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  • how to remove repeated record's from results linq to sql

    - by Sadegh
    hi, i want to remove repeated record's from results but distinct don't do this for me! why??? var results = (from words in _Xplorium.Words join wordFiles in _Xplorium.WordFiles on words.WordId equals wordFiles.WordId join files in _Xplorium.Files on wordFiles.FileId equals files.FileId join urls in _Xplorium.Urls on files.UrlId equals urls.UrlId where files.Title.Contains(query) || files.Description.Contains(query) orderby wordFiles.Count descending select new SearchResultItem() { Title = files.Title, Url = urls.Address, Count = wordFiles.Count, CrawledOn = files.CrawledOn, Description = files.Description, Lenght = files.Lenght, UniqueKey = words.WordId + "-" + files.FileId + "-" + urls.UrlId }).Distinct();

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  • Parsing XML wont display all items.

    - by Nauman A
    I have this code but the toast wont display any message what is wrong with my code.. I can get the value from link, linknext but title wont bring out any value. ( I am not very bright with writing code so please suggest anything you may feel like. final Button button = (Button) findViewById(R.id.Button01); button.setOnClickListener(new View.OnClickListener() { public void onClick(View v) { // Perform action on click try { URL url = new URL( "http://somelink.com=" + Link.setFirst_link); DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance(); DocumentBuilder db = dbf.newDocumentBuilder(); Document doc = db.parse(new InputSource(url.openStream())); doc.getDocumentElement().normalize(); NodeList nodeList = doc.getElementsByTagName("item"); /** Assign textview array lenght by arraylist size */ for (int i = 0; i < nodeList.getLength(); i++) { Node node = nodeList.item(i); Element fstElmnt = (Element) node; NodeList nameList = fstElmnt.getElementsByTagName("link"); Element nameElement = (Element) nameList.item(0); nameList = nameElement.getChildNodes(); String img = (((Node) nameList.item(0)).getNodeValue()); NodeList websiteList = fstElmnt.getElementsByTagName("linknext"); Element websiteElement = (Element) websiteList.item(0); websiteList = websiteElement.getChildNodes(); String nextlink = (((Node) websiteList.item(0)).getNodeValue()); Link.setFirst_link = nextlink; Drawable drawable = LoadImageFromWebOperations(img); imgView.setImageDrawable(drawable); NodeList titleList = fstElmnt.getElementsByTagName("title"); Element titleElement = (Element) titleList.item(0); websiteList = titleElement.getChildNodes(); String title = (((Node) titleList.item(0)).getNodeValue()); Context context = getApplicationContext(); CharSequence text = title; int duration = Toast.LENGTH_SHORT; Toast toast = Toast.makeText(context, text, duration); toast.show(); } } catch (Exception e) { System.out.println("XML Pasing Excpetion = " + e); } } }); /** Set the layout view to display */ } Here is the xml file <?xml version="1.0"?> <maintag> <item> <link>http://image.com/357769.jpg?40</link> <linknext>http://www.image.com</linknext> <title>imagename</title> </item> </maintag>

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  • Rails belongs_to issue in the views

    - by Jacobo Tibaquira
    Hi, Im having problems with an association in rails: Currently I have Post and User models, and the relationship is set this way: class User < ActiveRecord::Base attr_accessible :username, :name, :lastname has_many :posts end class Post < ActiveRecord::Base attr_accessible :title, :body belongs_to :user end However, in my app/views/posts/index.html.haml when Im trying to access the username for the post I get this error: undefined method `name' for nil:NilClass This is my view: - title "Posts" %table %tr %th Title %th Body %th Author - for post in @posts %tr %td= h post.title %td= h post.body %td= h post.user.name %td= link_to 'Show', post %td= link_to 'Edit', edit_post_path(post) %td= link_to 'Destroy', post, :confirm => 'Are you sure?', :method => :delete %p= link_to "New Post", new_post_path Any thoughts of what Im doing wrong will be appretiated

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  • Json to treeview (<ul>)

