter applied, just like it is possible from the "Save As" dialog in the Preview application. So far I've written the following code (using Python and pyObjC, but it isn't important for me): -- filter-pdf.py: begin from Foundation import * from Quartz import * import objc page_rect = CGRectMake (0, 0, 612, 792) fdict = NSDictionary.dictionaryWithContentsOfFile_("/System/Library/Filters/Blue \ Tone.qfilter") in_pdf = CGPDFDocumentCreateWithProvider(CGDataProviderCreateWithFilename ("test .pdf")) url = CFURLCreateWithFileSystemPath(None, "test_out.pdf", kCFURLPOSIXPathStyle, False) c = CGPDFContextCreateWithURL(url, page_rect, fdict) np = CGPDFDocumentGetNumberOfPages(in_pdf) for ip in range (1, np+1): page = CGPDFDocumentGetPage(in_pdf, ip) r = CGPDFPageGetBoxRect(page, kCGPDFMediaBox) CGContextBeginPage(c, r) CGContextDrawPDFPage(c, page) CGContextEndPage(c) -- filter-pdf.py: end Unfortunalte, the filter "Blue Tone" isn't applied, the output PDF looks exactly as the input PDF. Question: what I missed? How to apply a filter? Well, the documentation doesn't promise that such way of creating and using "fdict" should cause that the filter is applied. But I just rewritten (as far as I can) sample code /Developer/Examples/Quartz/Python/filter-pdf.py, which was distributed with older versions of Mac (meanwhile, this code doesn't work too): ----- filter-pdf-old.py: begin from CoreGraphics import * import sys, os, math, getopt, string def usage (): print ''' usage: python filter-pdf.py FILTER INPUT-PDF OUTPUT-PDF Apply a ColorSync Filter to a PDF document. ''' def main (): page_rect = CGRectMake (0, 0, 612, 792) try: opts,args = getopt.getopt (sys.argv[1:], '', []) except getopt.GetoptError: usage () sys.exit (1) if len (args) != 3: usage () sys.exit (1) filter = CGContextFilterCreateDictionary (args[0]) if not filter: print 'Unable to create context filter' sys.exit (1) pdf = CGPDFDocumentCreateWithProvider (CGDataProviderCreateWithFilename (args[1])) if not pdf: print 'Unable to open input file' sys.exit (1) c = CGPDFContextCreateWithFilename (args[2], page_rect, filter) if not c: print 'Unable to create output context' sys.exit (1) for p in range (1, pdf.getNumberOfPages () + 1): #r = pdf.getMediaBox (p) r = pdf.getPage(p).getBoxRect(p) c.beginPage (r) c.drawPDFDocument (r, pdf, p) c.endPage () c.finish () if __name__ == '__main__': main () ----- filter-pdf-old.py: end

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  • Subscribe through API .net C#

    - by Younes
    I have to submit subscription data to another website. I have got documentation on how to use this API however i'm not 100% sure of how to set this up. I do have all the information needed, like username / passwords etc. This is the API documentation: https://www.apiemail.net/api/documentation/?SID=4 How would my request / post / whatever look like in C# .net (vs 2008) when i'm trying to acces this API? This is what i have now, I think i'm not on the right track: public static string GArequestResponseHelper(string url, string token, string username, string password) { HttpWebRequest myRequest = (HttpWebRequest)WebRequest.Create(url); myRequest.Headers.Add("Username: " + username); myRequest.Headers.Add("Password: " + password); HttpWebResponse myResponse = (HttpWebResponse)myRequest.GetResponse(); Stream responseBody = myResponse.GetResponseStream(); Encoding encode = System.Text.Encoding.GetEncoding("utf-8"); StreamReader readStream = new StreamReader(responseBody, encode); //return string itself (easier to work with) return readStream.ReadToEnd(); Hope someone knows how to set this up properly. Thx!

