I am using a class to do some CRUD stuff on a database, this one (http://net.tutsplus.com/tutorials/php/real-world-oop-with-php-and-mysql) I am going to use jquery to check the if the username has been registered already.
I could just create a php file specifically for that task, but would just like to extend the class and create a method callled checkname().
How can I call this in jquery?
You can use jQuery to make ajax call to a php file following:
PHP [suppose, test.php]
<?php
class ABC extends XYZ {
public function checkname() {
if(isset($_POST) && !empty($_POST['name'])) {
echo json_encode(array('status' => 'done'));
}
}
}
$ins = new ABC();
$ins->checkname(); // calling the function checkname
?>
jQuery:
$.ajax({
url: 'test.php', // write the url correctly
type: 'post',
data: "name=XYZ&location=PQR"
}).done(function(response) {
console.log(response.status); // will log done
}).fail(function(jqXHR, textStatus) {
console.log("Failed: " + textStatus);
});
It is just an example.
You'll need to use jQuery's Ajax functionality to access a PHP page that calls that function. You can't directly call a PHP function via Ajax, something on the backend has to be set up.
Related
i'm new to codeigniter i can't not get data from the controller using the ajax request i think i do mistake in writing the url of the controller function in ajax call
here is the code of my ajax call
$(document).ready(function(){
$("#fname").focusout(function(){
// alert();
$.ajax({
url: "<?php echo base_url();?>/proposal/ajax_load",
type: 'POST',
success: function(result){
$("#div1").html(result);
}
});
});
});
Here is my controller
class Proposal extends CI_Controller {
public function ajax_load()
{
return ("Hello");
}
}
You are confuse between the meaning of [Return, Echo] in PHP,
Echo
echo — Output one or more strings
Return
return returns program control to the calling module. Execution
resumes at the expression following the called module's invocation.
and as long as the Ajax response callback is reading a server response [output], you must send an output to the server.
public function ajax_load()
{
echo "Hello";
}
Further reading :-
What is the difference between PHP echo and PHP return in plain English?
Difference between php echo and return in terms of a jQuery ajax call
a short and simple answer
in ajax_load() - Should be echo not return if you're getting a response via ajax.
I have a function that adds social buttons to my blog posts , but once i load more posts using ajax I cant figure out how can I call add_social_buttons() and pass the data to div.
I'm not really familiar with ajax , i tried this method :
$.ajax({
type:"POST",
url:"functions.php",
data: "social_sharing_buttons()",
success: function(data){
$('.pp').html(data);
}
but it seems that it tries to invoke some totally other function Fatal error: Call to undefined function add_action().
As far as I am aware, you can't. What you can do is have a handler file for your classes, so for example say we have this PHP class,
<?php
class Car {
function getCarType() {
return "Super Car";
}
}
?>
Then in your handler file,
<?php
require_once 'Car.php';
if(isset($_POST['getCarType'])) {
$car = new Car();
$result = $car->getCarType();
echo $result;
}
?>
You'd post your AJAX request to the handler, you could make specific handlers for each request or you could have a generic AJAX handler, however that file could get quite big and hard to maintain.
In your case you'd have in that data,
"getSocialButtons" : true
Then in your AJAX handler file,
if (isset($_POST['getSocialButtons'])) {
// Echo your function here.
}
Then you'd echo out the function within that if statement and using the success callback in your AJAX request do something like this.
document.getElementById("yourDivId").innerHTML = data
That is assuming you're using an ID. Adjust the JS function to suit you.
Try to call that function social_sharing_buttons() like this in function.php:
$.ajax({
type:"POST",
url:"functions.php",
data: {action: 'add'},
success: function(data){
$('.pp').html(data);
}
in functions.php
if(isset($_POST['action']) && !empty($_POST['action'])) {
if($_POST['action'] == 'add') {
echo social_sharing_buttons();
}
}
I have question about call to my module action via ajax.
I'd like call to class in my module via ajax. But best solution for me is call to clean class. Not extends Module.
I don't know hot can I make url without add article to database and add module to him.
I use JQuery instead mooTools but js framework is not important. Most important is call to php class by ajax.
I have ajax module. But if I call to ajax.php required is module id from tl_module table. I don't want use this table. (Ajax will be very often calling, I prefer to don't load all contao mechanism. It should be very fast).
Thanks in advance for answers.
I found the answer for Contao >3.x in a GitHub issuse(german)
At first do in your Front-end Template:
<script type="text/javascript">
var data = {};
data["REQUEST_TOKEN"] = "<?php echo REQUEST_TOKEN ?>";
$(document).ready(function(){
$("#trigger").click(function(event){
$.post(
'<?php echo \Contao\Environment::get('requestUri')?>',
data,
function(responseText) {
alert(responseText);
}
).fail(function( jqXhr, textStatus, errorThrown ){ console.log( errorThrown )});
event.preventDefault();
});
});</script>
Important is the
- data["REQUEST_TOKEN"] -> if you do not add it, the POST-request will not reach your module:
public function generate()
{
if ($_SERVER['REQUEST_METHOD']=="POST" && \Environment::get('isAjaxRequest')) {
$this->myGenerateAjax();
exit;
}
return parent::generate();
}
//do in frontend
protected function compile()
{
...
}
public function myGenerateAjax()
{
// Ajax Requests verarbeiten
if(\Environment::get('isAjaxRequest')) {
header('Content-Type: application/json; charset=UTF-8');
echo json_encode(array(1, 2, 3));
exit;
}
}
If you want to do the ajax via GET you do not need the reqest token but the jquery funktion $get();
I would suggest you to use Simple_Ajax extension.
