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  • How to Get Your Website on Top of Google

    Have you ever searched for something then just clicked the very first thing that came to the top? Well even if you haven't plenty of other people out there do this. So if they search for your site's category then you better be the one at the top of the list. If not first then let's try to make it on the first page at least.

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  • Vertically Center - with unknown height

    - by panthro
    I have two div's side by side. On the left is an image, on the right are inputs. The image varies depending on what the user uploads. How can I vertically centre align the image and the inputs? I would like the inputs to appear vertically centre to the image. Both the img and inputs have their own container: <div class="img-container"> <div class="data-container"> Fiddle: http://jsfiddle.net/vbLht/

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  • How can I get sessions to work if I'm using Google App Engine + Django 1.1?

    - by user341642
    Is there a way for me to get sessions working? I know Django has built in session management, and GAE has some tools for it if you're using their watered down version of Django 0.96, but is there a way to get sessions to work if you're trying to use GAE w/ Django 1.1 (i.e. use_library() call). I assume using a db-backed session doesn't work, and a file system backed one won't work b/c we don't have access to the filesystem if we deploy to the Google production servers. This kinda worked (as in didn't crap out) when I used SessionMiddleware backed by a local-memory backed cache and a non-persistent cache (i.e. setting SESSION_ENGINE to django.contrib.sessions.backends.cache). But the session never seems to persist in this case, no matter how I set the timeouts. A new session key is generated on every page reload. Maybe this is b/c the GAE assumes complete statelessness with each request and blows away my local cache? Apologies in advance, I'm pretty new to Python. Any suggestions would be greatly appreciated.

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  • Google Chrome Extensions For SEO

    With more than 12% of internet users now using Chrome, it has obviously become the browser of choice for many power users. As a result, a number of extremely useful plug-ins have been created for Chrome, including many which are highly relevant to SEO professionals. What follows is a sampling of the most useful.

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  • jQuery toggle() with unknown initial state

    - by Jason Morhardt
    I have a project that I am working on that uses a little image to mark a record as a favorite on multiple rows in a table. The data gets pulled from a DB and the image is based on whether or not that item is a favorite. One image for a favorite, a different image if not a favorite. I want the user to be able to toggle the image and make it a favorite or not. Here's my code: $(function () { $('.FavoriteToggle').toggle( function () { $(this).find("img").attr({src:"../../images/icons/favorite.png"}); var ListText = $(this).find('.FavoriteToggleIcon').attr("title"); var ListID = ListText.match(/\d+/); $.ajax({ url: "include/AJAX.inc.php", type: "GET", data: "action=favorite&ItemType=0&ItemID=" + ListID, success: function () {} }); }, function () { $(this).find("img").attr({src:"../../images/icons/favorite_not.png"}); var ListText = $(this).find('.FavoriteToggleIcon').attr("title"); var ListID = ListText.match(/\d+/); $.ajax({ url: "include/AJAX.inc.php", type: "GET", data: "action=favorite&ItemType=0&ItemID=" + ListID, success: function () {} }); } ); }); Works great if the initial state is not a favorite. But you have to double click to get the image to change if it IS a favorite initially. This causes the AJAX to fire twice and essentially make it a favorite then not a favorite before the image responds. The user thinks he's made it a favorite because the image changed, but in fact, it's not. Help anybody?

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  • Determining unknown content-types with the Html5 file api

    - by Jesse
    I'm working through a small file upload script (learning experience) and I noticed that when selecting microsoft office related files (.doc or .docx for example) the file objects do not have a type specified: For .doc files I had expected the type to be "application/msword" and along the same train of thought .docx to be "application/vnd.openxmlformats-officedocument.wordprocessingml.document". In the cases when the type cannot be determined is the correct course of action to look at the file extension and match that to the "expected" content / mime type? Sample script: <div id="fileUpload"> <input type="file" id="fileElem" style="display:none;" onchange="handleFiles(this.files)"/> <a href="#" id="fileSelect">Select some files</a> </div> <script type="text/javascript"> var fileSelect = document.getElementById("fileSelect"), fileElem = document.getElementById("fileElem"); fileSelect.addEventListener("click", function (e) { if (fileElem) { fileElem.click(); } e.preventDefault(); }, false); function handleFiles(files) { console.log(files); } </script>

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  • Create object of unknown class (two inherited classes)

    - by Paul
    I've got the following classes: class A { void commonFunction() = 0; } class Aa: public A { //Some stuff... } class Ab: public A { //Some stuff... } Depending on user input I want to create an object of either Aa or Ab. My imidiate thought was this: A object; if (/*Test*/) { Aa object; } else { Ab object; } But the compiler gives me: error: cannot declare variable ‘object’ to be of abstract type ‘A’ because the following virtual functions are pure within ‘A’: //The functions... Is there a good way to solve this?

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  • Get to the Top of Google With Blogging

    There are many websites out there all trying to get noticed by the search engines which will in turn help them to be found by users. Every website wants more traffic coming into it whether it is providing information or selling products.

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  • 1054 - Unknown column 'apa_calda' in 'where clause'

    - by sebastian
    Hi, I keep getting this error in mysql. Here is the query: SELECT user_id FROM detalii_contor WHERE tip_contor=apa_calda i want to use this query in a php file but it doesn't give any result. so i tried to write it in the sql command prompt. here is what i tried in the php file: $Q = "SELECT id_contor, den_contor FROM detalii_contor WHERE tip_contor='".$contor."'"; $Q = "SELECT id_contor, den_contor FROM detalii_contor WHERE tip_contor='$contor'"; even without "" or without '' i wanted to get $contor from a form, i also tried with $_POST['util'] and {$_POST['util']} i've also tried to set $contor the value i need, but no result. please help. thanks, Sebastian

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  • Google Indexing Gets You Noticed

    Getting attention on the online world is not as easy as it might seem and your website has to do a lot to get its share of visitors. The internet has so many websites and if someone wants to go through your website they eventually have to browse through various other businesses like yours.

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