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  • My SQL query is only returning I need the parent aswell

    - by sico87
    My sql query is only returning the children of the parent I need it to return the parent as well, public function getNav($cat,$subcat){ //gets all sub categories for a specific category if(!$this->checkValue($cat)) return false; //checks data $query = false; if($cat=='NULL'){ $sql = "SELECT itemID, title, parent, url, description, image FROM p_cat WHERE deleted = 0 AND parent is NULL ORDER BY position;"; $query = $this->db->query($sql) or die($this->db->error); }else{ //die($cat); $sql = "SET @parent = (SELECT c.itemID FROM p_cat c WHERE url = '".$this->sql($cat)."' AND deleted = 0); SELECT c1.itemID, c1.title, c1.parent, c1.url, c1.description, c1.image, (SELECT c2.url FROM p_cat c2 WHERE c2.itemID = c1.parent LIMIT 1) as parentUrl FROM p_cat c1 WHERE c1.deleted = 0 AND c1.parent = @parent ORDER BY c1.position;"; $query = $this->db->multi_query($sql) or die($this->db->error); $this->db->store_result(); $this->db->next_result(); $query = $this->db->store_result(); } return $query; }

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  • Adding a variable href attribute to a hyperlink using XSLT

    - by Pete
    I need to create links using dynamic values for both the URL and the anchor text. I have column called URL, which contains URL's, i.e., http://stackoverflow.com, and I have a column called Title, which contains the text for the anchor, i.e., This Great Site. I figure maybe all I need to do is nest this tags, along with a little HTML to get my desired result: <a href="> <xsl:value-of select="@URL">"> <xsl:value-of select="@Title"/> </xsl:value-of> </a> In my head, this would render: <a href=" http://stackoverflow.com"> This Great Site </a> I know there's more to it, but I haven't been able to clearly understand what more I need. I think this issue has been addressed before, but not in a way I could understand it. I'd be more than happy to improve the question's title to help noobs like myself find this item, and hopefully an answer. Any help is greatly appreciated.

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  • Problem in getting Http Response in chrome

    - by Bhaskasr
    Am trying to get http response from php web service in javascript, but getting null in firefox and chrome. plz tell me where am doing mistake here is my code, function fetch_details() { if (window.XMLHttpRequest) { xhttp=new XMLHttpRequest() alert("first"); } else { xhttp=new ActiveXObject("Microsoft.XMLHTTP") alert("sec"); } xhttp.open("GET","url.com",false); xhttp.send(""); xmlDoc=xhttp.responseXML; alert(xmlDoc.getElementsByTagName("Inbox")[0].childNodes[0].nodeValue); } I have tried with ajax also but am not getting http response here is my code, please guide me var xmlhttp = null; var url = "url.com"; if (window.XMLHttpRequest) { xmlhttp = new XMLHttpRequest(); alert(xmlhttp); //make sure that Browser supports overrideMimeType if ( typeof xmlhttp.overrideMimeType != 'undefined') { xmlhttp.overrideMimeType('text/xml'); } } else if (window.ActiveXObject) { xmlhttp = new ActiveXObject("Microsoft.XMLHTTP"); } else { alert('Perhaps your browser does not support xmlhttprequests?'); } xmlhttp.open('GET', url, true); xmlhttp.onreadystatechange = function() { if (xmlhttp.readyState == 4) { alert(xmlhttp.responseXML); } }; } // Make the actual request xmlhttp.send(null); I am getting xmlhttp.readyState = 4 xmlhttp.status = 0 xmlhttp.responseText = "" plz tell me where am doing mistake

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  • Apply a specific class to the first item in a series with PHP