    - by Pieter
    Hi I get the following data back from my WCF Data Service (I cut out the metadata) { "d" : [ {"CodeId": 6, "Title": "A Child Sub Item", "Parent":}, {"CodeId": 5, "Title": "Another Root Item", "Parent": -1}, {"CodeId": 4, "Title": "Child Item", "Parent": 2}, {"CodeId": 2, "Title": "Root Item", "Parent": -1} ] } I am trying to get this into a <ul> style tree with Parent = -1 as root and then the rest as sub items of their parent id's. Can anyone help me please, preferably in jQuery? I will use this in jstree if someone knows of a better way to do this. Thanks

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  • Styling an Input Field

    - by John
    Hello When I try to alter the CSS for the input field named "title" below, which is classed by "submissionfield", the position changes, but the height, length, and font of the field do not change. How could I make the height of the input field "title" 22 px, the length 550 px, and the font Times New Roman? The CSS below does not do it. Thanks in advance, John echo '<form action="http://www...com/.../submit2.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <div class="submissiontitle"><label for="title">Story Title:</label></div> <div class="submissionfield"><input name="title" type="title" id="title" maxlength="1000"></div> <div class="urltitle"><label for="url">Link:</label></div> <div class="urlfield"><input name="url" type="url" id="url" maxlength="500"></div> <div class="submissionbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; The CSS: .submissiontitle { position:absolute; width:100px; left:30px; top:200px; text-align: left; margin-bottom:3px; padding:0px; font-family:Arial, Helvetica, sans-serif; font-size: 12px; color:#000000; } .submissionfield { position:absolute; width:550px; left:50px; top:230px; text-align: left; margin-bottom:3px; padding:0px; font-family: "Times New Roman", Times, serif; font-size: 22px; color:#000000; } .urltitle { position:absolute; width:250px; left:30px; top:300px; text-align: left; margin-bottom:3px; padding:0px; font-family:Arial, Helvetica, sans-serif; font-size: 12px; color:#000000; } .urlfield { position:absolute; width:550px; left:30px; top:330px; text-align: left; margin-bottom:3px; padding:0px; font-family:Arial, Helvetica, sans-serif; font-size: 11px; color:#000000; } .submissionbutton { position:absolute; width:250px; left:30px; top:380px; text-align: left; margin-bottom:3px; padding:0px; font-family:Arial, Helvetica, sans-serif; font-size: 11px; color:#000000; } .submittitle { position:absolute; width:250px; left:30px; top:150px; text-align: left; margin-bottom:3px; padding:0px; font-family:Arial, Helvetica, sans-serif; font-size: 12px; color:#000000; }

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  • Generating 8000 text files from xml files