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  • how to make the user add and delete in android

    - by user3678019
    i have 1 activity .. and in this activity i have 2 web view next to each other , i would like to add , ADD and Delete Button that can add one more web view next to the last web view , and the delete wish will delete any of the web view the user choose . and i want to make the user but it in the order he want like webview 1 first then webview 2 second how can i do this this is mu main.xml <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" xmlns:tools="http://schemas.android.com/tools" android:layout_width="match_parent" android:layout_height="match_parent" tools:context="test.zezo.test.Main$PlaceholderFragment" > <HorizontalScrollView android:id="@+id/horizontalScrollView2" android:layout_width="match_parent" android:layout_height="match_parent" android:layout_alignParentLeft="true" android:layout_alignParentRight="true" android:layout_alignParentTop="true" > <LinearLayout android:layout_width="match_parent" android:layout_height="match_parent" android:orientation="horizontal" > <WebView android:id="@+id/webView1" android:layout_width="350dp" android:layout_height="match_parent" android:layout_alignParentLeft="true" /> <WebView android:id="@+id/webView22" android:layout_width="350dp" android:layout_height="match_parent" android:layout_toRightOf="@+id/webView1" android:layout_alignParentLeft="true" /> </LinearLayout> </HorizontalScrollView> and this is a part of my main.java webView = (WebView) findViewById(R.id.webView1); String url = "http://google.com"; webView.getSettings().setJavaScriptEnabled(true); webView.loadUrl(url); webView.setWebChromeClient(new WebChromeClient()); webView.setWebViewClient(new WebViewClient()); WebView webView22 = (WebView)findViewById(R.id.webView22); webView22.getSettings().setJavaScriptEnabled(true); webView22.loadUrl("google.com); webView22.setWebChromeClient(new WebChromeClient()); webView22.setWebViewClient(new WebViewClient()); so how can i do the ADD and DELETE and re Order Buttons to it and one more thing it should be save so when he reopen the app it will be the same as after he add or delete or re order

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  • Symfony 1.2 to 2.3 migration

    - by Bonswouar
    I've got a pretty big Symfony 1.2 project to migrate. First, I modified my .htaccess so I can have some pages handled by Symfony 2. What I'd like to do, to make the migration smoother, is to be able to render some SF2 action/templates/methods/... inside SF1. I added the autoloader to the SF1 app, so I can access to twig rendering methods and other stuff. But how can I call a SF2 action ? For example, if I want to migrate only the footer first, I would also need some php methods, not only rendering. That was previously in SF1 component, where should it be now ? If you've got any suggestion about the way of migrating, don't hesitate ! EDIT 1 : Apparently, the only way to do something like that is to render a full twig template, and/or in this template call some other partial twig templates with render(url, params). Here is my SF1 code to be able to render twig templates : public static function getTwig() { require_once __DIR__.'SF2_PATH/vendor/twig/extensions/lib/Twig/Extensions/Autoloader.php'; Twig_Autoloader::register(); $loader = new Twig_Loader_Filesystem( __DIR__.'SF2_PATH/sf2/src/VENDOR/BUNDLE/'); $twig = new Twig_Environment($loader, array( 'cache' => __DIR__.'SF2_PATH/sf2/app/cache/dev/twig', )); return $twig; } And so : $twig->loadTemplate('header.html.twig'); EDIT 2 : That doesn't seem to work, if in a twig template I try to render an other one with {{render(controller('BUNDLE:CONTROLER:ACTION', {})) }} for example Twig_Error : The function "controller" does not exist. And if I try to render the url Unknown tag name "render". I guess Symfony 2 twig functionalities are not loaded, how can I do that ? EDIT 3 : Ok, now I can do it, but I've got the following message... Twig_Error_Runtime An exception has been thrown during the rendering of a template ("Rendering a fragment can only be done when handling a master Request.") in ...

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  • multiple ajax requests with jquery

    - by Emil
    I got problems with the async nature of Javascript / JQuery. Lets say the following (no latency is counted for, in order to not make it so troublesome); I got three buttons (A, B, C) on a page, each of the buttons adds an item into a shopping cart with one ajax-request each. If I put an intentional delay of 5 seconds in the serverside script (PHP) and pushes the buttons with 1 second apart, I want the result to be the following: Request A, 5 seconds Request B, 6 seconds Request C, 7 seconds However, the result is like this Request A, 5 seconds Request B, 10 seconds Request C, 15 seconds This have to mean that the requests are queued and not run simultaneously, right? Isnt this opposite to what async is? I also tried to add a random get-parameter to the url in order to force some uniqueness to the request, no luck though. I did read a little about this. If you avoid using the same "request object (?)" this problem wont occure. Is it possible to force this behaviour in JQuery? This is the code that I am using $.ajax( { url : strAjaxUrl + '?random=' + Math.floor(Math.random()*9999999999), data : 'ajax=add-to-cart&product=' + product, type : 'GET', success : function(responseData) { // update ui }, error : function(responseData) { // show error } }); I also tried both GET and POST, no difference. I want the requests to be sent right when the button is clicked, not when the previous request is finnished. I want the requests to be run simultaneously, not in a queue.

