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  • How are you supposed to layout a page in VS2010 without using tables?

    - by CrustyApple
    I have been using .NET since beta and HTML since the days of HotDog pro & notepad, using table layout of course. I am FINALLY ready to use only div, li, CSS for the layout, but my question is, what is the proper way to layout pages in VS2010? When i use table layout its simple and i can visually see what im creating and where the elements are, such as the sample below - how should I do this using div's, etc in VS2010? <table width="300" border="0" cellpadding="5"> <tr> <td><img src="http://assets.devx.com/MS_Azure/azuremcau.jpg" alt="blah" width="70" height="70" /></td> <td><h2>This is some text to the right of the picture...</h2></td> </tr> <tr> <td colspan="2">Here some text underneath</td> </tr> </table>

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  • Is it possible to serve an ASPX page without it setting a cookie on your browser?

    - by Django Reinhardt
    Hi, we're in the process of trying to speed up the performance of our website by serving static content from a cookieless domain. That seems to be going well, but I have a new question: I know that it's "static content" that we're talking about when serving it from a cookieless domain, but we also have static content being served by ASPX pages, specifically images. For example: domain.com/resizeImages.aspx?src=images/image123.jpg&width=400&height=400 Pretty standard stuff, and although it's being served by managed code, it's still a static image. So my question is: Is it ok to serve the resizeImages.aspx image from our cookieless/static domain? And if so, how do I go about stopping ASP.NET from setting a ANONYMOUSASPX cookie every time I try? Thanks for any help!

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  • How do I get rid of the space between <img> elements in a row in a html page?

    - by Aperture
    I'm displaying 3 <img> in a row like this: <div style="width: 950px"> <img src='/UploadedImages/86.jpg' alt='' style="width: 300px; margin: 0px; padding: 0px; border: 1px solid Black" /> <img src='/UploadedImages/85.jpg' alt='' style="width: 300px; margin: 0px; padding: 0px; border: 1px solid Black" /> <img src='/UploadedImages/84.gif' alt='' style="width: 300px; margin: 0px; padding: 0px; border: 1px solid Black" /> </div> As you can see I have thin black borders around the images. My problem is that there are white spaces about 5px wide between the borders of neighbouring images and I have set the margin to be 0px but it does not work. So what is happening here?

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  • How can I close a SimpleModal dialog when the user clicks the Back button?

    - by Wally Hartshorn
    When a user clicks on a button or link, I use the SimpleModal jQuery plugin to display a dialog to overlay the entire page, preventing the user from clicking another button or link during the delay before the next page loads. (I'd like to avoid this, but that's an issue for another day.) After the next page displays, if the user clicks the Back button, the previous page still has the SimpleModal overlay displayed, preventing them from using the page. This is a problem. How can I cause the SimpleModal dialog to close automatically either when the leaves the page or when the user clicks the Back button to return to the page? I tried this without success: $("body").unload(function() { $.modal.close(); }); Thanks! Wally

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  • How can I load a web page inside of a jQuery dialog box?

    - by kielie
    Hi, I am designing a wordpress template, and I would like to use jQuery ui to display each of the calendar links in a dialog box, when they are clicked. What I want, is when someone clicks on a link in the wordpress calendar, the content will load in the dialog box, and display over the calendar. I would appreciate any help on this, thanx in advance!

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  • Submiting a Form without Refreshing the page with jQuery and Ajax Not updating MySQL database.

