Drupal 7 ajax call - ajax

I'm beginner with drupal, I use drupal 7.x and I would like to create a simple ajax call.
I have a template file and I have a button in it with a javascript function in onclick method.
<input type='button' onclick='doajax()'>
<div id='ajaxresponsediv'></div>
In the js file:
function doajax()
{
jQuery.get(url,function(data) {
jQuery("#ajaxresponsediv").html(data);
});
}
Ok it's a simple thing. But what do I have to code in .module file ? What will be the url variable in my code? How do I implement the callback function (for example I would like to print simply 'Hello' into ajaxresponsediv ) ? Do I have to register something in mymodule_menu() function? Thanks for the guidelines.

Yes, you need to register a menu. Hope the below examples help you.
https://clikfocus.com/blog/simple-ajax-example-for-drupal-6-and-7
https://www.drupal.org/docs/7/api/javascript-api/ajax-in-drupal-using-jquery

Related

How to call wordpress plugin function with ajax

i want to make a wordpress plugin.
But i want to learn, plugin admin panel how to click and run function.
same page if you click link call function but same page ajax.
like this
<a href='javascript:void(0); class="btn2"'>
function examplefunc() {
echo 'say hello':
}
this is easy example. same page if click btn2 class call examplefunc
How to make this thank you so much.
To call java script function in hyperlink then below is the simple example;
<script type="text/javascript">
function myFunction(myMessage) {
alert(myMessage);
}
</script>
My link
Simple WordPress Ajax Example: https://gist.github.com/devinsays/69a305053e35a10584f94d6011bba2d6

JavaScript code in view issue in Laravel

I put JavaScript code in a view file name product/js.blade.php, and include it in another view like
{{ HTML::script('product.js') }}
I did it because I want to do something in JavaScript with Laravel function, for example
var $path = '{{ URL::action("CartController#postAjax") }}';
Actually everything is work, but browser throw a warning message, I want to ask how to fix it if possible.
Resource interpreted as Script but transferred with MIME type text/html
Firstly, putting your Javascript code in a Blade view is risky. Javascript might contain strings by accident that are also Blade syntax and you definitely don't want that to be interpreted.
Secondly, this is also the reason for the browser warning message you get:
Laravel thinks your Javascript is a normal webpage, because you've put it into a Blade view, and therefore it's sent with this header...
Content-Type: text/html
If you name your file product.js and instead of putting it in your view folder you drop it into your javascript asset folder, it will have the correct header:
Content-Type: application/javascript
.. and the warning message will be gone.
EDIT:
If you want to pass values to Javascript from Laravel, use this approach:
Insert this into your view:
<script type="text/javascript">
var myPath = '{{ URL::action("CartController#postAjax") }}';
</script>
And then use the variable in your external script.
Just make sure that CartController#postAjax returns the content type of javascript and you should be good to go. Something like this:
#CartController.php
protected function postAjax() {
....
$contents = a whole bunch of javascript code;
$response = Response::make($contents, '200');
$response->header('Content-Type', 'application/javascript');
....
}
I'm not sure if this is what you're asking for, but here is a way to map ajax requests to laravel controller methods pretty easily, without having to mix up your scripts, which is usually not the best way to do things.
I use these kinds of calls to load views via ajax into a dashboard app.The code looks something like this.
AJAX REQUEST (using jquery, but anything you use to send ajax will work)
$.ajax({
//send post ajax request to laravel
type:'post',
//no need for a full URL. Also note that /ajax/ can be /anything/.
url: '/ajax/get-contact-form',
//let's send some data over too.
data: ajaxdata,
//our laravel view is going to come in as html
dataType:'html'
}).done(function(data){
//clear out any html where the form is going to appear, then append the new view.
$('.dashboard-right').empty().append(data);
});
LARAVEL ROUTES.PHP
Route::post('/ajax/get-contact-form', 'YourController#method_you_want');
CONTROLLER
public function method_you_want(){
if (Request::ajax())
{
$data = Input::get('ajaxdata');
return View::make('forms.contact')->with('data', $data);
}
I hope this helps you... This controller method just calls a view, but you can use the same method to access any controller function you might need.
This method returns no errors, and is generally much less risky than putting JS in your views, which are really meant more for page layouts and not any heavy scripting / calculation.
public function getWebServices() {
$content = View::make("_javascript.webService", $data);
return (new Response($content, 200))->header('Content-Type', "text/javascript");
}
return the above in a method of your controller
and write your javascript code in your webService view inside _javascript folder.
Instead of loading get datas via ajax, I create js blade with that specific data and base64_encode it, then in my js code, I decode and use it.