    - by Thomas
    I have a php script that echoes a series of uploaded images and I need to apply a class to the first image echoed in the series. Is this possible? For example, if I want to display 10 images and apply a class to only the first image, how would I go about it? Here is the code, I am working with: <div id="gallery"> <?php query_posts('cat=7'); while(have_posts()) { the_post(); $image_tag = wp_get_post_image('return_html=true'); $resized_img = getphpthumburl($image_tag,'h=387&w=587&zc=1'); $url = get_permalink(); $Price ='Price'; $Location = 'Location'; $title = $post->post_title; echo "<a href='$url'><img src='$resized_img' width='587' height='387' "; echo "rel=\"<div class='gallery_title'><h2>"; echo $title; echo "</h2></div>"; echo "<div class='pre_box'>Rate:</div><div class='entry'>\$"; echo get_post_meta($post->ID, $Price, true); echo "</div><div class='pre_box'>Location:</div><div class='entry'>"; echo get_post_meta($post->ID, $Location, true); echo "</div>\""; echo "'/></a>"; echo ""; } ?> On line 13, the code looks like this: echo "<a href='$url'><img src='$resized_img' width='587' height='387' "; For the first item only, I need it to look like this: echo "<a href='$url' class='show'><img src='$resized_img' width='587' height='387' ";

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  • Webview shouldoverrideurlloading doesn't work

    - by Zak
    I have this code in my app: public class Home extends Activity{ @Override public void onCreate(Bundle savedInstanceState){ super.onCreate(savedInstanceState); setContentView(R.layout.home); final ProgressDialog progressBar; if(isOnline()){ WebView webView = (WebView) findViewById(R.id.home_web); webView.setBackgroundColor(Color.parseColor(getString(R.color.colore_bg))); webView.getSettings().setJavaScriptEnabled(true); webView.getSettings().setBuiltInZoomControls(true); webView.getSettings().setPluginsEnabled(true); webView.setWebViewClient(new MyWebViewClient()); progressBar = ProgressDialog.show(this,getString(R.string.caricamento),getString(R.string.attendere)); webView.setWebViewClient(new WebViewClient(){ public void onPageFinished(WebView view, String url) { if (progressBar.isShowing()) { progressBar.dismiss(); } } }); webView.loadUrl("http://www.mysite.com/android.php"); }else{ Toast.makeText(this,getString(R.string.no_connessione),Toast.LENGTH_LONG).show(); } } private class MyWebViewClient extends WebViewClient { @Override public boolean shouldOverrideUrlLoading(WebView view, String url) { System.out.println("here"); if (Uri.parse(url).getHost().equals("mysite.com")) { // This is my web site, so do not override; let my WebView load the page return false; } // Otherwise, the link is not for a page on my site, so launch another Activity that handles URLs Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url)); startActivity(intent); return true; } } public boolean isOnline(){ ConnectivityManager cm=(ConnectivityManager)getSystemService(Context.CONNECTIVITY_SERVICE); NetworkInfo ni = cm.getActiveNetworkInfo(); if(ni==null){ return false; } return ni.isConnected(); } } The shouldOverrideUrlLoading doesn't work, neither print the system.out, it seems to be never called. How can I repair this? I need to open all the link (except the main page www.mysite.com/iphone.php) in the default browser

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  • Java Trying to get a line of source from a website

    - by dsta
    I'm trying to get one line of source from a website and then I'm returning that line back to main. I keep on getting an error at the line where I define InputStream in. Why am I getting an error at this line? public class MP3LinkRetriever { private static String line; public static void main(String[] args) { String link = "www.google.com"; String line = ""; while (link != "") { link = JOptionPane.showInputDialog("Please enter the link"); try { line = Connect(link); } catch(Exception e) { } JOptionPane.showMessageDialog(null, "MP3 Link: " + parseLine(line)); String text = line; Toolkit.getDefaultToolkit( ).getSystemClipboard() .setContents(new StringSelection(text), new ClipboardOwner() { public void lostOwnership(Clipboard c, Transferable t) { } }); JOptionPane.showMessageDialog(null, "Link copied to your clipboard"); } } public static String Connect(String link) throws Exception { String strLine = null; InputStream in = null; try { URL url = new URL(link); HttpURLConnection uc = (HttpURLConnection) url.openConnection(); in = new BufferedInputStream(uc.getInputStream()); Reader re = new InputStreamReader(in); BufferedReader r = new BufferedReader(re); int index = -1; while ((strLine = r.readLine()) != null && index == -1) { index = strLine.indexOf("<source src"); } } finally { try { in.close(); } catch (Exception e) { } } return strLine; } public static String parseLine(String line) { line = line.replace("<source", ""); line = line.replace(" src=", ""); line = line.replace("\"", ""); line = line.replace("type=", ""); line = line.replace("audio/mpeg", ""); line = line.replace(">", ""); return line; } }

