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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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  • 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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  • 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 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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  • Using formLabel in FormPanel for Localization

    - by user1782712
    I am trying to implement Localization within a panel, so to make the fieldLabel accessable from another file I am defining it like this Ext.define('Ext.app.detailForm', { extend: 'Ext.form.Panel', id: 'cscForm', // frame: true, uncomment this to show blue frame background title: 'CSC Details :', //trying cscCode: 'CSC Code', bodyPadding: 5, layout: 'anchor', // Specifies that the items will now be arranged in columns autoScroll: true, //width:1200, //height: 300, collapsible: true, fieldDefaults: { labelAlign: 'left', msgTarget: 'side' }, items: [{ columnWidth: 0.4, xtype: 'container', layout:'anchor', defaults: { labelWidth: 150 }, defaultType: 'textfield', items: [{ xtype: 'container', layout: 'hbox', items:[{ xtype: 'fieldset', columnWidth:1.5, layout: 'column', width:1050, defaultType: 'textfield', defaults: { labelWidth: 150, margin: '3 0 0 10' }, items: [{ fieldLabel: this.cscCode, name: 'CSCCode', width: 500 }, {..... but when I try to render this panel, the formLabel for cscCode does not display, is there some thing which I am doing wrong here? I am basically not able to access "this.cscCode"

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  • OpenGL and layouts

    - by Hnefi
    I'm using OpenGL to render a game view in my android application. The game is turn based and I wish to add some buttons to the interface. I'd prefer to use standard Android widgets, structured in an XML-generated layout (or, if I have to, a hardcoded layout) and put the OpenGL view in its own window as part of that layout. So in regards to this, I have 3 questions: 1: Is such a thing possible? I've done a few half-hearted tries, but have had no luck so far. 2: Is such a thing advisable? Does it carry a significant performance penalty, for example, over using OpenGL-based homebrew widgetry? 3: Is it possible to pass particular arguments to instances created in XML layouts? For example, my current OpenGL view has three arguments in its constructor; is it somehow possible for me to invoke that particular constructor with particular parameters when it's part of a layout?

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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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  • How to restrict bounds for Translate Animation for a view in Android?

    - by Kiran Parmar
    Hello All. Let me explain the scenario that I want to achieve:- Consider the below as the Layout I have inside a Parent_Linearlayout: [Linear Layout] (Fill_Parent, Wrap_Content) [ScrollView] Activity's setContentView is set to the Parent_Linearlayout In the application, when a condition is met, I want the Scrollview to be removed from the screen and instead put another View in its place.<br> I've been able to do this, When I remove the ScrollView, I'm applying translate Animation to it so that it seems as if the View has gone to the top, before removing it. But when the animation occurs, the ScrollView translates OVER the Linear layout present above it. How do I restrict it, so that the scrollview does not go over the linear layout, but disappears at the base of the Linearlayout. I want the linearlayout to always stay visible.. I've been trying to do this from quite some time, but I've not been able to get desired results.. Could someone kindly help me out here??

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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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  • Multiple Rails forks with separate designs and layouts

    - by mettadore
    I have a Rails project that is basically a simple web app for a membership-based organization. We've open sourced the code on Github for the web app so that others can use it, but have a licensed design/layout that the original organization is going to use. This layout cannot be open sourced. I was wondering if others have run into the situation where you have an open-source Rails app with a non-OS design. My initial thought is to put app/views in .gitignore, and to have anyone forking the code add their own views directory, perhaps including an app/views_default directory with a web-app-theme layout or something else to get people running. Is this the best option (realizing that there are other files such as JavaScript, CSS, etc that come with the layout that must also be ignored). Does anyone have some good thoughts or pointers on this?

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  • How do I (or should I?) access the service layer from a SiteMesh template (views/layouts/main.gsp) in Grails?

    - by knorv
    I need to create a toplist in the page footer on a site that I'm building. The footer is created in the default SiteMesh layout template (views/layouts/main.gsp). In order to create the toplist access to the database is needed, so I've encapsulated all logic needed for the toplist creation in a service class (services/FooService). Please note that while services are usually accessed from the controller layer, in this case the default layout template (views/layouts/main.gsp) is not generated from a controller. Can the layout view (views/layouts/main.gsp) access a service class? How? Is this the correct design decision? If not, what is a better encapsulation and how do I interact with said encapsulation from the layout view (views/layouts/main.gsp)?

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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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  • Spinner activity not working

    - by user1696863
    I'm trying to create an activity, RateCardActivity, which has a spinner in it. My layout file for RateCardActivity is rate_card. My RateCardActivity looks like the following. public class RateCardActivity { public void onCreate(Bundle bundle) { super.onCreate(bundle); setContentView(R.layout.rate_card); Spinner spinner = (Spinner) findViewById(R.id.select_city); ArrayAdapter<CharSequence> adapter = ArrayAdapter.createFromResource(this, R.array.select_city, android.R.layout.simple_spinner_item); adapter.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item); spinner.setAdapter(adapter); } } The layout file rate_card is: <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" xmlns:custom="http://schemas.android.com/apk/res/com.olacabs.customer" android:layout_width="fill_parent" android:layout_height="fill_parent" android:orientation="vertical" > <TextView android:id="@+id/textView1" android:layout_width="fill_parent" android:layout_height="wrap_content" android:background="@android:color/darker_gray" android:gravity="center" android:paddingBottom="4dp" android:paddingTop="4dp" android:text="@string/rate_card" android:textColor="@color/white" android:textSize="20dp" custom:customFont="litera_bold.ttf" /> <Spinner android:id="@+id/select_city" android:layout_width="fill_parent" android:layout_height="wrap_content" /> </LinearLayout> The RateCardActivity is called from another activity using an intent (I'm sure there is nothing wrong with that part of the code as when I substitute RateCardActivity with another activity, the application works fine). When I try to open the RateCardActivity in the application in emulator, the application crashes and I got the message "The application has stopped unexpectedly. Please try again later." I can't seem to understand what I'm doing wrong, and want to know how to correct this?

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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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  • Can I embed a .png image into an html page?

    - by Ole Jak
    So I have a .png file. I am using windows OS. How can I embed my png file/image into (blank by default) file.html so that when you open that file in any browser you see that image, but the file is not anyhow linked to it - it is ebbeded into it? Step by step instructions would be nice.

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