    - by Ray
    Hi all, i need to generate the same number of text files as the xml files i have. Within the text files, i need the title and maybe some other tags of it. I can generate text files with the elements i wanted but not all xml files can be generated. Only some of them are generated. Something might be wrong with my parser so help out please thanks. This is my code. Please have a look and give me suggestions. Thanks in advance. import java.io.File; import javax.xml.parsers.DocumentBuilder; import javax.xml.parsers.DocumentBuilderFactory; import org.w3c.dom.*; import java.io.*; public class AccessingXmlFile1 { public static void main(String argv[]) { try { //File file = new File("C:\\MyFile.xml"); // create a file that is really a directory File aDirectory = new File("C:/Documents and Settings/I2R/Desktop/test"); // get a listing of all files in the directory String[] filesInDir = aDirectory.list(); System.out.println(""+filesInDir.length); // sort the list of files (optional) // Arrays.sort(filesInDir); //////////////////////////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////////////////////////// // have everything i need, just print it now for ( int a=0; a<filesInDir.length; a++ ) { String xmlFile = filesInDir[a]; String newLine = System.getProperty("line.separator"); File file = new File(xmlFile); DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance(); DocumentBuilder db = dbf.newDocumentBuilder(); Document document = db.parse(file); document.getDocumentElement().normalize(); //System.out.println("Root element " + document.getDocumentElement().getNodeName()); NodeList node = document.getElementsByTagName("metadata"); System.out.println("Information of Xml File"); System.out.println(xmlFile.substring(0, xmlFile.length() - 4)); //////////////////////////////////////////////////////////////////////////////////// String titleStoreText = ""; String descriptionStoreText = ""; String collectionStoreText = ""; String textToWrite = ""; //////////////////////////////////////////////////////////////////////////////////// for (int i = 0; i < node.getLength(); i++) { Node firstNode = node.item(i); if (firstNode.getNodeType() == Node.ELEMENT_NODE) { Element element = (Element) firstNode; NodeList titleElementList = element.getElementsByTagName("title"); Element titleElement = (Element) titleElementList.item(0); NodeList title = titleElement.getChildNodes(); //////////////////////////////////////////////////////////////////////////////////// if(titleElement == null) titleStoreText = " There is no title for this file."+ newLine; else titleStoreText = titleStoreText+((Node) title.item(0)).getNodeValue() + newLine; //titleStoreText = titleStoreText+((Node) title.item(0)).getNodeValue()+ newLine; //////////////////////////////////////////////////////////////////////////////////// System.out.println("Title : " + titleStoreText); NodeList collectionElementList = element.getElementsByTagName("collection"); Element collectionElement = (Element) collectionElementList.item(0); NodeList collection = collectionElement.getChildNodes(); //////////////////////////////////////////////////////////////////////////////////// if(collectionElement == null) collectionStoreText = " There is no collection for this file."+ newLine; else collectionStoreText = collectionStoreText+((Node) collection.item(0)).getNodeValue() + newLine; //collectionStoreText = collectionStoreText+((Node) collection.item(0)).getNodeValue()+ newLine; //////////////////////////////////////////////////////////////////////////////////// System.out.println("Collection : " + collectionStoreText); NodeList descriptionElementList = element.getElementsByTagName("description"); Element descriptionElement = (Element) descriptionElementList.item(0); NodeList description = descriptionElement.getChildNodes(); //////////////////////////////////////////////////////////////////////////////////// if(descriptionElement == null) descriptionStoreText = " There is no description for this file."+ newLine; else descriptionStoreText = descriptionStoreText+((Node) description.item(0)).getNodeValue() + newLine; //descriptionStoreText = descriptionStoreText+((Node) description.item(0)).getNodeValue() + newLine; //////////////////////////////////////////////////////////////////////////////////// System.out.println("Description : " + descriptionStoreText); //////////////////////////////////////////////////////////////////////////////////// textToWrite = "=====Title=====" + newLine + titleStoreText + newLine + "=====Collection=====" + newLine + collectionStoreText + newLine + "=====Description=====" + newLine + descriptionStoreText;// + newLine + "=====Subject=====" + newLine + subjectStoreText; //////////////////////////////////////////////////////////////////////////////////// } } ///////////////////////////////////////////write to file part is here///////////////////////////////////////// Writer output = null; File file2 = new File(xmlFile.substring(0, xmlFile.length() - 4)+".txt"); output = new BufferedWriter(new FileWriter(file2)); output.write(textToWrite); output.close(); System.out.println("Your file has been written"); //////////////////////////////////////////////////////////////////////////////////// } } catch (Exception e) { e.printStackTrace(); } } }

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  • Getting started with workflows in sharepoint 2010

    - by Thomas Stock
    Hi, I'm a beginning sharepoint developer asked to implement the following scenario in sharepoint 2010. We're a bit lost on the best approach to get started.. I'm really struggling to find the best practise solution. This is the flow: A user can make a request with a title and a description. A mail gets sent to the representative with a link to a form. A representative can approve or reject the request. If approved: A mail gets sent to Board with a link to form If rejected: A mail gets sent to the user with the message that it has been rejected. when the request was approved by the representative, the board can approve or reject the request. A mail gets sent to the user and the representative with the descision of the board. So the list has the following fields: Request title Request description Representative approval Representative description Board approval Board description The user should see the following form: Request title (editable) Request description (editable) The representative should see the following form: Request title (read-only) Request description (read-only) Representative approval (editable) Representative description (editable) The Board should see the following form: Request title (read-only) Request description (read-only) Representative approval (read-only) Representative description (read-only) Board approval (editable) Board description (editable) My questions: What tool is most appropriate for making the forms? Infopath? SPD? VS2010? How do I handle rights to make sure only the board can access the board edit form? What kind of workflow do I use? When do I start the workflow(s)? What do I use to develop the workflow(s)? How do I handle rights when showing the listview with all requests? How can I build the links in the mails sent to the different groups. Thanks in advance for any advice.