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  • Spring security - same page to deliver different content based on user role

    - by Ramesh
    Hello, i tried to search for any previous post related to my issue but couldnt find any. I have a scenario where in page handles 3 different scenarios and one of them not working. This page returns different content depending on if the user is authenticated or anonymous. localhost:8080/myApp/muUrl?test=authenticatedContent - used for Scenario 1 & 2 localhost:8080/myApp/muUrl?test=anonymousContent - used for Scenario 3 Scenario: 1) Authenticated user accesing the page url - the user gets displayed correct information. Works fine 2) Anonymous user accesing page URL with parameters that requires authentication - If anonymous, there is second level of check on the content they are accessing. for example, based on the GET parameters, there is custom logic to determine if the user has to be authenticated. In which case the page gets redirected to login page (WORKS fine). 3) Anonymous user accessing page URL with parameters that doesnt need authentication - in this case i get the SAvedRequest and redirect to the URL which is taking me to an infinite loop. Am i missing something very obvious or is there a way in AuthenticationProcessFilterEntryPoint to say "DON'T redirect to LOGIN page but process it" ? thanks.

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  • Any one point me how to customize facebook share

    - by Venkat
    I am trying to share my own custom url, image, title and description using Facebook and twitter. I am having lot of images and videos in my website. So i want to make my content viral on social websites. I am trying to keep share options for both facebook and twitter for everything individually. If some one share one image i want that image in the sharing thumbnail and url will be the page url with my own title, description. Based on the url i will point the user to that pic in my website. I tried in the below way. Facebook share: <a href="javascript:;" onclick="window.open('http://www.facebook.com/share.php?u=your_page_url','facebook share','resizable=yes,width=700,height=500,scrollbars=yes,status=yes')"><img alt="facebook" src="yourimage.jpg" /></a> Twitter share: <a href="javascript:;" onclick="window.open('https://twitter.com/share','twitter share','resizable=yes,width=700,height=500,scrollbars=yes,status=yes')"><img alt="twitter" src="yourimage.jpg" /></a>

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  • Trying to use jquery ui in google chrome extension in the content level

    - by user135697
    The problem is that the scope of the content script is on the web page that your plugin is suppose to be used at. So the css background:url(images/ui-bg_inset-hard_100_fcfdfd_1x100.png) becomes url('http://webpageforplugin/images/ui-bg_inset-hard_100_fcfdfd_1x100.png') in order for this to work as far as i understood i need to have it to point to: url('chrome-extension://extensionId/images/ui-bg_inset-hard_100_fcfdfd_1x100.png') So i tried to haxorz the document.styleSheets var ss = document.styleSheets; for (var i=0; i<ss.length; i++) { var found=-1, x,i; var rules = ss[i].cssRules || ss[i].rules; for (var j=0; j<rules.length; j++) { if ('.ui-helper-hidden'==rules[j].selectorText){ found=i; break; } } if (found>-1){ for (var j=0; j<rules.length; j++) { if (x=rules[j].style.background){ if ((i=x.indexOf('url'))!=-1) rules[j].style.background = x.replace('http://page/images/','chrome-extension://extensionId/images/'); } } break; } }; I feel that i'm missing the obvious. That there must be an easier way. Even if i manage to change this how will i get the extension id to build the string. Btw this doesn't work, the icons are not properly fetched. (I hardcoded the extension id) Any ideas?

