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
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I want to update my MVC3 view Using Ajax,
My Code is here Here
Please tell me how i'll add Ajax in my code to update a content of HTML table
The basic outline would be:
Client Side
Call a JavaScript function from your update button.
Within the JavaScript function, use jQuery to make an AJAX call to
an action method on the server.
On success, use the returned JSON to generate the updated HTML and update your containing div, probably using a jQuery selector and the html() method.
Server Side
Create an action method with a return type of JsonResult.
Retrieve whatever data it is you need from wherever.
Return the data using return Json(myData) (you may find it easier to use the JSON.NET Newtonsoft library to get more control over serialising your objects into JSON).
Have a Google into these various steps, completely overhaul your code and see how you get on.
I want to create a page with will be filled with dynamic info using Ajax (JQuery). The info will come from various GETs I need to do in other URLs.
I'll be using Sinatra + JQuery to to that, but as my WEB experience is almost null and don't have any idea how do to it right.
The requisites for this are:
Each time a GET completes, a new line of information should appear on the page.
If the GET could not be complete, a default info appear on the page.
My idea so far is to do something like this:
Have my controller performing each GET inside a thread.
Each time a thread ends, with success or not, I inform the view of the result and render a partial
I'll have as many partial as I need (for each GET I must do)
The first time I load the page I fill in the default info, them I update via AJAX with the successful GET responses
This does not seem the correct approach, so I'm asking someone that already did something similar or has more experience on this some help.
You start off with a simple get('/'){} route that holds the default message (or any other GET route). Then you have your other GET routes that you want to display on your default route. In Sinatra you can check whether a request is an xhr-request or not with a request.xhr? If you have an xhr request you return a json value to your view, otherwise reject the request or render a view with proper html. This is on your sinatra backend. In your views you can use JQuery or any other JS library or plain JS to handle asynchronous data requests. You can use the ajax function in JQuery to request data from your routes and then add them to your DOM. It's as simple as that :)
Now you will have to investigate on the JQuery site how to make ajax requests and how to append data to existing DOMs. That's all there is to it.
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.
What is the best way to address an AJAX script that sends data to POST in codeigniter? Right now I am loading a view with the AJAX through $this->load->view('AJAX', $data); however there is no UI or user actions in the view. It's simply running through the script and returning POST data a little after the script loads. I receive the POST data in my model where I input the values into the DB and output some other values based on the data.
I need to open a real view, set metatags and re-direct the user to another website afterwards.
How do I address this?
The problem I'm facing is that I cannot open up another view because the AJAX view is the one that's in focus but I need this AJAX view to be temporary that basically does it's thing and sends to POST.
Is there any convention that I can lookup/research to what I'm describing? Let me know what kind of clarification is needed if any.
Some people like to write "ajax" controllers and post to them exclusively, but you don't need to do that. You can handle the request in the same controller that handles the non-ajax request. Personally, I exclusively return json, but you can return chunks of HTML if that works better for you.
Your exact problem is vague (actual code would help clarify), but I think you are on the wrong track. Don't use a view for processing anything ever. Use your Controller layer, this is for handling input and requests.
Example of controller method responding to either ajax or non-ajax request:
function edit_user()
{
$data['status'] = $this->user_model->save();
if ($this->input->is_ajax_request())
{
// return json string with our update status
// Something like: {"status":true}
echo json_encode($data);
exit;
}
// Load the non ajax view with the same data
$this->load->view('users/edit', $data)
}
$this->input->is_ajax_request() is a function of the Input class that reads $_SERVER['HTTP_X_REQUESTED_WITH'] and checks if it's value is XMLHttpRequest. This should only be true if it's an "ajax" request.
You can make life easier by wrapping this in a class or function. No matter what you decide to do, don't use the view layer for processing data.
I think my problem is, how do I address javascript without a view? how do I call the script and/or where do I put the JS code in the controller? I felt it was the wrong direction to address the code in a view but I didn't see how else to do it.
Whenever possible, you should put javascript code in a .js file and use a <script> tag to load it, in an HTML document. The only other exception is putting it in a "view" file (a file that's only purpose is to construct your final HTML output). In other words, follow the same rules of HTML as to where to put javascript, and follow the usual conventions of MVC of where HTML belongs (in the view). Javascript code does not belong in your controller. Javascript is not processing your data, it is sending the data to the server.
I need to open a real view, set metatags and re-direct the user to another website afterwards.
If you want to load a view, then redirect (after a certain amount of time I assume), you can do it with javascript or a <meta> tag (but don't use a meta tag, use js).
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.