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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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  • Connection

    - by pepersview
    Hello, I would like to ask you about NSURLConnection in objective-c for iPhone. I have one app that needs to connect to one webservice to receive data (about YouTube videos), Then I have all the things that I need to connect (Similar to Hello_Soap sample code in the web). But now, my problem is that I create a class (inherits from NSObject) named Connection and I have implemented the methods: didReceiveResponse, didReceiveData, didFailWithError and connectionDidFinishLoading. Also the method: -(void)Connect:(NSString *) soapMessage{ NSLog(soapMessage); NSURL *url = [NSURL URLWithString:@"http://....."]; NSMutableURLRequest *theRequest = [NSMutableURLRequest requestWithURL:url]; NSString *msgLength = [NSString stringWithFormat:@"%d", [soapMessage length]]; [theRequest addValue: @"text/xml; charset=utf-8" forHTTPHeaderField:@"Content-Type"]; [theRequest addValue: msgLength forHTTPHeaderField:@"Content-Length"]; [theRequest setHTTPMethod:@"POST"]; [theRequest setHTTPBody: [soapMessage dataUsingEncoding:NSUTF8StringEncoding]]; NSURLConnection *theConnection = [[NSURLConnection alloc] initWithRequest:theRequest delegate:self]; if( theConnection ) { webData = [[NSMutableData data] retain]; } else { NSLog(@"theConnection is NULL"); } } But when from my AppDelegate I create one Connection object: Connection * connect = [[Connection alloc] Init:num]; //It's only one param to test. [connect Connect:method.soapMessage]; And I call this method, when this finishes, it doesn't continue calling didReceiveResponse, didReceiveData, didFailWithError or connectionDidFinishLoading. I'm trying to do this but I can't for the moment. The thing I would like to do is to be able to call this class "Connection" each time that I want to receive data (after that to be parsed and displayed in UITableViews). Thank you.

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  • How do i make an installer instead of a crash (.NET)

    - by acidzombie24
    My situation is Using .NET 3.5 Using SDL.NET Need to make a friendly installer or warning system. Chances are the user will be on XP (.NET 1.1). If possible can i do something to let the user know he needs to update to 3.5? Maybe have a yes/no dialog which downloads and install the .NET runtimes for him? Now how do i detect if the user has sdl.net installed (chances are its in program files/sdldotnet) and let them know they need sdl.net runtime and have a yes/no dialog that brings them to http://sourceforge.net/projects/cs-sdl/files/ The problem i have mostly is how to make the app not outright crash and how to download 3.5 .NET runtime if it is possible.

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  • Best toolkit for new Web application (odd requirements)

    - by ireadit
    I need to write a new application, and have no experience with any new technology, framework, or language. Here are the requirements: HTML front end (best if it's cross-browser friendly) Web deployable, but also ideally want to be able to install as standalone on a desktop SQL Server database Ideally, would like to use a good (and easy) AJAX toolkit with widgets Ideally, would like to be able to write in ASP.Net but later (or concurrently) also write in Java. This is a big concern, as I would like to not have to rewrite the whole thing. Is there a toolkit I can use that makes this cross-platform requirement easier? All suggestions and comments are much appreciated. Thank you.

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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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  • ASP.NET MVC: Returning a view with querystring intact

    - by ajbeaven
    I'm creating a messaging web app in ASP.NET and are having some problems when displaying an error message to the user if they go to send a message and there is something wrong. A user can look through profiles of people and then click, 'send a message'. The following action is called (url is /message/create?to=username) and shows them a page where they can enter their message and send it: public ActionResult Create(string to) { ViewData["recipientUsername"] = to; return View(); } On the page that is displayed, the username is entered in to a hidden input field. When the user clicks 'send': [AcceptVerbs(HttpVerbs.Post)] public ActionResult Create(FormCollection collection, string message) { try { //do message stuff that errors out } catch { ModelState.AddModelErrors(message.GetRuleViolations()); //adding errors to modelstate } return View(); } So now the error message is displayed to the user fine, however the url is changed in that it no longer has the querystring (/message/create). Again, this would be fine except that when the user clicks the refresh button, the page errors out as the Create action no longer has the 'to' parameter. So I'm guessing that I need to maintain my querystring somehow. Is there any way to do this or do I need to use a different method altogether?