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  • Problem with Xpath PHP

    - by user294597
    Im tring to access some links through Google using xpath. The below does works fine and all the links are shown. $query = $xpath->evaluate("/html/body//a"); for ($x=0 ; $x < $query -> length; $x++) { $href=$query->item($x); $url=$href->getAttribute('href'); echo $url."<br>"; } But when i try the below xpath nothing is shown..Im sure that the xpath is correct coz its evaluated and the result is shown in xpather.. /html/body[@id='gsr']/div[@id='cnt']/div[@id='res']/div[1]/ol/li/div//cite for ($x=0 ; $x < $query -> length; $x++) { $href=$query->item($x); $url=$href->getAttribute('cite'); echo $url."<br>"; } can some one please tell me what i am doin wrong? any help will be much appreciated

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  • Why are Facebook Likes Insisting on using Wrong Product Image...?

    - by Joan Kent
    Firstly, I'm not a web developer so please be patient. I have read the other posts but I think i have everything covered. My website http://www.joaniesgifts.co.uk includes the like button on the product pages. However, I've found that certain product pages are using the incorrect image when a user likes the page. For example - http://www.joaniesgifts.co.uk/terramundi-money-pots/terramundi-money-pot-holiday-fund I think this may have been down to an original incorrect setup which is now corrected. However, the problem remains... The only thing I have to go on :- if i use the facebook url linter (developers.facebook.com/tools/debug) on the above product page, I receive the following error :- Object at URL 'http://www.joaniesgifts.co.uk/terramundi-money-pot-holiday-fund' of type '213689662010141:product' is invalid because the domain 'www.joaniesgifts.co.uk' is not allowed for the application id '213689662010141' which owns the specified object type. If you are the owner of this application, you can verify your configured 'Site Domain' at developers.facebook.com/apps/213689662010141. (I have verified my site's domain) Everything else appears fine except it is also showing the wrong image!! However, under Raw Open Graph Document Information it has the correct link! If I then click graph api - graph.facebook.com/10150450766583352 it again shows the wrong image was linked! I've no idea what else to do - can you help me? Kind Regards, Joan PS Graph API shows the incorrect image after a scrape only minutes ago { "url": "http://www.joaniesgifts.co.uk/terramundi-money-pot-holiday-fund", "type": "website", "title": "Terramundi Money Pot - Holiday Fund", "image": [ { "url": "http://www.joaniesgifts.co.uk/index.php?route=product\u00252Fproduct\u00252Fcaptcha" } ], "updated_time": "2011-11-11T18:54:38+0000", "id": "10150450766583352" }

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  • How to display data in spinner from json

    - by user1605913
    I am retrieving values from a json url and then storing it in a string variable. Now I want to display that value in a spinner. I have created an array list in my strings.xml file. The xml file contains following code: <string name="credit_card_title">Card Type</string> <string-array name="credit_card"> <item >Select</item> <item >Visa</item> <item >MC</item> <item >Amex</item> <item >Discover</item> my spinner code is: <Spinner android:id="@+id/crdtcrd_crdtype" android:layout_width="match_parent" android:layout_height="wrap_content" android:entries="@array/credit_card" android:prompt="@string/credit_card_title" /> After retriving the value from the json url I am storing it in variable name String cardtype Now how can I display the value of cardtype in the Spinner crtdcrd_crdtype.... the json url is: http://mygogolfteetime.com/iphone/login/[email protected]/123456 From this URL I have to retrieve the value of cardtype and after retrieving the value i have to display it in the spinner.. There are different values for cardype like visa, mc, amex and discover All these values are in my strings.xml file and after retrieving the value I have to display it in Spinner.. Help needed still not able to find the solution.. Thanks in advance...