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  • iphone how to customize UITableViewCell cell.textlabel and cell.detailtextlabel programmatically

    - by Prerak
    Hi! In my iPhone App In table view cell I want to dispaly one main title and 5 subtitlessubtitles suppose item1 as main title and item2 , item3 , item4 , item5 and item6 as subtitles, for that i have saperate two arrays for passing the values in table view cell one for cell.textLabel.text= second for cell.detailTextLabel.text now I want the flexibility to make item2 as maintitle and want to add item1 to subtitle How can I set title and subtitle programmatically from single array? and any of them as Please Help and Suggest, Thank You.

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  • Breadcrumbs in Fusebox 4/5

    - by Jordan Reiter
    I'm wondering if anyone has come up with a clean way to generate a breadcrumbs trail in Fusebox. Specifically, is there a way of keeping track of "where you are" and having that somehow generate the breadcrumbs for you? So, for example, if you're executing /index.cfm?fuseaction=Widgets.ViewWidget&widget=1 and the circuit structure is something like /foo/bar/widgets/ then somehow the system automatically creates an array like: [ { title: 'Foo', url: '#self#?fuseaction=Foo.Main' }, { title: 'Bar', url: '#self#?fuseaction=Bar.Main' }, { title: 'Widgets', url: '#self#?fuseaction=Widgets.Main' }, { title: 'Awesome Widget', url: '' } ] Which can then be rendered as Foo Bar Widgets Awesome Widget Right now it seems the only way to really do this is to create the structure for each fuseaction in a fuse of some kind (either the display fuse or a fuse dedicated to creating the crumbtrail).

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  • PHP write file... need help

    - by Jordan Pagaduan
    <?php $title = $_POST['title']; $filename = $title , ".php"; $fh = fopen($filename, 'w') or die ("can't open file"); $stringData = $title; fwrite($fh, $stringData); $stringData = $blog; fwrite($fh, $stringData); fclose($fh); ?> This is only a sample. What is the correct code for that?

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  • Help needed in AdventureWorks in a sql query.

    - by vaibhav
    I was just playing with adventureworks database in sqlserver. I got stuck in a query. I wanted to Select all titles from HumanResources.Employee which are either 'Male' or 'Female' but not both. i.e if title Accountant is Male and Female both I want to leave that title. I need only those titles where Gender is either Male or Female. I have done this till yet. select distinct(title) from humanresources.employee where gender='M' select distinct(title) from humanresources.employee where gender='F' Probably a join between these two queries, would work. But If you have any other solution, please let me know. It is not a homework. :) Thanks in advance.

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  • How to add a WHERE clause on the second table of a 1-to-1 join in Fluent NHibernate?

    - by daddywoodland
    I'm using a legacy database that was 'future proofed' to keep track of historical changes. It turns out this feature is never used so I want to map the tables into a single entity. My tables are: CodesHistory (CodesHistoryID (pk), CodeID (fk), Text) Codes (CodeID (pk), CodeName) To add an additional level of complexity, these tables hold the content for the drop down lists throughout the application. So, I'm trying to map a Title entity (Mr, Mrs etc.) as follows: Title ClassMap - Public Sub New() Table("CodesHistory") Id(Function(x) x.TitleID, "CodesHistoryID") Map(Function(x) x.Text) 'Call into the other half of the 1-2-1 join in order to merge them in 'this single domain object Join("Codes", AddressOf AddTitleDetailData) Where("CodeName like 'C.Title.%'") End Sub ' Method to merge two tables with a 1-2-1 join into a single entity in VB.Net Public Sub AddTitleDetailData(ByVal m As JoinPart(Of Title)) m.KeyColumn("CodeID") m.Map(Function(x) x.CodeName) End Sub From the above, you can see that my 'CodeName' field represents the select list in question (C.Title, C.Age etc). The problem is that the WHERE clause only applies to the 'CodesHistory' table but the 'CodeName' field is in the 'Codes' table. As I'm sure you can guess there's no scope to change the database. Is it possible to apply the WHERE clause to the Codes table?