In this case you dont need to use Database and you can do pretty much anything you can do normally with Jquery ajax calls.
It works with Contao 2.11 and you can call your php class with it.
I find it much easier to use than ajax.php .
You can get it from : https://contao.org/de/extension-list/view/simple_ajax.de.html
Copy SimpleAjax.php to Contao's root folder.
Go to [CONTAO ROOT FOLDER]/system/modules and create a php file like following :
class AjaxRequestClass extends System
{
public function AjaxRequestMethod()
{
if ($this->Input->post('type') == 'ajaxsimple' )
{
// DO YOUR STUFF HERE
exit; // YOU SHOULD exit; OTHERWISE YOU GET ERRORS
}
}
}
Create a folder called config with a php file like following ( You can hook you class to TL_HOOKS with class name - class method, simple_ajax will execute you method whenever a ajax call is made ):
$GLOBALS['TL_HOOKS']['simpleAjax'][] = array('AjaxRequestClass','AjaxRequestMethod'); // Klassenname - Methodenname
Now you can easily make ajax calls with simply posting data to SimpleAjax.php:
$.ajax({
type: "POST",
url: "SimpleAjax.php",
data: { type: "ajaxsimple" },
success: function(result)
{
//DO YOUR STUFF HERE
}
I am writing a javascript which will post hostname of the site to a php page and get back response from it, but I don't know how to assign the hostname to adrs in url and not sure that code is correct or not.And this needs to done across server
javascript:
function ursl()
{
$.ajax({
url: 'http://example.com/en/member/track.php?adrs=',
success: function (response)
if (response)=='yes';
{
alert("yes");
}
});
}
track.php
$url=$_GET['adrs'];
$sql="SELECT * FROM website_ad where site='$url'";
$res=mysqli_query($link,$sql);
if(mysqli_num_rows($res)==0)
{
echo"no";
}
else
{
echo"yes";
}
Your ajax function should be written thusly:
$.ajax({
url: 'http://example.com/en/member/track.php?adrs=' + window.location.hostname,
success: function (response) {
if (response === 'yes') {
$.getScript('http://example.com/en/pop.js', function () {
// do anything that relies on this new script loading
});
}
}
});
window.location.hostname will give you the host name. You are passing it to the ajax url by concatenating it. Alternatively, as katana314 points out, you could pass the data in a separate parameter. Your ajax call would then look like this:
$.ajax({
url: 'http://example.com/en/member/track.php?adrs=',
data: {adrs: window.location.hostname},
success: function (response) {
if (response === 'yes') {
$.getScript('http://example.com/en/pop.js', function () {
// do anything that relies on this new script loading
});
}
}
});
I'm not sure what you intend response to be, but this code assumes it is a string and will match true if the string is 'yes'. If response is meant to be something else, you need to set your test accordingly.
$.getScript() will load your external script, but since it's asynchronous you'll have to put any code that is dependent on that in the callback.
In this type of GET request, the variable simply comes after the equals sign in the URL. The most basic way is to write this:
url: 'http://example.com/en/member/track.php?adrs=' + valueToAdd,
Alternatively, JQuery has a more intuitive way of including it.
$.ajax({
url: 'http://example.com/en/member/track.php',
data: { adrs: valueToAdd }
// the rest of the parameters as you had them.
Also note that you can't put a script tag inside a script. You will need some other way to run the Javascript function mentioned; for instance, wrap its contents in a function, load that function first (with a script tag earlier in the HTML), and then call it on success.
And for the final puzzle piece, you can retrieve the current host with window.location.host
You'll need to change this line to look like so:
url: 'http://example.com/en/member/track.php?adrs='+encodeURIComponent(document.URL)
The full success function should look like so:
success: function (response){
if (response==="yes"){
//do your thing here
}
}
That should solve it...
The website I'm developing is structured in this way:
It has a function that switches the module for the homepage content when $_GET['module'] is set, example:
function switchmodules()
{
$module=$_GET['module'];
switch($module)
{
case 'getnews': news(); break;
default: def();
}
}
As you can see, this switch calls another function for each module.
For example I wrote getNews() to work in this way:
function getNews()
{
$id=$_GET['id'];
if(!id)
{
//CODE TO LIST ALL NEWS
}
if(isset($id))
{
//CODE TO GET ONLY 1 NEWS BY ID
}
}
So, as you can see I'm not using an unique file for each module; all operations of a module are part of an unique function in which an action is switched again to change the result.
If I want to get a news from database I should use an url like this: ?index.php&module=getnews&id=1
My question now is:
With jQuery and $.ajax() method is there a way to get a news (for example) using this structure based on functions switched by a get? Or do I have to change everything and make a file for each function?
you can simply call the same url via $.ajax(). the only thing which should be changed for axjax calls is that you don't output your layout, but only the news itsself.
in php you can check for an ajax request like
if (isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest') {
// dont output layout
}
If this code is in say 'test.php' you can do:
<script>
$.get(
'/test.php',
{
module: 'news',
id: 1
},
function( data ) {
alert( data );
}
);
</script>
And js will send GET request to test.php with needed params. Then your server script will decide how to process request.
jQuery
$(document).ready( function() {
var form = '#my_awesome_form';
$(form + ' input[type=submit]').click(function(e) {
e.preventDefault();
$.ajax({
type: "GET",
url: 'index.php',
data: $(form).serialize(),
success: function( response ) {
alert('DONE!');
}
});
});
});
HTML
<form id="my_awesome_form" // OTHERS PROPERTIES //>
// LOT OF FIELDS HERE //
<input type="submit" value="Send">
</form>