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  • wxHaskell on OS X

    - by Bill
    I want to use wxHaskell on OS X (Snow Leopard, MacBook Pro). I was able to install the library successfully and the script below: module Main where import Graphics.UI.WX main :: IO () main = start hello hello :: IO () hello = do f <- frame [text := "Hello!"] quit <- button f [text := "Quit", on command := close f] set f [layout := widget quit] does result in a window being displayed with a single button, as specified. However, nothing happens when I click the button - I don't even get the visual response of the button turning blue to indicate that it's been depressed (haha, no pun intended). I've heard that you have to run a package called "macosx-app" on wxHaskell binaries to get them to run, but I can't find this anywhere. It's not on my machine or (as far as I can tell) in the WX or wxHaskell distros. Anyone know what I need to do to get this to work?

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  • problem with dropdown in jsp

    - by Deven
    hello friends i am having three drop downs and button in my web page naming 1st 2nd and 3rd in which the value of 2nd depends on 1st and value of 3rd depends on 2dn its done using ajax now when i click on the button it calls one function of ajax which takes value from database and then select value at 1st dropdown and then fire on change event using fireEvent() function which is used to set values in 2nd dropdown and then the function called at button now select value at 2nd dropdown and then fire on change event using fireEvent() function which is used to set values in 3rd dropdown The main problem is that when i click on button it set value in 1st dropdown but 2nd and 3rd dropdown shows blank values

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  • How to instantiate JQuery UI widget by string?

    - by limcheekin
    Hi there, Do you know how to instantiate JQuery UI widget by string? Let's illustrate it with some sample code. Given the html link element below: <a id="testLink" href="#">Test Link</a> Normally, we can make it into button using code below: $('#testLink').button(); What if I want to instantiate the button with string, for example: var widget='button'; $('#testLink').[widget](); Of course the code block above is not working (It is just for illustration purpose only), otherwise you will not see this question. Please advice. Thanks, Chee Kin

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  • problem in adding image to JFrame

    - by firestruq
    Hi, I'm having problems in adding a picture into JFrame, something is missing probebly or written wrong. here are the classes: main class: public class Tester { public static void main(String args[]) { BorderLayoutFrame borderLayoutFrame = new BorderLayoutFrame(); borderLayoutFrame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); borderLayoutFrame.setSize(600,600); borderLayoutFrame.setVisible(true); } } public class BorderLayoutFrame extends JFrame implements ActionListener { private JButton buttons[]; // array of buttons to hide portions private final String names[] = { "North", "South", "East", "West", "Center" }; private BorderLayout layout; // borderlayout object private PicPanel picture = new PicPanel(); // set up GUI and event handling public BorderLayoutFrame() { super( "Philosofic Problem" ); layout = new BorderLayout( 5, 5 ); // 5 pixel gaps setLayout( layout ); // set frame layout buttons = new JButton[ names.length ]; // set size of array // create JButtons and register listeners for them for ( int count = 0; count < names.length; count++ ) { buttons[ count ] = new JButton( names[ count ] ); buttons[ count ].addActionListener( this ); } add( buttons[ 0 ], BorderLayout.NORTH ); // add button to north add( buttons[ 1 ], BorderLayout.SOUTH ); // add button to south add( buttons[ 2 ], BorderLayout.EAST ); // add button to east add( buttons[ 3 ], BorderLayout.WEST ); // add button to west add( picture, BorderLayout.CENTER ); // add button to center } // handle button events public void actionPerformed( ActionEvent event ) { } } I'v tried to add the image into the center of layout. here is the image class: public class PicPanel extends JPanel { Image img; private int width = 0; private int height = 0; public PicPanel() { super(); img = Toolkit.getDefaultToolkit().getImage("table.jpg"); } public void paintComponent(Graphics g) { super.paintComponents(g); if ((width <= 0) || (height <= 0)) { width = img.getWidth(this); height = img.getHeight(this); } g.drawImage(img,0,0,width,height,this); } } Please your help, what is the problem? thanks BTW: i'm using eclipse, which directory the image suppose to be in?