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  • Nested preference screens lose theming

    - by stealthcopter
    I have a preference screen for my application and in the manifest I have given it a theme using: android:theme="@android:style/Theme.Light.WallpaperSettings" However when I nest another preference screen inside this one such as: <?xml version="1.0" encoding="utf-8"?> <PreferenceScreen xmlns:android="http://schemas.android.com/apk/res/android" android:title="@string/setting_title" android:key="..."> <PreferenceCategory android:title="@string/title_themes" > <PreferenceScreen android:title="@string/title_themes_opt" > <ListPreference android:key="Setting_BG" android:title="@string/setting_bg" android:summary="@string/setting_bg_summary" android:entries="@array/bg_titles" android:defaultValue="0" android:entryValues="@array/bg_values" /> </PreferenceScreen> </PreferenceCategory> </PreferenceScreen> The nested preference screen loses the theme of the parent. How can this be prevented? Thanks in advance.

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  • pywinauto: taking more than one app windows

    - by Denis Barmenkov
    I have a GUI application which can create many similar windows on desktop. All windows have same title. I have to enumerate all dialogs with same title and make some tests against each of such dialogs. If I call: dialog = app['Window Name'] pywinauto returns a WindowSpecification object which is useful along with accessing controls by name. When I call: dialogs = app.windows_(title='Window Name') pywinauto returns me a list of HwndWrapper instances which are not so useful. How to obtain a list of windows with specified title but as WindowSpecification objects?

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  • Cleanup ActiveRecord field

    - by Beer Brother
    I have model Article it has field title with some text that may contain some "magic" patterns. In some cases i need to process text in title and other cases i don't, but in last case i need to get string w/o that patterns. For example i have title value like "Something **very** interesting" and when i call @article.title i need to get cleaned up string like "Something very interesting", but when i call @article.title_raw i need get original string. The problem also is that i have working application and i cannt do "revolution" but what way to go... -- Excuse me for my bad English.

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  • issue when outputting an http:// address in a sub-array with json and php

    - by Patrick
    Im trying to achieve an output like this {"status":"ok","0":{"id":"11","title":"digg","url":"http://www.digg.com"}} but instead i am getting this {"status":"ok","0":{"id":"11","title":"digg","url":"http:\/\/www.digg.com"}} this is the php code im using to generate the json $links = array('id'=>'11','title'=>'digg','url'=>"http://www.digg.com"); $msg = array('status'=>'ok',$links); echo json_encode($msg); any idea what is causing this?

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  • combine lines from 2 prints to single line and insert into mysql database

    - by bleomycin
    Hello everyone i currently have this: import feedparser d = feedparser.parse('http://store.steampowered.com/feeds/news.xml') for i in range(10): print d.entries[i].title print d.entries[i].date How would i go about making it so that the title and date are on the same line? Also it doesn't need to print i just have that in there for testing, i would like to dump this output into a mysql db with the title and date, any help is greatly appreciated!

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  • Trouble parsing self closing XML tags using SAX parser

    - by sandesh
    Hi, I am having trouble parsing self closing XML tags using SAX. I am trying to extract the link tag from the Google Base API.I am having reasonable success in parsing regular tags. Here is a snippet of the xml <entry> <id>http://www.google.com/base/feeds/snippets/15802191394735287303</id> <published>2010-04-05T11:00:00.000Z</published> <updated>2010-04-24T19:00:07.000Z</updated> <category scheme='http://base.google.com/categories/itemtypes' term='Products'/> <title type='text'>En-el1 Li-ion Battery+charger For Nikon Digital Camera</title> <link rel='alternate' type='text/html' href='http://rover.ebay.com/rover/1/711-67261-24966-0/2?ipn=psmain&amp;icep_vectorid=263602&amp;kwid=1&amp;mtid=691&amp;crlp=1_263602&amp;icep_item_id=170468125748&amp;itemid=170468125748'/> . . and so on I can parse the updates and published tags, but not the link and category tag. Here is my startElement and endElement overrides public void startElement(String uri, String localName, String qName, Attributes attributes) throws SAXException { if (qName.equals("title") && xmlTags.peek().equals("entry")) { insideEntryTitle = true; } xmlTags.push(qName); } public void endElement(String uri, String localName, String qName) throws SAXException { // If a "title" element is closed, we start a new line, to prepare // printing the new title. xmlTags.pop(); if (insideEntryTitle) { insideEntryTitle = false; System.out.println(); } } declaration for xmltags.. private Stack<String> xmlTags = new Stack<String>(); Any help guys? this is my first post here.. I hope I have followed posting rules! thanks a ton guys..