How to create a simple ajax form in Drupal 7?

I want to create a simple form in Drupal7 with only one field. When hitting the submit button I need to call an ajax function with the field value and update a DIV on the page without reload.
I'm not familiar with Drupal form API, and I tried to implement the examples form the form api documentation but I always get errors like this:
Notice: Undefined index: #title_display form_pre_render_conditional_form_element() függvényben (/var/www/nmtest/includes/form.inc 2986 sor).
Notice: Undefined index: #value theme_textarea() függvényben /var/www/nmtest/includes/form.inc 3727 sor).
I can do it by creating a custom page and simply use native PHP but this is not an "elegant" way.
I want something similar like the one here, but without form valitation, page reload or so: Create a VERY simple form in Drupal
I only want one button that calls an ajax function. Shoud I do this with forms API or just use plain old native PHP and jQuery?
Edit: Looks like I don't even need a real ajax call, because I only need to change an SRC of an IMG tag.
Here's what I did in native PHP. I want to achive this trough forms api:
<script>
function getstar() {
starpoints = jQuery('#starpoints').val();
rand = Math.floor(Math.random() * 9999) + 1;
src = "test.php?star=" + starpoints + '&rand=' + rand;
jQuery('#starimg').attr("src",src);
return false;
}
</script>
<form>
<input type="text" name="starpoints" id="starpoints" />
<input type="submit" onclick="return getstar();" />
</form>
<img src="/test.php?star=5" id="starimg" />
Edit: Ok, I managed to render a form with the Drupal API. Now the only thing is that I want to insert the placeholder for the IMG via my module, but I don't know how to do it. The form renders fine, but I want to add a HTML block after it. Currently I do it by javascript:
jQuery(document).ready(function() {
jQuery('<div><br /><img id="starimg" src="" /></div>').insertAfter('#drawstar-form');
});
What hook should I catch to render this HTML block after the form? I tried drupal_get_form but that returns an array and I cannot attach the HTML block.
The Form API provides everything that you need to do this without writing a single line of JavaScript code. See http://drupal.org/node/752056 and http://api.drupal.org/api/examples/ajax_example!ajax_example.module/function/ajax_example_simplest/7 and .

AJAX replace content in DIV not by ID or class

Can some one help me..
I have a script who generates code like this >>
<div id="imageok">img1.png</div>
<div id="imageok">img22.jpg</div>
<div id="imageok">img333.gif</div>
In AJAX the name of the image is called imgname its equal to content of the div.
I want to replace the text inside the div but I can't use the id because its the same for each div.. I tried to search for div by it content (I have the name of the image in the AJAX)
Perhaps its something like..
GET element where content =="img1.png" > replace
I just can't write it correct..
Please help me .
I think what you are trying to do can be accomplished more 'neatly' with jQuery.
Have you used jQuery before? If yes, then you can do something like this:
// get content into a variable called 'result'
$('div').each(function(i) {
if ($(this).text() == 'img1.png') {
$(this).text(result);
}
});
I haven't tested this code. Hope it helps.

jQuery Ajax don't remember generated code?

I just started using Ajax with jQuery and PHP. I have a working code (below) which inserts some HTML code to a HTML container (div called nav sub).
Next time I try to run a similar code to the one below on my generated HTML, jQuery don't seem to find it. I guess it don't update it self about it when it's added.
$(".nav.top a").click(function(){
var a_class = $(this).parent().attr("class");
$(".nav.sub").html("loading...");
$(".nav.sub").load("<?php echo get_bloginfo('url'); ?>/?addmod_ajax=1",{button: a_class});
return false;
});
Let's say the generated code looks like this:
<div class="nav sub">
My new generated button, forgotten by jQuery?
</div>
And the new container looks like this:
<div class="settings"><?php # AJAX ?></div>
Is it some way to use jQuery and Ajax on HTML code generated with jQuery?
I figured it out. The solution is to use "live".
$('.nav.top a').live( 'click', function() {
var a_class = $(this).parent().attr("class");
$(".nav.sub").html("loading...");
$(".nav.sub").load("<?php echo get_bloginfo('url'); ?>/?addmod_ajax=1",{button: a_class});
return false;
});

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