when i use this code,the jquery code in the #paylasimlar does not run.i search a lot.some say
use .ajax some say do not use bind.load.but none of them works.what can i do if i want codes
run in both places?
$('#paylasimlar').load('profile.php #paylasimlar');
(shortly i submit an information and i want to load the section that these information are shown,and there are comment sections below informations whixh work with jquery)
$.load() requires a callback function, and runs the function when the item gets added to the DOM.
So generally you do:
$('#paylasimlar').load(function(){
do Something
});
is this what you are looking for? If not you may want to look at $.ajax() or $.get()
I have to load one index.php file inside another index file. Include() didn't help me at all but load() did!I have no "do something" as mentioned above.
It did at least loaed the second index.php inside the other one, but well not peoperly(the page jumps out)!
My question is that could 'not having "do something"' be the reason?
Related
I have a partial view that I have included on my _Layout.cshtml. It simply has a javascript function that changes an image based on the state of my system. I don't need to reload any data, I don't need to go to the code of the controller for anything, I simply need to reload that partial view.
I tried many of the examples that I found here but couldn't get any of them to work. I felt as if they were too complex for what I was doing anyway. Any guidance would be appreciated.
Thanks,
Steve
If the partial is loaded into the layout directly then there's no straightforward way to refresh it, because it's basically a part of the complete rendered page.
Your best bet is to render the partial using $.load or whatever equivalent you have available by hitting a controller method and rendering the result into a container (like a div). You would have to do this within a script that is loaded with the layout itself, by observing document.ready or something like that. Once you have that in place then it's trivial to keep reloading or refreshing the contents by hitting the controller method as many times as you need. For example in jQuery:
$(document).ready(function () {
RefreshPartial();
window.setInterval(RefreshPartial, 10000);
});
function RefreshPartial() {
$('#container').load('/some/controller/endpoint', {parameters});
}
This will call the controller method, and set the inner contents of the element identified with #container. You can call RefreshPartial as many times as you want.
Partial views only exist on the server. The only way to "refresh" the partial is to go back to the server to get it again.
Obviously, you must be doing something in the partial that needs refreshing. Whatever that is, should be callable from javascript to do the refresh.
I put a Ajax link using the following code:
echo chtml::ajaxLink('GO', 'http://localhost/index.php?r=user/delete', array('method'=>'POST'));
But, regardless of giving the second parameter as URL i,e 'http://localhost/index.php?r=user/delete'. It generates link with the current URL in the browser not the URL I just specified.
What is the issue? How could I create AJAX link? Google several hours but can't solve the issue.
Any kind of help is highly appreciated.
First of all, you should always try and create normalized urls.
But i think your doubt lies in the # that is generated/appended. If you go and check the source of yii ajaxLink you'll see this:
public static function ajaxLink($text,$url,$ajaxOptions=array(),$htmlOptions=array())
{
if(!isset($htmlOptions['href']))
$htmlOptions['href']='#';
$ajaxOptions['url']=$url;
$htmlOptions['ajax']=$ajaxOptions;
self::clientChange('click',$htmlOptions);
return self::tag('a',$htmlOptions,$text);
}
so if you don't set the href property of the a tag in the htmloptions array, the # will be appended.
You should also understand that yii uses jquery, so if you check out the source of the page, you'll see at the bottom, how jquery is used to carry out an ajax request, your actual url that is called will also be seen in that script. So the third option/parameter in ajaxLink is for options for jquery's ajax function. You can create better ajax links using this option.
Regardless of where(which controller) your url points to in your project, the action associated with that url will be called.
So anyway, you can modify your code like this if you want the url to be shown and not a # :
echo CHtml::ajaxLink('GO', 'http://localhost/index.php?r=user/delete',
array('type'=>POST), //there are various other options for jquery ajax
array('href'=>'http://localhost/index.php?r=user/delete'));
To make better ajax links i would suggest going through jquery's ajax documentation. There is an option for a success function, that you can use to let the user know that the operation was completed.
Hope this helps, don't hesitate to leave comments if i haven't answered your question completely.
Have you tried:
echo CHtml::ajaxLink('GO', array('/user/delete'), array('method'=>'POST'));
as the ajaxLink documentation suggests...? Look also at the normalizeUrl method.
Using these methods, which in turn are using createUrl, is usually better since it will take care to create a valid url for your site.