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  • Android App - disappearance of app GUI

    - by Radek Šimko
    I'm trying to create a simple app, whose main task is to open the browser on defined URL. I've created first Activity: public class MyActivity extends Activity { @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); //setContentView(R.layout.main); Intent myIntent = new Intent(Intent.ACTION_VIEW, Uri.parse("http://my.url.tld")); startActivity(myIntent); } Here's my AndroidManifest.xml: <manifest ...> <application ...> <activity android:name=".MyActivity" ...> <intent-filter> <action android:name="android.intent.action.MAIN"/> <category android:name="android.intent.category.LAUNCHER"/> </intent-filter> </activity> This code is fully functional, but before it opens the browser, it displays a black background - blank app GUI. I didn't figured out, how to go directly do the browser (without displaying the GUI). Anyone knows?

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  • css menu <ul><li> dinamically centered or width of buttons that covers the whole page

    - by Tony Stark
    I am building a home page for my minecraft server. Probably in the following 4-6 months I will opend my second and this is why I am in trouble. My first site is 1000 pixel wide, and the second will be 1200. First big difference. My menus are dinamically generated by my php code. It checks on my databases if there is another button or it is over. These buttons can be added or removed directly online. Another big issue is the browser compatibility. In a survey I did on our previous server I had a lot of users using: chrome, internet explorer, safari and firefox. That means that I must find a solution that is compatible with most browsers. What do I have to do? I came up with this CSS, which is touch compatible, it allows menus to be swapped to the left and it is enough to set 1 parameter to fix it for every page width. Sadly it is left aligned. body, nav, ul, li, a {margin: 0; padding: 0;} body {font-family: Verdana,"Helvetica Neue", Helvetica, Arial, sans-serif; } a {text-decoration: none;} .container { max-width: 900px; margin: 0px auto 0px auto; } .toggleMenu { display: none; background: #666; padding: 10px 15px; color: #999999; } .nav { border: 1px solid #424242; background-color: #121212; filter: progid:DXImageTransform.Microsoft.gradient(startColorstr='#686868', endColorstr='#121212'); background-image: -moz-linear-gradient(#686868, #121212); background-image: -webkit-gradient(linear, left top, left bottom, from(#686868), to(#121212)); background-image: -webkit-linear-gradient(#686868, #121212); background-image: -o-linear-gradient(#686868, #121212); background-image: -ms-linear-gradient(#686868, #121212); background-image: linear-gradient(#686868, #121212); -moz-box-shadow: 0 1px 1px #777, 0 1px 0 #666 inset; -webkit-box-shadow: 0 1px 1px #777, 0 1px 0 #666 inset; box-shadow: 0 1px 1px #777, 0 1px 0 #666 inset; list-style: none; *zoom: 1; position: relative; } .nav:before,.nav:after { content: " "; display: table; } .nav:after { clear: both; } .nav ul { list-style: none; width: 11em; z-index: 1; background-color: #121212; -moz-box-shadow: 0 -1px rgba(255,255,255,.3); -webkit-box-shadow: 0 -1px 0 rgba(255,255,255,.3); box-shadow: 0 -1px 0 rgba(255,255,255,.3); } .nav a { padding: 10px 