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  • Automatic login into asp.net site from flash movie

    - by IrfanRaza
    Hello friends, I have a landing page designed as a flash movie. Please visit http://ivautoinc.com. The movie contains login button. For now when you click on this button I am redirecting to asp.net site login page. What I need is if you click on login button, the login form which is designed within flash will be shown on the same movie. The users will provide username and password. As soon as they press OK button they should see the page from my asp.net site as it is displayed after loggin in. Can anybody help me? Thanks for sharing your valuable time.

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  • User control or custom control for this design?

    - by m0s
    If I want to create a button control like that, is it supposed to be a User Control or a Custom Control? I am not sure maybe something else? Ideally I want to be able style/animate the main button and the inner button separately; also obviously Ill need to process their events separately. These buttons will be created at run-time and Ill need to set the icons dynamically.

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  • icon as an image

    - by Ken
    Hi All, is it possible to use icon file as a image for a button in visual basic? f.e. I have 3 buttons that need to have 3 icons when you click the button the icon of the button needs to be the icon of the form is this posible? btnIcon1 = my.resources.ICO1 btnIcon2 = my.resources.ICO2

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  • Hiding/blocking tabs using windows forms in c#

    - by Audel
    The thing is that i have a 'log in window' and a 'mainwindow' that is called after pressing the log in button or the "VISITANT" button If pressing the log in button, the whole system will come out, and if i press the VISITANT button, one tab should disappear or be blocked or something. private void visitant(object sender, EventArgs e) { mainwindow menu = new mainwindow(); menu.Show(); //mainwindow.tabPage1.Enabled = false; //attempt1 //mainwindow.tabPage1.Visible = false; //attempt1 //System.Windows.Forms.tabPage1.Enabled = false;//attempt2 //System.Windows.Forms.tabPage1.Visible = false;//attempt2 this.Hide(); } the errors i get for using the attempt1 are Error 1 'System.mainwindow.tabPage1' is inaccessible due to its protection level' Error 2 An object reference is required for the non-static field, method, or property 'System.mainwindow.tabPage1' and the one i get for using the attempt2 is Error 1 The type or namespace name 'tabPage1' does not exist in the namespace 'System.Windows.Forms' (are you missing an assembly reference?) as you probably have guessed "tabPage1" is the tab i need to hide when pressing the visitant button. I can't think of any more details, I will be around to provide any extra information Thanks in advance.

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  • How to achieve table like rows within container using CSS

    - by Barry
    I'm helping an artist maintain her website and have inherited some pretty outdated code. Have moved lots of redundant common code to include files and am now working on moving from inline styles to more CSS-driven styles. For the gallery pages, e.g. http://artistsatlaketahoe.com/abstract.html, a lot of inline styling is used to force the current layout. My preference would be to replace this entirely with CSS that presents the following table-like layout within the "content" div: [image] [image descriptives and purchase button] [image] [image descriptives and purchase button] [image] [image descriptives and purchase button] I'd like to middle-align the image descriptives & purchase button relative to the image if possible. And then apply some padding above and below each row to stop using tags for vertical spacing. Any ideas how to create a div that I can use to get this kind of layout? Thanks!

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  • Using Javascript, How do I bring an already existing open window to the front on top of other window

    - by user268249
    The question was fairly descriptive but I'll describe it further. Basically, I have Window1, clicking a button link opens window2. Clicking a button in window2 opens window3, clicking a button in window3 should bring window2 back to the front of the screen on top of window2. I'm not sure how this is exactly done, however I have used and played around with focus(), opener and other various methods and I cannot seem to get it to work properly. Thanks for any help in advance!