    - by HEEEEEEELP
    I'm a newbie to JQuery and have a problem, when I click submit button on the form everything says registration was successful but my MYSQL database was not updated everything worked fine until I tried to add the JQuery to the picture. Can someone help me fix this problem so my database is updated? Thanks Here is the JQuery code. $(function() { $(".save-button").click(function() { var address = $("#address").val(); var address_two = $("#address_two").val(); var city_town = $("#city_town").val(); var state_province = $("#state_province").val(); var zipcode = $("#zipcode").val(); var country = $("#country").val(); var email = $("#email").val(); var dataString = 'address='+ address + '&address_two=' + address_two + '&city_town=' + city_town + '&state_province=' + state_province + '&zipcode=' + zipcode + '&country=' + country + '$email=' + email; if(address=='' || address_two=='' || city_town=='' || state_province=='' || zipcode=='' || country=='' || email=='') { $('.success').fadeOut(200).hide(); $('.error').fadeOut(200).show(); } else { $.ajax({ type: "POST", url: "http://localhost/New%20Project/home/index.php", data: dataString, success: function(){ $('.success').fadeIn(200).show(); $('.error').fadeOut(200).hide(); } }); } return false; }); }); Here is the PHP code. if (isset($_POST['contact_info_submitted'])) { // Handle the form. // Query member data from the database and ready it for display $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.*, contact_info.* FROM users INNER JOIN contact_info ON contact_info.user_id = users.user_id WHERE users.user_id=3"); $user_id = mysqli_real_escape_string($mysqli, htmlentities('3')); $address = mysqli_real_escape_string($mysqli, htmlentities($_POST['address'])); $address_two = mysqli_real_escape_string($mysqli, htmlentities($_POST['address_two'])); $city_town = mysqli_real_escape_string($mysqli, htmlentities($_POST['city_town'])); $state_province = mysqli_real_escape_string($mysqli, htmlentities($_POST['state_province'])); $zipcode = mysqli_real_escape_string($mysqli, htmlentities($_POST['zipcode'])); $country = mysqli_real_escape_string($mysqli, htmlentities($_POST['country'])); $email = mysqli_real_escape_string($mysqli, strip_tags($_POST['email'])); //If the table is not found add it to the database if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"INSERT INTO contact_info (user_id, address, address_two, city_town, state_province, zipcode, country, email) VALUES ('$user_id', '$address', '$address_two', '$city_town', '$state_province', '$zipcode', '$country', '$email')"); } //If the table is in the database update each field when needed if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE contact_info SET address = '$address', address_two = '$address_two', city_town = '$city_town', state_province = '$state_province', zipcode = '$zipcode', country = '$country', email = '$email' WHERE user_id = '$user_id'"); } if (!$dbc) { // There was an error...do something about it here... print mysqli_error($mysqli); return; } } Here is the XHTML code. <form method="post" action="index.php"> <fieldset> <ul> <li><label for="address">Address 1: </label><input type="text" name="address" id="address" size="25" class="input-size" value="<?php if (isset($_POST['address'])) { echo $_POST['address']; } else if(!empty($address)) { echo $address; } ?>" /></li> <li><label for="address_two">Address 2: </label><input type="text" name="address_two" id="address_two" size="25" class="input-size" value="<?php if (isset($_POST['address_two'])) { echo $_POST['address_two']; } else if(!empty($address_two)) { echo $address_two; } ?>" /></li> <li><label for="city_town">City/Town: </label><input type="text" name="city_town" id="city_town" size="25" class="input-size" value="<?php if (isset($_POST['city_town'])) { echo $_POST['city_town']; } else if(!empty($city_town)) { echo $city_town; } ?>" /></li> <li><label for="state_province">State/Province: </label> <?php echo '<select name="state_province" id="state_province">' . "\n"; foreach($state_options as $option) { if ($option == $state_province) { echo '<option value="' . $option . '" selected="selected">' . $option . '</option>' . "\n"; } else { echo '<option value="'. $option . '">' . $option . '</option>'."\n"; } } echo '</select>'; ?> </li> <li><label for="zipcode">Zip/Post Code: </label><input type="text" name="zipcode" id="zipcode" size="5" class="input-size" value="<?php if (isset($_POST['zipcode'])) { echo $_POST['zipcode']; } else if(!empty($zipcode)) { echo $zipcode; } ?>" /></li> <li><label for="country">Country: </label> <?php echo '<select name="country" id="country">' . "\n"; foreach($countries as $option) { if ($option == $country) { echo '<option value="' . $option . '" selected="selected">' . $option . '</option>' . "\n"; } else if($option == "-------------") { echo '<option value="' . $option . '" disabled="disabled">' . $option . '</option>'; } else { echo '<option value="'. $option . '">' . $option . '</option>'."\n"; } } echo '</select>'; ?> </li> <li><label for="email">Email Address: </label><input type="text" name="email" id="email" size="25" class="input-size" value="<?php if (isset($_POST['email'])) { echo $_POST['email']; } else if(!empty($email)) { echo $email; } ?>" /><br /><span>We don't spam or share your email with third parties. We respect your privacy.</span></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="contact_info_submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form>

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  • Using the filename for GET data and making the PHP page output as a JPG extension?

    - by Rob
    Alright, currently I'm using GD to create a PNG image that logs a few different things, depending on the GET data, for example: http://example.com/file.php?do=this will log one thing while http://example.com/file.php?do=that will log another thing. However, I'd like to do it without GET data, so instead http://example.com/dothis.php will log one thing, and http://example.com/dothat.php will log the other. But on top of that, I'd also like to make it accessible via the JPG file extension. I've seen this done but I can't figure out how. So that way http://example.com/dothis.JPG will log one thing, while http://example.com/dothat.JPG logs the other. The logging part is simple, of course. I simple need to know how to use filenames in place of the GET data and how to set the php file to be accessible via a jpg file extension.