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  • Calling a jQuery plugin inside itself

    - by Real Tuty
    I am trying to create a comet like thing. I have a plugin that collects data from a php page. The problem is that i dont know how to call the plugin inside itself. If it were a function i could go like this: function j () {setTimeout(j(), 1000);}, but i am using a jQuery plugin. Here is my plugin code: (function($) { $.fn.watch = function(ops) { var $this_ = this, setngs = $.extend({ 'type' : 'JSON', 'query' : 'GET', 'url' : '', 'data' : '', 'wait' : 1000 }, ops); if (setngs.type === '') { return false; } else if (setngs.query === '') { return false; } else if (setngs.url === '') { return false; } else if (setngs.wait === '') { return false; } else if (setngs.wait === 0) { setngs.wait = 1000; } var xhr = $.ajax({ type : setngs.query, dataType : setngs.type, url : setngs.url, success : function(data) { var i = 0; for (i = 0; i < data.length; i++) { var html = $this_.html(), str = '<li class="post" id="post-' + data[i].id + '"><div class="inner"><div class="user">' + data[i].user + '</div><div class="body">' + data[i].body + '</div></div></li>'; $this_.html(str + html); } setTimeout($this_, 1000); } }); }; })(jQuery); where it says setTimeout($this_, 1000); this is where im having trouble. I don't know what to call the plugin as. $this_ is what I thought might work but I am wrong. That is what i need to replace. Thanks for your help.

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  • JQuery SelectList Option Changed doesn't refresh

    - by Jean-Philippe
    Hi. I have this select list: <select url="/Admin/SubCategories" name="TopParentId" id="ParentList"> <option value="">Please select a parent category</option> <option value="1" selected="selected">New Store</option> <option value="2">Extensions</option> <option value="3">Remodel</option> <option value="4">Bespoke Signage</option> <option value="5">Tactical Signage</option> <option value="6">Size Chart</option> <option value="7">Contact Info</option> </select> As you can see the option 1 is marked as selected. When I change the selection, I use this code to do an ajax call to get some values to populate a new select list: $("#ParentList").unbind("change"); $("#ParentList").change(function() { var itemId = $(this).val(); var url = $(this).attr("url"); var options; $.getJSON(url, itemId, function(data) { var defaultoption = '<option value="0">Please select a sub-category</option>'; options += defaultoption; $.each(data, function(index, optionData) { var option = '<option value="' + optionData.valueOf + '">' + optionData.Text + '</option>'; options += option; }); $("#SubParentList").html(options); }); }); My problem is that whenever I change the selection, the itemId is always the id of option 1, because it is marked as selected. It doesn't pick up the value of the option it is being changed too. Can someone please enlighten me of their knowledge. Regards, Jean-Philippe

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  • UIDocumentInteractionController change title of presented view

    - by whatdoesitallmean
    I am using the UIDocumentInteractionController to display pdf files. My files are stored in the filesystem using encoded filenames that are not user-friendly. I do have access to a more friendly name for the file, but the file is not stored with this name (for uniqueness reasons). When I load the document into the UIDocumentInteractionController then the view displays the unfriendly 'real' filename in the titlebar. Is there any way to change the displayed title as presented by the UIDocumentInteractionController? Thanks.

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  • need help with mvc & routes

    - by geoff
    I'm very new to MVC and I'm trying to get a new site set up using it. For SEO reasons we need to make the url of a page something like "Recruiter/4359/John_Smith" or basically {controller}/{id}/{name}. I have that working when I create the url in the code behind like so... //r is a recruiter object that is part of the results for the view r.Summary = searchResult.Summary + "... &lt;a href=\"/Recruiter/" + r.Id + "/" + r.FirstName + "_" + r.LastName + "\"&gt;Read More&lt;/a&gt;" But when I am using the collection of results from a search in my view and iterating through them I am trying to create another link to the same page doing something like <%=Html.ActionLink<RecruiterController>(x => x.Detail((int)r.Id), r.RecruiterName)%> but that doesn't work. When I use that code in the view it gives me a url in the form of /Recruiter/Detail/4359 I was told by a coworker that I should use the Html.ActionLink to create the link in both the view and the controller so that if the route changes in the future it will automatically work. Unfortunately he wasn't sure how to do that in this case. So, my problems are... How can I make the Html.ActionLink work in the view to create a url like I need (like r.Summary above)? How do I use the Html.ActionLink in a controller instead of hardcoding the link like I have above?