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  • WPF Buttons Style

    - by Polaris
    I have WPF Form which has many buttons with the same code. Appearance of all buttons must be the same For example, code for one of these buttons <Button x:Name="btnAddRelative" Width="120" Click="btnAddRelative_Click" > <Button.Content> <StackPanel Orientation="Horizontal"> <Image Height="26" HorizontalAlignment="Left"> <Image.Source> <BitmapImage UriSource="images/add.png" /> </Image.Source> </Image> <TextBlock Text=" Add Relative" Height="20" VerticalAlignment="Center"/> </StackPanel> </Button.Content> </Button> How can I create one style and use it for all my buttons. All buttons has the same png image, only their text different. How can I do this. I tried to do this with Style object in Resource Section: <UserControl.Resources> <Style TargetType="Button" x:Key="AddStyle"> <Setter Property="Content"> <Setter.Value> <StackPanel Orientation="Horizontal"> <Image Height="26" HorizontalAlignment="Left"> <Image.Source> <BitmapImage UriSource="images/add.png" /> </Image.Source> </Image> <TextBlock Text=" " Height="20" VerticalAlignment="Center"/> </StackPanel> </Setter.Value> </Setter> </Style> </UserControl.Resources> But this code not work. Can any body know how can I do this?

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  • Making a login Using JavaScript, Will Incorporate PHP Later

    - by TIMOTHY
    not sure why my code wont work, im teaching myself javascript i know php moderatly and i also know the intelligence of using java to hold a password and username, but at the moment i just want the script to work. <html> <head> <title>34webs</title> <link rel="stylesheet" href="css/lightbox.css" type="text/css" media="screen" /> <link rel="stylesheet" type="text/css" media="screen" href="main.css"> <script type="text/javascript" src="js/prototype.js"></script> <script type="text/javascript" src="js/scriptaculous.js?load=effects,builder"></script> <script type="text/javascript" src="js/lightbox.js"></script> <script type="javascript" > function logintry (){ var usern = document.logn.username.value; var passw = document.logn.password.value; if(usern == blue && passw == bluee){ alert('password is correct!'); }else{ alert('password is wrong'); } } </script> </head> <body> <div id="bod"> <div id="nav"> <p id="buttonhead">34 web</p> <a href="#" class="button">HOME</a> <a href="#" class="button">NEWS</a> <a href="#" class="button">DOWNLOADS</a> <a href="#" class="button">ERA</a> <a href="#" class="button">IE BROWSER</a> <a href="#" class="button">DRIVERS</a> <form name="logn"> <table style="margin:0 auto; background:#0174DF; color:white;"> <tr> <td> Username<br> <input type="text" name="username" value=""> </td> </tr> <tr> <td> Password<br> <input type="password" name="password" value=""> </td> </tr> <tr> <td style="text-align:center;"> <input type="button" name="Submit" value="submit" onclick= javascript: logintry()> </td> </tr> </table> </form> </div>

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  • Given a start and end point, how can I constrain the end point so the resulting line segment is horizontal, vertical, or 45 degrees?

    - by GloryFish
    I have a grid of letters. The player clicks on a letter and drags out a selection. Using Bresenham's Algorithm I can create a line of highlighted letters representing the player's selection. However, what I really want is to have the line segment be constrained to 45 degree angles (as is common for crossword-style games). So, given a start point and an end point, how can I find the line that passes through the start point and is closest to the end point? Bonus: To make things super sweet I'd like to get a list of points in the grid that the line passes through, and for super MEGA bonus points, I'd like to get them in order of selection (i.e. from start point to end point).

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  • Why is the dash so unresponsive, and is there a way to fix this?

    - by Jon
    I just upgraded to 12.04. When I press the super key to open the dash, there's a lag of 1-3 seconds before it displays, with no other programs running. (This is similar, but not identical, to the issue described in Dash application search unresponsive at startup about 11.10.) At login time, this lag is up to 10 seconds, and sometimes the dash doesn't respond at all to the super key. In contrast, the launcher Kupfer immediately responds to its hotkey, in milliseconds, and responds to my typing an application name also in fractions of a second. Is there a way to load the dash in memory or a RAM disk of some sort to make it more responsive?