I had the same issue(or maybe similar).
I've used renderPartial to load view and later in that view i was using ajaxLink and it was not working.
What i have found, that when using renderPartial, there was no jquery script for ajax action.
What you have to do is to add 4th argument(true) in renderPartial function to generate jquery script.
See the documentation: http://www.yiiframework.com/doc/api/1.1/CController/#renderPartial-detail
Hope it helps and saves time to figure it out.
I have searched on this matter but havent been able to find an answer that work for me.
On my website I have one page with a menu and a div. When clicking on a menu link the information in the div changes (ajax). But then Cufón stop working in that div. I have read about Cufon.refresh but can't get it to work. Which code should I use and where do I put it?
My links look like this:test
I have read about Cufon.refresh at this page but I don't understand how to use it: https://github.com/sorccu/cufon/wiki/API
Cufon.refresh also does not work for me. A workaround is to use Cufon.replace again.
Based on the JS you are using, you can edit the trigger function to work like the below and see if that works. Essentially, it will refresh the Cufon call after you have updated the content on that element.
if (xmlhttp.readyState == 4) if (xmlhttp.status == 200)
{
document.getElementById(ajaxdestination).innerHTML =xmlhttp.responseText;
Cufon.refresh('#ajaxdestination');
}
I am trying to use CodeIgniter and jQuery-ui dialog to create a modal window with form to update user information.
The process should be like:
1. Press a button on a view page.
2. A modal window pops up.
3. Inside the window is a form that a user can fill.
4. If the user filled something before, the information should be shown in corresponding field
5. Click the update button on the modal window to save the changes to database.
Can anyone provide a good sample of this process?
I used ajax to pass the data but it didn't work when I was trying to update the data to the database. It would be nice if an example of how to pass data from ajax to php and how php handle that.
Thanks,
Milo
well the jquery bit for post(), get(), ajax() works the same in any measure you would normally use it.. key difference here is with CI you can't post directly to a file-name file-location due to how it handles the URI requests. That said your post URL would be the similar to how you would access a view file normally otherwise
ie: /viewName/functionName (how you've done it with controllers to view all along. post, get, ajax doesnt have to end in a extension. I wish I had a better example then this but I can't seem to find one at the moment..
url = '/home/specialFunction';
jQuery.get(url, function(data) {
jQuery("#div2display").html(data);
});
in the case of the above you notice despite it not being a great example that. you have the url with 2 parameters home and specialFunction
home in this case is the controller file for home in the control folder for the home file in views the specialFunction is a "public function" within the class that makes the home controller file. similar to that of index() but a separate function all together. Best way I have found to handle it is through .post() and a callback output expected in JSON cause you can form an array of data on the php side json_encode it and echo out that json_encode and then work with that like you would any JSON output. or if your just expecting a sinlge output and not multiples echoing it out is fine but enough of the end run output thats for you to decide with what your comfortable doing currently. Hopefully all around though this gives you some clairity and hopefully it works out for you.
i am using cakephp in my project. in this at one section i need to update the particular row onclick of the image. using ajax. i used mootools as javascript library. so please help me how could i do this.
thanks in advance
Simply speaking:
Create a CakePHP controller action that performs the row update.
Determine the URL of the controller action you just created. (ie. /controllername/actionname)
Determine if you need to do a GET or POST request to this URL for it to work.
Put code in your view that attaches an "onclick" event that makes and AJAX (GET/POST) request to the above controller.
CakePHP has a javascript helper that traditionally produced Prototype code, but in v1.3 it is now able to produce code for other Javascript frameworks (such as Mootools, jQuery, etc.)
However, many suggest writing your javascript in javascript (eg. actually using the Mootools framwork), rather than writing your javascript in PHP (like using CakePHP's helper to produce Mootools code).
Either way, in your view you need to have something like: <?php echo $js->link(.. or <script>Moo.. or <a onclick="Moo.. to attach your Javascript to that link.
You may also wish for your controller action to return some sort of response indicating whether or not the row update failed or succeeded. In that case you need to make sure the CakePHP controller action you are calling has a view that outputs this. JSON seems to be the ideal format for this (eg. { success: true }), but you need to remember to turn off Cake's debug output. This response can be captured into a variable by your Mootools code where you can decide what to do with it (eg. displaying an error).
As i know most programmer work with protype.js library.
i am giving you link see
go to there