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  • How to change the sorting of a view using hook_views_pre_view()?

    - by RD
    I've got the following: function view_sorter_views_pre_view(&$view) { // don't need $items if ($view->name == 'MOST_RECENT') { $insert = array(); $insert[order] = 'DESC'; //SORT ORDER $insert[id] = 'title'; $insert[table] = 'node'; $insert[field] = 'title'; $insert[override] = array(); $insert[override][button] = 'Override'; $insert[relationship] = 'none'; unset ($view->display['default']->display_options['sorts']['title']); $view->display['default']->display_options['sorts']['title'] = $insert; } } Basically, I'm just changing the sort order... but this does not appear on the view when opening it. Any idea why?

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  • jQuery toggle caching / cookies

    - by user1706680
    I’m using the jQuery toggle function for my navigation. By default the Authors toggle is visible, the Archives toggle is closed. http://jsfiddle.net/TeDFs/4/ My problem is that when the users switches to another page the toggles reset themselves. For example, when the user closes the Authors toggle and opens the Archive toggle and then navigates to another page the default settings are loaded. I read that it’s possible to store the settings via cookies but I’m absolutely new to jQuery and it would be great if somebody could help me out! HTML <div id="authors" class="widget"> <h2 class="widget-title-visible">Authors</h2> <div class="toggle"> <div class="submenu"> <ul> <li>Name 1</li> <li>Name 2</li> <li>Name 3</li> <li>Name 3</li> </ul> </div> </div> <div id="archives" class="widget"> <h2 class="widget-title">Archiv</h2> <div class="toggle hidden"> <div class="submenu"> <ul> <li>November 2012</li> <li>Oktober 2012</li> </ul> </div> </div> </div>? jQuery function toggleWidgets() { jQuery('.widget-title').addClass('plus'); jQuery('.widget-title-visible').addClass('minus'); jQuery('.widget-title').click(function() { $(this).toggleClass('plus').toggleClass('minus').next().toggle(180); }); jQuery('.widget-title-visible').click(function() { $(this).toggleClass('minus').toggleClass('plus').next().toggle(180); }); } jQuery(document).ready(function() { toggleWidgets(); } )? CSS .hidden{ display:none; } .plus { background: url(http://moargh.de/daten/sidebar_arrows.png) 0 5px no-repeat; padding: 0 0 0 12px; } .minus { background: url(http://moargh.de/daten/sidebar_arrows.png) 0 -10px no-repeat; padding: 0 0 0 12px; }

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  • How do I write this SQL statement to get the ad and posting? (PHP/MySQL)

    - by ggfan
    I am a little confused on the logic of how to write this SQL statement. When a user clicks on a tag, say HTML, it would display all the posts with HTML as its tag. (a post can have multiple tags) I have three tables: Posting--posting_id, title, detail, etc tags--tagID, tagname postingtag--posting_id, tagID I want to display all the title of the post and the date added. global $dbc; $tagID=$_GET['tagID']; //the GET is set by URL //part I need help with. I need another WHERE statment to get to the posting table $query = "SELECT p.title,p.date_added, t.tagname FROM posting as p, postingtag as pt, tags as t WHERE t.tagID=$tagID"; $data = mysqli_query($dbc, $query); echo '<table>'; echo '<tr><td><b>Title</b></td><td><b>Date Posted</b></td></tr>'; while ($row = mysqli_fetch_array($data)) { echo '<tr><td>'.$row['title'].'</td>'; echo '<td>'.$row['date_added'].'</td></tr>'; } echo '</table>'; }

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