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  • Need to have an aspx page with a Feature in SharePoint 2010

    - by camit90
    I have added a custom button to the server ribbon in SharePoint (I have used a feature with Farm scope, so that the button is visible throughout the various site collections). For the elements of the feature, I have added a CustomUIExtension through which I want to load an aspx page on the click of the button. <CommandUIHandler Command="Test_Button" CommandAction="javascript: function demoCallback(dialogResult, returnValue) { SP.UI.Notify.addNotification('Operation Successful!'); SP.UI.ModalDialog.RefreshPage(SP.UI.DialogResult.OK); } var options = { url: '/_layouts/CustomPage.aspx', tite: 'Custom Page', dialogReturnValueCallback: demoCallback }; SP.UI.ModalDialog.showModalDialog(options);" /> I have added the CustomPage.aspx and its corresponding code behind class to the 14 hive (inside 14/TEMPLATE/LAYOUTS). However when I install the feature and click the button, I get an error saying "Cannot load CustomPage". I understand that I haven't deployed the assembly, but shouldn't the aspx page be compiled Just In Time?

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  • How to limit the number of the same Activity on the stack for an Android application

    - by johnrock
    Is this possible in an Android app? I want to make it so that no matter how many times a user starts activityA, when they hit the back button they will never get more than one occurence of activityA. What I am finding in my current code is that I have only two options: 1. I can call finish() in activityA which will prevent it from being accessible via the back button completely, or 2. I do not call finish(), and then if the user starts activityA (n) times during their usage, there will be (n) instances when hitting the back button. Again, I want to have activityA accessible by hitting the back button, but there is no reason to keep multiple instances of the same activity on the stack. Is there a way to limit the number of instances of an activity in the queue to only 1?

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  • Magento: Customer Comment on order page required field

    - by Shamim Ahmed
    I am using whiteOrderComment module for customer comment on order review page. but in this section text-area field required option not working. I did little bit change on /checkout-onepage-review-button.phtml like this <script type="text/javascript"> function validate(){ if(document.getElementById("whiteOrderComment").value == ""){ alert('Required'); }else{ review.save(); } </script> <button type="submit" title="<?php echo $this->__('Place Order') ?>" class="button btn-checkout" onclick="validate();"><span><span><?php echo $this->__('Place Order') ?></span></span></button> but in this page javascript not working. can you please give any better idea, how can i make this text-area field required. thanks

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  • Why is my asp.net mvc form POSTing instead of GETting?

    - by quakkels
    My code is straightforward enough: <% using(Html.BeginForm(FormMethod.Get)) %> <% { %> Search for in Screen Name and Email: <%: Html.TextBox("keyword", Request.QueryString["keyword"]) %> <button type=submit>Search</button> <% } %> The issue I'm running into is that when I submit this form, the values are not added to the querystring. Instead, it appears that the form is submitting by a post request. When I look at the generated HTML, I have this: <form action="/find/AdminMember/MemberList" method="post"> Search for in Screen Name and Email: <input id="keyword" name="keyword" type="text" value="" /> <button type=submit>Search</button> </form> Does anyone know why? This seems pretty simple and straighforward to me.

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  • Submit form with POST data in Android app

    - by datguywhowanders
    I've been searching the web for a way to do this for about a week now, and I just can't seem to figure it out. I'm trying to implement an app that my college can use to allow users to log in to various services on the campus with ease. The way it works currently is they go to an online portal, select which service they want, fill in their user name and pwd, and click login. The form data is sent via post (it includes several hidden values as well as just the user name and pwd) to the corresponding login script which then signs them in and loads the service. I've been trying to come at the problem in two ways. I first tried a WebView, but it doesn't seem to want to support all of the html that normally makes this form work. I get all of the elements I need, fields for user and pwd as well as a login button, but clicking the button doesn't do anything. I wondered if I needed to add an onclick handler for it, but I can't see how as the button is implemented in the html of the webview not using a separate android element. The other possibility was using the xml widgets to create the form in a nice relative layout, which seems to load faster and looks better on the android screen. I used EditText fields for the input, a spinner widget for the service select, and the button widget for the login. I know how to make the onclick and item select handlers for the button and spinner, respectively, but I can't figure out how to send that data via POST in an intent that would then launch a browser. I can do an intent with the action url, but can't get the POST data to feed into it. Anyone have any suggestions?