15px; color:#999999; text-transform: uppercase; font: bold 11px Arial, Helvetica; text-decoration: none; text-shadow: 0 1px 0 #000; *zoom: 1; } .nav a:hover{ color:#000000; background-color: #B2B2B2; filter: progid:DXImageTransform.Microsoft.gradient(startColorstr='#D3D3D3', endColorstr='#B2B2B2'); background-image: -moz-linear-gradient(#D3D3D3, #B2B2B2); background-image: -webkit-gradient(linear, left top, left bottom, from(#D3D3D3), to(#B2B2B2)); background-image: -webkit-linear-gradient(#D3D3D3, #B2B2B2); background-image: -o-linear-gradient(#D3D3D3, #B2B2B2); background-image: -ms-linear-gradient(#D3D3D3, #B2B2B2); background-image: linear-gradient(#D3D3D3, #B2B2B2); } /*Delimitazione di ogni tab | HOME | */ .nav li { position: relative; border-right: 1px solid #424242; -moz-box-shadow: 1px 0 0 #686868; -webkit-box-shadow: 1px 0 0 #686868; box-shadow: 1px 0 0 #686868; } .nav > li { float: left; border-top: 1px solid #424242; z-index: 200; } .nav > li > .parent { background-image: url("../downArrow.png"); background-repeat: no-repeat; background-position: center right; } .nav > li li > .parent { background-image: url("../rightArrow.png"); background-repeat: no-repeat; background-position: center right; } .nav > li > a { display: block; } .nav li ul { position: absolute; left: -9999px; z-index: 100; } /* freccetta che indica un sottomenu nell'ultimo tab */ .nav > li:last-child li > .parent{ background-image: url("../leftArrow.png"); background-repeat: no-repeat; background-position: left; } /*flip subsubmenu*/ .nav li.last.hover > ul { left:auto; right: 0; } .nav > li.hover > ul { left: 0; } .nav li li.hover > ul { left: 100%; top: 0; } /* Spostare il 2^ sottomenu a sinistra */ .nav li.last li.hover ul { left:auto; right: 100%; top:0; } .nav li li a { display: block; background-color: #686868; -moz-box-shadow: 0 -1px rgba(255,255,255,.3); -webkit-box-shadow: 0 -1px 0 rgba(255,255,255,.3); box-shadow: 0 -1px 0 rgba(255,255,255,.3); z-index:100; border-top: 1px solid #686868; } .nav li li li a { background-color: #686868; -moz-box-shadow: 0 -1px rgba(255,255,255,.3); -webkit-box-shadow: 0 -1px 0 rgba(255,255,255,.3); box-shadow: 0 -1px 0 rgba(255,255,255,.3); z-index:200; border-top: 1px solid #686868; } .nav li li li li a { display: block; background-color: #686868; -moz-box-shadow: 0 -1px rgba(255,255,255,.3); -webkit-box-shadow: 0 -1px 0 rgba(255,255,255,.3); box-shadow: 0 -1px 0 rgba(255,255,255,.3); z-index:300; border-top: 1px solid #686868; } .nav li li li li a { background-color: #686868; -moz-box-shadow: 0 -1px rgba(255,255,255,.3); -webkit-box-shadow: 0 -1px 0 rgba(255,255,255,.3); box-shadow: 0 -1px 0 rgba(255,255,255,.3); z-index:400; border-top: 1px solid #686868; } @media screen and (max-width: 768px) { .active { display: block; } .nav > li { float: none; } .nav > li > .parent { background-position: 95% 50%; } .nav li li .parent { background-image: url("../downArrow.png"); background-repeat: no-repeat; background-position: 95% 50%; } .nav ul { display: block; width: 100%; } .nav > li.hover > ul , .nav li li.hover ul { position: static; } } My girlfriend (who adapted this code) is really busy for school and cannot help me. Leaving the borders on the whole square (page width), is it possible to make buttons cover the page width dinamically? Or is it possible to center the buttons? Thank you very much!