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  • overwrite parameters passed by querystring

    - by opensas
    I have the following problem I have a web framework built with classic asp that saves the page state in hidden textboxes, and then issues a submit to itself. Before submitting, we have a javascript functions that saves the action in a hidden "action" input, and then performs the submit. The page loads the state from those hidden texts, reads the action issued, reads extra parameters, like the id of the record to edit, and then builds the page accordingly. I'd like to make a url link to automatically start the page with "edit" action on a "x" id. So I was thinking about building the following url, for example http://myapp/user?action=edit&id=23 the problem is that when the page auto-submits, que url string keeps the parameters. I'd like to achieve the following: when the user clicks on http://myapp/user?action=edit&id=23 my page should receive the posted values action=edit and id=23 but the url should be just http://myapp/user and both parameters should be kept in the hidden texts... (I wonder if I make myself clear...) thanks a lot saludos sas ps: I have a couple of ideas about how to solve it, but I'll post them as answers...

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  • UIAlertView Will not show

    - by John
    I have a program that is requesting a JSON string. I have created a class that contains the connect method below. When the root view is coming up, it does a request to this class and method to load up some data for the root view. When I test the error code (by changing the URL host to 127.0.0.1), I expect the Alert to show. Behavior is that the root view just goes black, and the app aborts with no alert. No errors in debug mode on the console, either. Any thoughts as to this behavior? I've been looking around for hints to this for hours to no avail. Thanks in advance for your help. Note: the conditional for (error) is called, as well as the UIAlertView code. - (NSString *)connect:(NSString *)urlString { NSString *jsonString; UIApplication *app = [UIApplication sharedApplication]; app.networkActivityIndicatorVisible = YES; NSError *error = nil; NSURLResponse *response = nil; NSURL *url = [[NSURL alloc] initWithString:urlString]; NSURLRequest *req = [NSURLRequest requestWithURL:url cachePolicy:NSURLRequestReloadIgnoringCacheData timeoutInterval:10]; NSData *_response = [NSURLConnection sendSynchronousRequest: req returningResponse: &response error: &error]; if (error) { /* inform the user that the connection failed */ //AlertWithMessage(@"Connection Failed!", message); UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Oopsie!" message:@"Unable to connect! Try later, thanks." delegate:nil cancelButtonTitle:@"OK" otherButtonTitles: nil]; [alert show]; [alert release]; } else { jsonString = [[[NSString alloc] initWithData:_response encoding:NSUTF8StringEncoding] autorelease]; } app.networkActivityIndicatorVisible = NO; [url release]; return jsonString; }

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  • Parsing HTML: Call to a member function > children() on a non-object

    - by sm56d
    Hello all, I was just helped with this question but I can't get it to move to the next block of HTML. $html = file_get_html('http://music.banadir24.com/singer/aasha_abdoo/247.html'); $urls = $html->find('table[width=100%] table tr'); foreach($urls as $url){ $song_name = $url->children(2)->plaintext; $url = $url->children(6)->children(0)->href; } It returns the list of the names of the first album (Deesco) but it does not continue to the next album (The Best Of Aasha)? It just gives me this error: Notice: Trying to get property of non-object in C:\wamp\www\test3.php on line 26 Fatal error: Call to a member function children() on a non-object in C:\wamp\www\test3.php on line 28 Why is this and how can I get it to continue to the next table element? I appreciate any help on this! Please note: This is legal as the songs are not bound by copyright and they are available to download freely, its just I need to download a lot of them and I can't sit there clicking a button all day. Having said that, its taken me an hour to get this far.

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  • Catch hibernate exceptions with errorpage handler?