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  • Jquery class changing works, but doesn't effect another function like it should

    - by Qwibble
    So I have content boxes that close and expand when you click an arrow. The content box has two classes for telling whether it is open or closed (.box_open, .box_closed). A hover function is assigned to box_open so when it is open and you hover over the header, the arrow appears. However, I don't want this to happen when the box is closed, as I want to arrow to remain visible when the box is closed. When the box closes, the box_open class is removed, but the function assigned to that class still works. Here's the jquery code for the two functions. You can also see them in the head of the demo below. // Display Arrow on Box Header Hover $(".box_open").hover( function () { $(this).find('a').show(); }, function () { $(this).find('a').hide(); } ); // Open and Close Boxes: $(".box_header a").click( function () { $(this).parent().next('.box_border').stop().toggle(); $(this).parent().toggleClass("box_open"); $(this).parent().toggleClass("box_closed"); return false; } ); Can anyone take a look at what the problem is? Here's the demo url: Demo Url

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  • Another answer to the CAPTCHA problem?

    - by Xeoncross
    Most sites at least employ server access log checking and banning along with some kind of bot prevention measure like a CAPTCHA (those messed-up text images). The problem with CAPTCHAs is that they poss a threat to the user experience. Luckily they now come with user friendly features like refresh and audio versions. Anyway, like linux vs windows, it isn't worth the time of a spammer to customize and/or build a script to handle a custom CAPTCHA example that only pertains to one site. Therefore, I was wondering if there might be better ways to handle the whole CAPTCHA thing. In A Better CAPTCHA Peter Bromberg mentions that one way would be to convert the image to HTML and display it embedded in the page. On http://shiflett.org/ Chris simply asks users to type his name into an input. Examples like this are ways to simplifying the CAPTCHA experience while decreasing the value for spammers. Does anyone know of more good examples I could use or see any problem with the embedded image idea?

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  • Simplest PHP Routing framework .. ?

    - by David
    I'm looking for the simplest implementation of a routing framework in PHP, in a typical PHP environment (Running on Apache, or maybe nginx) .. It's the implementation itself I'm mostly interested in, and how you'd accomplish it. I'm thinking it should handle URL's, with the minimal rewriting possible, (is it really a good idea, to have the same entrypoint for all dynamic requests?!), and it should not mess with the querystring, so I should still be able to fetch GET params with $_GET['var'] as you'd usually do.. So far I have only come across .htaccess solutions that puts everything through an index.php, which is sort of okay. Not sure if there are other ways of doing it. How would you "attach" what URL's fit to what controllers, and the relation between them? I've seen different styles. One huge array, with regular expressions and other stuff to contain the mapping. The one I think I like the best is where each controller declares what map it has, and thereby, you won't have one huge "global" map, but a lot of small ones, each neatly separated. So you'd have something like: class Root { public $map = array( 'startpage' => 'ControllerStartPage' ); } class ControllerStartPage { public $map = array( 'welcome' => 'WelcomeControllerPage' ); } // Etc ... Where: 'http://myapp/' // maps to the Root class 'http://myapp/startpage' // maps to the ControllerStartPage class 'http://myapp/startpage/welcome' // maps to the WelcomeControllerPage class 'http://myapp/startpage/?hello=world' // Should of course have $_GET['hello'] == 'world' What do you think? Do you use anything yourself, or have any ideas? I'm not interested in huge frameworks already solving this problem, but the smallest possible implementation you could think of. I'm having a hard time coming up with a solution satisfying enough, for my own taste. There must be something pleasing out there that handles a sane bootstrapping process of a PHP application without trying to pull a big magic hat over your head, and force you to use "their way", or the highway! ;)

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  • Experts help needed on libcurl programming in sending HTTP HEAD Request.