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  • Is there an easy way to stream a m3u in iPhone?

    - by marty
    I can have a UIWebView with the .m3u file opened, which will go to the webview with a play button displayed, and that automatically goes to the quicktime player and starts playing the stream. But when I press the done button, it goes back to the UIWebView with a little play button in the middle, and from there you can go back to the previous screen (it was selected from a tableview). So I just want it to automatically load the quicktime player in the view. How can I do that?

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  • Friday Fun: E7 (Mission to Save Earth)

    - by Asian Angel
    It has been another long week at work and you should take a few minutes to relax and have some fun. In this week’s game you journey to E7 in an attempt to find and destroy the deadly bomb that is aimed at planet Earth. Can you survive the journey across the planet and complete your mission? Latest Features How-To Geek ETC Learn To Adjust Contrast Like a Pro in Photoshop, GIMP, and Paint.NET Have You Ever Wondered How Your Operating System Got Its Name? Should You Delete Windows 7 Service Pack Backup Files to Save Space? What Can Super Mario Teach Us About Graphics Technology? Windows 7 Service Pack 1 is Released: But Should You Install It? How To Make Hundreds of Complex Photo Edits in Seconds With Photoshop Actions Super-Charge GIMP’s Image Editing Capabilities with G’MIC [Cross-Platform] Access and Manage Your Ubuntu One Account in Chrome and Iron Mouse Over YouTube Previews YouTube Videos in Chrome Watch a Machine Get Upgraded from MS-DOS to Windows 7 [Video] Bring the Whole Ubuntu Gang Home to Your Desktop with this Mascots Wallpaper Hack Apart a Highlighter to Create UV-Reactive Flowers [Science]

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  • notications pop up in user side

    - by user2931015
    i try to show notification as a pop up. like when admin login and through his account he send notification to user i add this html in admin form like this.. <asp:Button ID="notic" runat="server" Text="Send" onclick="Button1_Click" /> <br /> <input class="add_message" type="text" value="type your message" name="add_message"></input> <input type="button" value="add message" onclick="sNotify.addToQueue($('.add_message').attr('value'))"/> Then when admin click on button then notification send to users account like when any user login then he/she able to see pop ups in user form I call this java script in page load like this .. ClientScript.RegisterStartupScript(GetType(), "Javascript", "javascript:sNotify.addToQueue($('.add_message').attr('value'))();", true); it works like when i login as a admin and click on button then notification in his own page .. but i want to show this notifications in user form. so how to solve it?

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  • ViewPager and Fragment Pager adapter implementation