    - by membersound
    I'm catching all exceptions within: java.lang.Throwable /page.xhtml java.lang.Error /page.xhtml But what If I get eg a hibernate ex: org.hibernate.exception.ConstraintViolationException Do I have to define error-pages for EVERY maybe occuring exception? Can't I just say "catch every exception"? Update Redirect on a 404 works. But on a throwable does not! <servlet> <servlet-name>Faces Servlet</servlet-name> <servlet-class>javax.faces.webapp.FacesServlet</servlet-class> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>Faces Servlet</servlet-name> <url-pattern>*.jsf</url-pattern> <url-pattern>*.xhtml</url-pattern> </servlet-mapping> <error-page> <exception-type>java.lang.Throwable</exception-type> <location>/error.xhtml</location> </error-page> <error-page> <error-code>404</error-code> <location>/error.xhtml</location> </error-page>

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  • PHP Form - Edit & Delete via Text File Db

    - by Jax
    hi, I pieced together the script below from various tutorials, examples, etc... Right now the script currently: Saves Id, Name, Url with a "|" delimiter to a text file Db like: 1|John|http://www.john.com| 2|Mark|http://www.mark.com| 3|Fred|http://www.fred.com| But I'm having a hard time trying to make the "UPDATE" and "DELETE" buttons work. Can someone please post code which will: let me update/save any changed data for that row (for UPDATE button) let me delete that row (for DELETE button) PLEASE copy n paste the code below and try for yourself. I would like to keep the output format of the script below too. thanks D- $file = "data.txt"; $name = $_POST['name']; $url = $_POST['url']; $data = file('data.txt'); $i = 1; foreach ($data as $line) { $line = explode('|', $line); $i++; } if (isset($_POST['submits'])) { $fp = fopen($file, "a+"); fwrite($fp, $i."|".$name."|".$url."|\n"); fclose($fp); } ? '); } ?

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  • Android - BitmapFactory.decodeByteArray - OutOfMemoryError (OOM)

    - by Bob Keathley
    I have read 100s of article about the OOM problem. Most are in regard to large bitmaps. I am doing a mapping application where we download 256x256 weather overlay tiles. Most are totally transparent and very small. I just got a crash on a bitmap stream that was 442 Bytes long while calling BitmapFactory.decodeByteArray(....). The Exception states: java.lang.OutOfMemoryError: bitmap size exceeds VM budget(Heap Size=9415KB, Allocated=5192KB, Bitmap Size=23671KB) The code is: protected Bitmap retrieveImageData() throws IOException { URL url = new URL(imageUrl); InputStream in = null; OutputStream out = null; HttpURLConnection connection = (HttpURLConnection) url.openConnection(); // determine the image size and allocate a buffer int fileSize = connection.getContentLength(); if (fileSize < 0) { return null; } byte[] imageData = new byte[fileSize]; // download the file //Log.d(LOG_TAG, "fetching image " + imageUrl + " (" + fileSize + ")"); BufferedInputStream istream = new BufferedInputStream(connection.getInputStream()); int bytesRead = 0; int offset = 0; while (bytesRead != -1 && offset < fileSize) { bytesRead = istream.read(imageData, offset, fileSize - offset); offset += bytesRead; } // clean up istream.close(); connection.disconnect(); Bitmap bitmap = null; try { bitmap = BitmapFactory.decodeByteArray(imageData, 0, bytesRead); } catch (OutOfMemoryError e) { Log.e("Map", "Tile Loader (241) Out Of Memory Error " + e.getLocalizedMessage()); System.gc(); } return bitmap; } Here is what I see in the debugger: bytesRead = 442 So the Bitmap data is 442 Bytes. Why would it be trying to create a 23671KB Bitmap and running out of memory?

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  • If attacker has original data, and encrypted data, can they determine the passphrase?