    - by Mani
    Hi all, I need clarifications on using libcurl for the following: I need to send an http HEAD request shown as below :: HEAD /mshare/3/30002:12:primary/stream_xNKNVH.mpeg HTTP/1.1 Host: 192.168.70.1:8080 Accept: */* User-Agent: Kreatel_IP-STB getcontentFeatures.dlna.org: 1 The code I wrote (shown below) , sends the HEAD Request in slightly different way: curl_global_init(CURL_GLOBAL_ALL); CURL* ctx = NULL; const char *url = "http://192.168.70.1:8080/mshare/3/30002:12:primary/stream_xNKNVH.mpeg" ; char *returnString; struct curl_slist *headers = NULL; ctx = curl_easy_init(); headers = curl_slist_append(headers,"Accept: /"); headers = curl_slist_append(headers,"User-Agent: Kreatel_IP-STB");\ headers = curl_slist_append(headers,"getcontentFeatures.dlna.org: 1"); headers = curl_slist_append(headers,"Pragma:"); headers = curl_slist_append(headers,"Proxy-Connection:"); curl_easy_setopt(ctx,CURLOPT_HTTPHEADER , headers ); curl_easy_setopt(ctx,CURLOPT_NOBODY ,1 ); curl_easy_setopt(ctx,CURLOPT_VERBOSE, 1); curl_easy_setopt(ctx,CURLOPT_URL,url ); curl_easy_setopt(ctx,CURLOPT_NOPROGRESS ,1 ); curl_easy_perform(ctx); curl_easy_cleanup(ctx); curl_global_cleanup(); The code shown above sends the HEAD Request in slightly different form (shown below) HEAD http://192.168.70.1:8080/mshare/3/30002:12:primary/stream_xNKNVH.mpeg HTTP/1.1 Host: 192.168.70.1:8080 Proxy-Connection: Keep-Alive Accept: */* User-Agent: Kreatel_IP-STB getcontentFeatures.dlna.org: 1 Can any one , share the appropriate code ?

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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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  • Pulling specific entries from RSS feed [PHP]

    - by n0s
    So, I have an RSS feed with variations of each item. What I want to do is just get entries that contain a specific section of text. For example: <item> <title>RADIO SHOW - CF64K - 05-20-10 + WRAPUP </title> <link>http://linktoradioshow.com</link> <comments>Radio show from 05-20-10</comments> <pubDate>Thu, 20 May 2010 19:12:12 +0200</pubDate> <category domain="http://linktoradioshow.com/browse/199">Audio / Other</category> <dc:creator>n0s</dc:creator> <guid>http://otherlinktoradioshow.com/</guid> <enclosure url="http://linktoradioshow.com/" length="13005" /> </item> <item> <title>RADIO SHOW - CF128K - 05-20-10 + WRAPUP </title> <link>http://linktoradioshow.com</link> <comments>Radio show from 05-20-10</comments> <pubDate>Thu, 20 May 2010 19:12:12 +0200</pubDate> <category domain="http://linktoradioshow.com/browse/199">Audio / Other</category> <dc:creator>n0s</dc:creator> <guid>http://otherlinktoradioshow.com/</guid> <enclosure url="http://linktoradioshow.com/" length="13005" /> </item> I only want to display the results that contain the string CF64K. While it's probably really simple regex, I can't seem to wrap my head around getting it right. I always get seem to only be able to display the string 'CF64K', and not the stuff that surrounds it. Thanks in advance.

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  • NoClassDefFoundError: HttpClient 4 (APACHE)

    - by Pujan Srivastava
    Hello All, I am using HC APACHE. I have added both httpcore-4.0.1.jar and httpclient-4.0.1.jar in the classpath of netbeans. I am getting error: java.lang.NoClassDefFoundError: org/apache/http/impl/client/DefaultHttpClient My Code is as follows. Please help. import org.apache.http.client.HttpClient; import org.apache.http.client.ResponseHandler; import org.apache.http.client.methods.HttpGet; import org.apache.http.impl.client.BasicResponseHandler; import org.apache.http.impl.client.DefaultHttpClient; public class HttpClientManager { public HttpClient httpclient; public HttpClientManager() { this.init(); } public void init() { try { httpclient = new DefaultHttpClient(); } catch (Exception e) { e.printStackTrace(); } } public void getCourseList() { String url = "http://exnet.in.th/api.php?username=demoinst&ha=2b62560&type=instructor"; HttpGet httpget = new HttpGet(url); ResponseHandler<String> responseHandler = new BasicResponseHandler(); try { String responseBody = httpclient.execute(httpget, responseHandler); System.out.println(responseBody); } catch (Exception e) { } } }