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  • How can I properly handle 404s in ASP.NET MVC?

    - by Brian
    I am just getting started on ASP.NET MVC so bear with me. I've searched around this site and various others and have seen a few implementations of this. EDIT: I forgot to mention I am using RC2 Using URL Routing: routes.MapRoute( "Error", "{*url}", new { controller = "Errors", action = "NotFound" } //404s ); The above seems to take care of requests like this (assuming default route tables setup by initial MVC project): "/blah/blah/blah/blah" Overriding HandleUnknownAction() in the controller itself: //404s - handle here (bad action requested protected override void HandleUnknownAction(string actionName) { ViewData["actionName"] = actionName; View("NotFound").ExecuteResult(this.ControllerContext); } However the previous strategies do not handle a request to a Bad/Unknown controller. For example, I do not have a "/IDoNotExist", if I request this I get the generic 404 page from the web server and not my 404 if I use routing + override. So finally, my question is: Is there any way to catch this type of request using a route or something else in the MVC framework itself? OR should I just default to using Web.Config customErrors as my 404 handler and forget all this? I assume if I go with customErrors I'll have to store the generic 404 page outside of /Views due to the Web.Config restrictions on direct access. Anyway any best practices or guidance is appreciated.

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  • Rails - strip xml import from whitespace and line break

    - by val_to_many
    Hey folks, I am stuck with something quite simple but really annoying: I have an xml file with one node, where the content includes line breaks and whitspaces. Sadly I can't change the xml. <?xml version="1.0" encoding="utf-8" ?> <ProductFeed> ACME Ltd. Fooproduct Foo Root :: Bar Category I get to the node and can read from it without trouble: url = "http://feeds.somefeed/feed.xml.gz" @source = open((url), :http_basic_authentication=>["USER", "PW"]) @gz = Zlib::GzipReader.new(@source) @result = @gz.read @doc = Nokogiri::XML(@result) @doc.xpath("/ProductFeed/Vendors/Vendor").each do |manuf| vendor = manuf.css("Name").first.text manuf.xpath("//child::Product").each do |product| product_name = product.css("Name").text foocat = product.css("Category").text puts "#{vendor} ---- #{product_name} ---- #{foocat} " end end This results in: ACME Ltd. ---- Fooproduct ---- Foo Root :: Bar Category Obviously there are line breaks and tab stops or spaces in the string returned by product.css("Category").text. Does anyone know how to strip the result from linebreaks and taps or spaces right here? Alternatively I could do that in the next step, where I do a find on 'foocat' like barcat = Category.find_by_foocat(foocat) Thanks for helping! Val

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  • Django and json request

    - by Hulk
    In a template i have the following code <script> var url="/mypjt/my_timer" $.post(url, paramarr, function callbackHandler(dict) { alert('got response back'); if (dict.flag == 2) { alert('1'); $.jGrowl("Data could not be saved"); } else if(dict.ret_status == 1) { alert('2'); $.jGrowl("Data saved successfully"); window.location = "/mypjt/display/" + dict.rid; } }, "json" ); </script> In views i have the following code, def my_timer(request): dict={} try: a= timer.objects.get(pk=1) dict({'flag':1}) return HttpResponse(simplejson.dumps(dict), mimetype='application/javascript') except: dict({'flag':1}) return HttpResponse(simplejson.dumps(dict), mimetype='application/javascript') My question is since we are making a json request and in the try block ,after setting the flag ,cant we return a page directly as return render_to_response('mypjt/display.html',context_instance=RequestContext(request,{'dict': dict})) instead of sending the response, because on success again in the html page we redirect the code Also if there is a exception then only can we return the json request. My only concern is that the interaction between client and server should be minimal. Thanks..

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  • Losing URI segments when paginating with CodeIgniter

    - by Danny Herran
    I have a /payments interface where the user should be able to filter via price range, bank, and other stuff. Those filters are standard select boxes. When I submit the filter form, all the post data goes to another method called payments/search. That method performs the validation, saves the post values into a session flashdata and redirects the user back to /payments passing the flashdata name via URL. So my standard pagination links with no filters are exactly like this: payments/index/20/ payments/index/40/ payments/index/60/ And if you submit the filter form, the returning URL is: payments/index/0/b48c7cbd5489129a337b0a24f830fd93 This works just great. If I change the zero for something else, it paginates just fine. The only issue however is that the << 1 2 3 4 page links wont keep the hash after the pagination offset. CodeIgniter is generating the page links ignoring that additional uri segment. My uri_segment config is already set to 3: $config['uri_segment'] = 3; I cannot set the page offset to 4 because that hash may or may not exists. Any ideas of how can I solve this? Is it mandatory for CI to have the offset as the last segment in the uri? Maybe I am trying an incorrect approach, so I am all ears. Thank you folks.

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  • NSString writeToFile operation couldn't be completed

    - by Chonch
    Hey, I have an xml file in my application bundle. I want to copy it to the documents folder at installation and then, every time the app launches, get the newest version of this file from the Internet. I use this code: // Check if the file exists in the documents folder NSString *documentsPath = [NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES) objectAtIndex:0]; if (![[NSFileManager defaultManager] fileExistsAtPath:[documentsPath stringByAppendingPathComponent:@"fileName.xml"]]) // If not, copy it there (from the bundle) [[NSFileManager defaultManager] copyItemAtPath:[[[NSBundle mainBundle] resourcePath] stringByAppendingPathComponent:@"OriginalFile.xml"] toPath:[documentsPath stringByAppendingPathComponent:@"fileName.xml"] error:nil]; // Get the newest version of the file from the server NSURL *url=[[NSURL alloc] initWithString:@"http://www.sitename.com/webservice.asmx/webserviceName"]; NSString *results = [[NSString alloc] initWithContentsOfURL:url]; // Replace the current version with the newest one, only if it is valid if (results != nil) [results writeToFile:[documentsPath stringByAppendingPathComponent:@"fileName.xml"] atomically:NO encoding:NSStringEncodingConversionAllowLossy error:nil]; The problem is that the writeToFile command always returns NO and the file's contents remain identical to the original file I included in my app bundle. I checked the value of results and it's correct. I also made sure that the app does perform the writeToString command, but still, it always returns NO. Can anybody tell me what I'm doing wrong? Thanks,

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