    - by Brad Cupit
    If an attacker has several distinct items (for example: e-mail addresses) and knows the encrypted value of each item, can the attacker more easily determine the secret passphrase used to encrypt those items? Meaning, can they determine the passphrase without resorting to brute force? This question may sound strange, so let me provide a use-case: User signs up to a site with their e-mail address Server sends that e-mail address a confirmation URL (for example: https://my.app.com/confirmEmailAddress/bill%40yahoo.com) Attacker can guess the confirmation URL and therefore can sign up with someone else's e-mail address, and 'confirm' it without ever having to sign in to that person's e-mail account and see the confirmation URL. This is a problem. Instead of sending the e-mail address plain text in the URL, we'll send it encrypted by a secret passphrase. (I know the attacker could still intercept the e-mail sent by the server, since e-mail are plain text, but bear with me here.) If an attacker then signs up with multiple free e-mail accounts and sees multiple URLs, each with the corresponding encrypted e-mail address, could the attacker more easily determine the passphrase used for encryption? Alternative Solution I could instead send a random number or one-way hash of their e-mail address (plus random salt). This eliminates storing the secret passphrase, but it means I need to store that random number/hash in the database. The original approach above does not require this extra table. I'm leaning towards the the one-way hash + extra table solution, but I still would like to know the answer: does having multiple unencrypted e-mail addresses and their encrypted counterparts make it easier to determine the passphrase used?

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  • Why does my Perl CGI program fail with "Software error: ..."?

    - by kiran
    When I try to run my Perl CGI program, the returned web page tells me: Software error: For help, please send mail to the webmaster (root@localhost), giving this error message and the time and date of the error. Here is my code in one of the file: #!/usr/bin/perl use lib "/home/ecoopr/ecoopr.com/CPAN"; use CGI; use CGI::FormBuilder; use CGI::Session; use CGI::Carp (fatalsToBrowser); use CGI::Session; use HTML::Template; use MIME::Base64 (); use strict; require "./db_lib.pl"; require "./config.pl"; my $query = CGI-new; my $url = $query-url(); my $hostname = $query-url(-base = 1); my $login_url = $hostname . '/login.pl'; my $redir_url = $login_url . '?d=' . $url; my $domain_name = get_domain_name(); my $helpful_msg = $query-param('m'); my $new_trusted_user_fname = $query-param('u'); my $action = $query-param('a'); $new_trusted_user_fname = MIME::Base64::decode($new_trusted_user_fname); ####### Colin: Added July 12, 2009 ####### my $view = $query-param('view'); my $offset = $query-param('offset'); ####### Colin: Added July , 2009 ####### #print $session-header; #print $new_trusted_user; my $helpful_msg_txt = qq[]; my $helpful_msg_div = qq[]; if ($helpful_msg)

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  • ASP.NET MVC twitter/myspace style routing

    - by Astrofaes
    Hi guys, This is my first post after being a long-time lurker - so please be gentle :-) I have a website similar to twitter, in that people can sign up and choose a 'friendly url', so on my site they would have something like: mydomain.com/benjones I also have root level static pages such as: mydomain.com/about and of course my homepage: mydomain.com/ I'm new to ASP.NET MVC 2 (in fact I just started today) and I've set up the following routes to try and achieve the above. public static void RegisterRoutes(RouteCollection routes) { routes.IgnoreRoute("{resource}.axd/{*pathInfo}"); routes.IgnoreRoute("content/{*pathInfo}"); routes.IgnoreRoute("images/{*pathInfo}"); routes.MapRoute("About", "about", new { controller = "Common", action = "About" } ); // User profile sits at root level so check for this before displaying the homepage routes.MapRoute("UserProfile", "{url}", new { controller = "User", action = "Profile", url = "" } ); routes.MapRoute("Home", "", new { controller = "Home", action = "Index", id = "" } ); } For the most part this works fine, however, my homepage is not being triggered! Essentially, when you browser to mydomain.com, it seems to trigger the User Profile route with an empty {url} parameter and so the homepage is never reached! Any ideas on how I can show the homepage?

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