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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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  • Show progress during a long Ajax call

    - by kousen
    I have a simple web site (http://www.kousenit.com/twitterfollowervalue) that computes a quantity based on a person's Twitter followers. Since the Twitter API only returns followers 100 at a time, a complete process may involve lots of calls. At the moment, I have an Ajax call to a method that runs the Twitter loop. The method looks like (Groovy): def updateFollowers() { def slurper = new XmlSlurper() followers = [] def next = -1 while (next) { def url = "http://api.twitter.com/1/statuses/followers.xml?id=$id&cursor=$next" def response = slurper.parse(url) response.users.user.each { u -> followers << new TwitterUser(... process data ...) } next = response.next_cursor.toBigInteger() } return followers } This is invoked from a controller called renderTTFV.groovy. I call the controller via an Ajax call using the prototype library: On my web page, in the header section (JavaScript): function displayTTFV() { new Ajax.Updater('ttfv','renderTTFV.groovy', {}); } and there's a div in the body of the page that updates when the call is complete. Everything is working, but the updateFollowers method can take a fair amount of time. Is there some way I could return a progress value? For example, I'd like to update the web page on every iteration. I know ahead of time how many iterations there will be. I just can't figure out a way to update the page in the middle of that loop. Any suggestions would be appreciated.

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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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  • how to dispaly image in grid view reading imageUrl from xml using sax parser in android

    - by Pramod kuamr
    thanks for answer but i am able to read xml file from url but i need if in xml imageUrl is there so show in grid view ..this is my xml file and read URL <?xml version="1.0" encoding="UTF-8"?> <channels> <channel> <name>ndtv</name> <logo>http://a3.twimg.com/profile_images/670625317/aam-logo--twitter.png</logo> <description>this is a news Channel</description> <rssfeed>ndtv.com</rssfeed> </channel> <channel> <name>star news</name> <logo>http://a3.twimg.com/profile_images/740897825/AndroidCast-350_normal.png</logo> <description>this is a newsChannel</description> <rssfeed>starnews.com</rssfeed> </channel> </channels>

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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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  • 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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  • 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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  • PHP & cUrl - POST problems in while loop.

    - by Max Hoff
    I have a while loop, with cUrl inside. (I can't use curl_multi for various reasons.) The problem is that cUrl seems to save the data it POSTS after each loop traversal. For instance, if parameter X is One the first loop through, and if it's Two the second loop through, cUrl posts: "One,Two". It should just POST "Two".(This is despite closing and unsetting the curl handle.) Here's a simplified version of the code (with unecessary info stripped out): while(true){ // code to form the $url. this code is local to the loop. so the variables should be "erased" and made new for each loop through. $ch = curl_init(); $userAgent = 'Googlebot/2.1 (http://www.googlebot.com/bot.html)'; curl_setopt($ch,CURLOPT_USERAGENT, $userAgent); curl_setopt($ch,CURLOPT_URL,$url); curl_setopt($ch,CURLOPT_POST,count($fields)); curl_setopt($ch,CURLOPT_POSTFIELDS,$fields_string); curl_setopt($ch, CURLOPT_FAILONERROR, true); curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true); curl_setopt($ch, CURLOPT_AUTOREFERER, true); curl_setopt($ch, CURLOPT_RETURNTRANSFER,true); curl_setopt($ch, CURLOPT_TIMEOUT, 10); $html = curl_exec($ch); curl_close($ch); unset($ch); $dom = new DOMDocument(); @$dom->loadHTML($html); $xpath = new DOMXPath($dom); $resultTable = $xpath->evaluate("/html/body//table"); // $resultTable is 20 the first time through the loop, and 0 everytime thereafter becauset he POSTing doesn't work right with the "saved" parameters. What am I doing wrong here?