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  • How can I change the link format in will_paginate for page_cache in Ruby on Rails?

    - by jaehyun
    I want to use page_cache with will_paginate. There are good information on this page below. http://railsenvy.com/2007/2/28/rails-caching-tutorial#pagination http://railslab.newrelic.com/2009/02/05/episode-5-advanced-page-caching I wrote routes.rb looks like: map.connect '/products/page/:page', :controller => 'products', :action => 'index' But, links of url are not changed to '/products/page/:page' which are in will_paginate helper. They are still 'products?page=2' How can i change url format is in will_paginate?

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  • How to remove back button action while page is loading.

    - by user133611
    Hi All In my project i have a list when select on a item it will take to next controller using PushViewController. When i go there i will get data using libxml parser. But when i am clicking on the back button while loading it is showing exception. So i want to disable back button action till loading of data is completed how can i handle it. Thank You

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  • Is there a way to dynamically define the height and width that are going to appear in a page?

    - by Starx
    For so many time, I have encountered problems with managing image having abnormally long height or width. If I fixed their height and widht, they will appear streched? If I fixed their width, and if the height of the image is very long then also it will mess up the overall website. If I fixed their height, and if the width of the image is very long then also it will mess up the overall website. How is the best way to fix this?

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  • jQuery Dynamic Page Loading wont work, not sure why any ideas?

    - by Luke
    Live demo here <- http://webcallonline.exoflux.co.uk/html/ $(function() { var url = $(this).attr("href"); $("nav").delegate("a", "click", function(event) { event.preventDefault(); window.location.hash = $(this).attr('href'); $("#main").slideUp('slow', function(){ $("#main").load(url + " #main", function() { $("#main").slideDown('slow'); }); }); }); $(window).bind('hashchange', function(){ newHash = window.location.hash.substring(1); }); $(window).trigger('hashchange'); }); Does anyone have any ideas?

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  • Why is my web page left-aligned on iPad?

    - by Andrew
    I recently built a site and centered it using margin: 0 auto. I also wrapped elements in a .wrapper class with a width set to 960px and then had the parent element extend across the whole browser. When I view the Brands screen on an iPad though, the site is left-aligned and does not extend across the whole window. Any thoughts to why this might be happening, and how to correct it? See below for a screenshot:

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  • How do you automatically refresh part of a page automatically using Javascript or AJAX?

    - by Ryan
    $messages = $db->query("SELECT * FROM chatmessages ORDER BY datetime DESC, displayorderid DESC LIMIT 0,10"); while($message = $db->fetch_array($messages)) { $oldmessages[] = $message['message']; } $oldmessages = array_reverse($oldmessages); ?> <div id="chat"> <?php for ($count = 0; $count < 9; $count++) { echo $oldmessages[$count]; } ?> <script language="javascript" type="text/javascript"> <!-- setInterval( "document.getElementById('chat').innerHTML='<NEW CONTENT OF #CHAT>'", 1000 ); --> </script> </div> I'm trying to create a PHP chatroom script but I'm having a lot of trouble getting it to AutoRefresh The content should automatically update to , how do you make it do that? I've been searching for almost an hour

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  • How to load the whole content of a page into another window?

    - by Cristian Castiblanco
    I'm building a Dashboard and for some stupid reasons my boss wants to load it in a frame on the homepage (yes a frame, he still lives in the 1990s). Anyway, sometimes the dashboard needs some room so that it can show all charts correctly, so I want to add a feature to load the content of the dashboard into a new window. The problem is that, if the user has had some interaction with the dashboard, it will contain modal dialogs, new images, etc... so I want to load all the dashboard content into a new window without reloading its content. Of course, the user should be able to continue browsing the dashboard without problems. How can I do that? I'm using jQuery as my JavaScript framework.

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  • What is the best way to create a debugging web page for a computation in Java?

    - by Shooshpanchick
    I'm developing a website that uses some complex computations (NLP-related). My customer wants to have "debugging" webpages for some of these computations where he can run them with arbitrary input and see all the intermediate results that occur during computation. Before this request all of the computations were encapsulated in beans and intermediate results were logged into general log. What is the best way to capture all these results on Java level to render them as webpage?

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