    - by Rohit Deshmukh
    So I am trying to implement sliding views/fragments using viewpager and fragment pager adapter. convert_home is my main xml file that has android.support.v4.view.PagerTitleStrip and temperature.xml and velocity.xml are my two other views. I have no clue where I am going wrong. package app.converto; import android.os.Bundle; import android.support.v4.app.Fragment; import android.support.v4.app.FragmentActivity; import android.support.v4.app.FragmentManager; import android.support.v4.app.FragmentPagerAdapter; import android.support.v4.view.ViewPager; import android.view.LayoutInflater; import android.view.Menu; import android.view.View; import android.view.ViewGroup; import android.widget.TextView; public class ConverTo extends FragmentActivity { SectionsPagerAdapter mSectionsPagerAdapter; ViewPager mViewPager; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); mSectionsPagerAdapter = new SectionsPagerAdapter(getSupportFragmentManager()); mViewPager.setAdapter(mSectionsPagerAdapter); setContentView(R.layout.converto_home); mViewPager = (ViewPager) findViewById(R.id.pager); } @Override public boolean onCreateOptionsMenu(Menu menu) { getMenuInflater().inflate(R.menu.converto_home, menu); return true; } public class SectionsPagerAdapter extends FragmentPagerAdapter { public SectionsPagerAdapter(FragmentManager fm) { super(fm); } @Override public Fragment getItem(int i) { switch(i){ case 0: Fragment1 fragment = new Fragment1(); return fragment; case 1: Fragment2 fragment2 = new Fragment2(); return fragment2; } defaultFragment fragment3 = new defaultFragment(); return fragment3; } @Override public int getCount() { return 2; } // // @Override // public CharSequence getPageTitle(int position) { // switch (position) { // case 0: return getString(R.string.velocity); // case 1: return getString(R.string.temperature); // case 2: return getString(R.string.distance); // } // return null; // } } public static class Fragment1 extends Fragment{ public Fragment1(){ } @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); } @Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) { //return inflater.inflate(R.layout.temperature, container, false); View view = inflater.inflate(R.layout.temperature, container, false); TextView textView = (TextView) view.findViewById(R.id.sample); textView.setText(getArguments().getString("title")); return view; } } public static class Fragment2 extends Fragment{ public Fragment2(){ } @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); } @Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) { //return inflater.inflate(R.layout.velocity, container, false); View view = inflater.inflate(R.layout.temperature, container, false); TextView textView = (TextView) view.findViewById(R.id.sample); textView.setText(getArguments().getString("title")); return view; } } public static class defaultFragment extends Fragment{ public defaultFragment(){ }//end constructor @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); } @Override public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) { // return inflater.inflate(R.layout.temperature, container, false); View view = inflater.inflate(R.layout.temperature, container, false); TextView textView = (TextView) view.findViewById(R.id.sample); textView.setText(getArguments().getString("title")); return view; }//end oncreate }//end default fragment }

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  • How can I ignore an http request without clearing the browser?

    - by Timid Developer
    To prevent duplicate requests (i.e. pressing F5 right after clicking a command button), I've setup my page base class to ignore the request if it's detected as a duplicate. When I say 'ignore' I mean Response.End() Now I thought I've seen this work before, where there's an issue, I just Response.End() and the users page just does nothing. I don't know the exact circumstance in which this worked, but I'm unable to repeat it now. Now when I call Response.End(), I just get an empty browser. More specifically, I get this html. <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=Content-Type content="text/html; charset=utf-8"></HEAD> <BODY></BODY></HTML> I setup the following test app to confirm the problem is not elsewhere in my app. Here it is: Add the following to an aspx form <asp:Label ID="lbl" Text="0" runat="server" /><br /> <asp:Button ID="btnAdd1" Text="Add 1" runat="server" /><br /> <asp:Button ID="btnAdd2" Text="Add 2" runat="server" /><br /> <asp:Button ID="btnAdd3" Text="Add 3" runat="server" /><br /> And here's the code behind file using System; namespace TestDupRequestCancellation { public partial class _Default : System.Web.UI.Page { protected void Page_Init(object sender, EventArgs e) { btnAdd1.Click += btnAdd1_Click; btnAdd2.Click += btnAdd2_Click; btnAdd3.Click += btnAdd3_Click; } protected void Page_Load(object sender, EventArgs e) { if (!IsPostBack) CurrentValue = 0; else if (Int32.Parse(lbl.Text) != CurrentValue) Response.End(); } protected void Page_PreRender(object sender, EventArgs e) { lbl.Text = CurrentValue.ToString(); } protected int CurrentValue { get { return Int32.Parse(Session["CurrentValue"].ToString()); } set { Session["CurrentValue"] = value.ToString(); } } void btnAdd3_Click(object sender, EventArgs e) { CurrentValue += 3; } void btnAdd2_Click(object sender, EventArgs e) { CurrentValue += 2; } void btnAdd1_Click(object sender, EventArgs e) { CurrentValue += 1; } } } When you load the page, clicking any button does what is expected, but if you press F5 at any time after pressing one of the buttons, it will detect it as a duplicate request and call Response.End() which promptly ends the task. Which leaves the user with an empty browser. Is there anyway to leave the user with the page as it was, so they can just click a button? Also; please note that this code is the simplest code I could come up with to demonstrate my problem. It's not meant to demonstrate how to check for dup requests.

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