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  • parser 2.1 and 2.2

    - by yaniv
    hi i using the follwing Code to retrive XML element text using getElementsByTagName this code success in 2.2 and Failed in 2.1 any idea ? URL metafeedUrl = new URL("http://x..../Y.xml") URLConnection connection ; connection= metafeedUrl.openConnection(); HttpURLConnection httpConnection = (HttpURLConnection)connection ; int resposnseCode= httpConnection.getResponseCode() ; if (resposnseCode == HttpURLConnection.HTTP_OK) { InputStream in = httpConnection.getInputStream(); DocumentBuilderFactory dbf ; dbf = DocumentBuilderFactory.newInstance(); DocumentBuilder db = dbf.newDocumentBuilder(); // Parse the Earthquakes entry Document dom = db.parse(in); Element docEle = dom.getDocumentElement(); //ArrayList<Album> Albums = new ArrayList<Album>(); /* Returns a NodeList of all descendant Elements with a given tag name, in document order.*/ NodeList nl = docEle.getElementsByTagName("entry"); if (nl!=null && nl.getLength() > 0) { for (int i = 0; i < nl.getLength(); i++) { Element entry = (Element)nl.item(i); /* Now on every property in Entry **/ Element title =(Element)entry.getElementsByTagName("title").item(0); *Here i Get an Error* String album_Title = title.getTextContent(); Element id =(Element)entry.getElementsByTagName("id").item(0); String album_id = id.getTextContent(); //

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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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  • UISplitView's DetailView not changing its content

    - by Knodel
    I have an iPad app with a UISplitView. In the Root View I have a two level UITableView (it takes its content from a plist). In the Detail View I have a UIWebView. - (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath { //Get the dictionary of the selected data source. NSDictionary *dictionary = [self.tableDataSource objectAtIndex:indexPath.row]; //Get the children of the present item. NSArray *Children = [dictionary objectForKey:@"Children"]; RootViewController *rvController = [[RootViewController alloc] init]; rvController.CurrentTitle = [dictionary objectForKey:@"Title"]; rvController.CurrentLevel = +1; //Push the new table view on the stack [self.navigationController pushViewController:rvController animated:YES]; rvController.tableDataSource = Children; NSString *name = [NSString stringWithFormat:@"%@.rtfd.zip", rvController.CurrentTitle ]; NSString* resourcePath = [[NSBundle mainBundle] resourcePath]; NSString* sourceFilePath = [resourcePath stringByAppendingPathComponent:name]; NSURL* url = [NSURL fileURLWithPath:sourceFilePath isDirectory:NO]; NSURLRequest* request = [NSURLRequest requestWithURL:url]; [detailViewController.webView loadRequest:request]; } That's how I want the UIWebView's content to be changed. If I name the .rtfd.zip file the same as a first level cell of the UITableView, it works, UIWebView's content changes. But this doesn't work with the second level UITableView. What am I doing wrong? Thanks in advance!

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  • Yii user rights extension

    - by jailed abroad
    i have installed the yii rights extension and here is my code after installation, database tables are created after installation. 'modules'=>array( // uncomment the following to enable the Gii tool 'rights'=>array( 'superuserName'=>'Admin', // Name of the role with super user privileges. 'authenticatedName'=>'Authenticated', //// Name of the authenticated user role. 'userIdColumn'=>'id',// Name of the user id column in the database. 'userNameColumn'=>'username', // Name of the user name column in the database. 'enableBizRule'=>true, // Whether to enable authorization item business rules. 'enableBizRuleData'=>false, //Whether to enable data for business rules. 'displayDescription'=>true, // Whether to use item description instead of name. ' // Key to use for setting success flash messages. 'flashErrorKey'=>'RightsError', / Key to use for setting error flash messages. // 'install'=>true, // Whether to install rights. 'baseUrl'=>'/rights', // Base URL for Rights. Change if module is nested. 'layout'=>'rights.views.layouts.main', // Layout to use for displaying Rights. 'appLayout'=>'application.views.layouts.main', //Application layout. 'cssFile'=>'rights.css', // Style sheet file to use for Rights. ' 'install'=>false, // Whether to enable installer. 'debug'=>false, ), 'gii'=>array( 'class'=>'system.gii.GiiModule', 'password'=>'1234', // If removed, Gii defaults to localhost only. Edit carefully to taste. 'ipFilters'=>array('127.0.0.1','::1'), ), ), But when i type url http://localhost/rightsTest/index.php/rights then it says There must be at least one superuser! I have tried many things but unable to find answer. Thanks for your help.

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