I am using Yii2 and socket.io for sending notifications. I have a client.js file which should emit an id of newly connected user to server.
The code responsible for this in my client.js is:
socket.emit('AddId', '<?php $model->id ?>');//of course it does not work but I need smth like this.
So how to pass a Yii2 model attribute to socket.io client.js?
Client.js code:
$( document ).ready(function() {
var socket = io.connect('http://localhost:8890');
socket.emit('AddId', '<?php $model->id ?>');
});
If you want send info about an user you could do this way
$( document ).ready(function() {
var socket = io.connect('http://localhost:8890');
socket.emit('AddId', '<?php echo Yii::$app->user->id ?>');
});
In Yii2 inside my index.php view I wrote:
$test = 'My text';//here could be any text data, model attributes etc
<script>
var test = "<?php echo $test ?>";
</script>
After this update I was able to use a variable (var test) in my notification.js.
notification.js is added also to AppAsset.php as following:
public $js = [
'js/notification.js'
];
notification.js file:
$( document ).ready(function() {
var socket = io.connect('http://localhost:8890');
socket.emit('AddId', test);//test is a variable (var test) from index.php file
});
Related
I've been struggling for a few days with this issue and I really hope you can help me out.
I've created a plugin, which is located in:
'/wp-content/plugins/my-cool-plugin'.
My plugin allows users to post a custom post type via a form on a public page, basically anyone should be able to post something.
Using jQuery, I listen to when my frontend form is submitted and using Ajax I pass the data from the form to a php file to process it into a post.
This file is located at:
'/wp-content/plugins/my-cool-plugin/inc/processor.php'.
Below is the content of my processor file:
$var1= $_POST['some'];
$var2= $_POST['data'];
$new_post = array(
'post_type' => 'my_custom_post',
'post_status' => 'publish',
'mcp_1' => $var1,
'mcp_2' => $var2
);
$post_id = wp_insert_post( $new_post, $wp_error );
if ($wp_error == 'false'){
$post_url = get_permalink( $post_id );
echo $post_url;
}else {
// some sort of error
}
When I test my form, it results in the following error:
Call to undefined function wp_insert_post() on line ... which is the following line:
$post_id = wp_insert_post( $new_post, $wp_error );
Do I need to include something since I'm not in the WordPress 'scope' anymore?
Or is there another (much better) way for inserting custom posts from a front end form?
Why are you running the file out of wordpress scope? That is not the best practive. Instead you could run it in wordpress scope and user wordpress native ajax.
add_action('wp_ajax_yourplugin_create_post', 'yourplugin_create_post');
add_action('wp_ajax_nopriv_yourplugin_create_post', 'yourplugin_create_post');
function yourplugin_create_post() {
// your code here
}
Then you would need your ajax url to be passed from php to js:
function your_plugin_ajaxurl() {
?>
<script type="text/javascript">
var yourPluginAjaxUrl = "<?php echo admin_url('admin-ajax.php'); ?>";
</script>
<?php
}
add_action('wp_head','your_plugin_ajaxurl');
Then you can use your ajax request but you would need to indicate action:yourplugin_create_post and url = yourPluginAjaxUrl
Try adding
require(dirname(__FILE__) . '/wp-load.php');
It took me some time to process Nick's answer, but I finally got it to work! Like Nick said, I dropped using the process file because is was out of the scope of WordPress. I moved my post creation from my proces file to a new function in the plugin init file (my-cool-plugin.php), as Nick suggested. This resulted in the following new function:
add_action('wp_ajax_coolplugin_create_post', 'coolplugin_create_post');
add_action('wp_ajax_nopriv_coolplugin_create_post', 'coolplugin_create_post');
function coolplugin_create_post() {
$var1 = $_POST['some'];
$var2 = $_POST['data'];
$new_post = array(
'post_type' => 'my_custom_post',
'post_status' => 'publish'
'post_title' => 'Some title'
);
$post_id = wp_insert_post( $new_post, $wp_error );
// check if there is a post id and use it to add custom meta
if ($post_id) {
update_post_meta($post_id, 'mcp_1', $var1);
update_post_meta($post_id, 'mcp_2', $var2);
}
if ($wp_error == false){
$post_url = get_permalink( $post_id );
echo $post_url;
}else {
// some sort of error
}
}
I also had to change the way I inserted my custom values into the newly created post, because the wp_insert_post() function only accepts default post parameters (see the wp_insert_post documentation for these parameters).
Next to my insert/create post function I also had to make some adjustments to my javascript file, which retrieves the filled in data from my form. Therefore (as Nick suggested) I needed to pass my Ajax URL from PHP to JS by adding the following function to my-cool-plugin.php like this:
function your_plugin_ajaxurl() { ?>
<script type="text/javascript">
var coolPluginAjaxUrl = "<?php echo admin_url('admin-ajax.php'); ?>";
</script>
<?php }
add_action('wp_head','your_plugin_ajaxurl');
By adding the coolPluginAjaxUrl variable to the head I'm able to use the URL in my javascript to post the data to when my form is submitted, like this:
$( '#form' ).on( 'submit', function(e) {
var request;
e.preventDefault();
var val_one = $( '#val-one' ).val();
var val_two = $( '#val-two' ).val();
var formData = {
action: 'coolplugin_create_post',
some: val_one,
data: val_two,
};
request = $.ajax({
type: 'POST',
url: coolPluginAjaxUrl,
data: formData,
});
});
The formData holds the coolplugin_create_post action defined in PHP and the request is posted to the coolPluginAjaxUrl URL, defined in the head.
Thanks Nick for pointing me into the right direction and I hope that my solution will also help others. Please note that I've stripped my code of several security measures for others to easily understand how the code works.
I'm currently doing an application which uses ajax in getting user inputs from form fields then concatenate it in a string, this works fine but I'm getting a
"POST 500 (Internal Server Error")
and cannot pass the data. How do I pass a string from view to controller with ajax and access it in a controller.
Here are my codes:
the script
<script>
// $(document).ready(function(){
function saveData(){
var con = document.getElementById('tiks').value;
// var dataString = {'conid':con};
alert(con);
var cct = $("input[name=csrf_token_name]").val();
var base_url = <?php base_url(); ?>;
$.ajax({
url: base_url + "welcome/newShortestPath",
type: "POST",
data: {'conid':con, 'csrf_token_name':cct},
success:function(){
alert("SUCCESS KALUY-I "+con);
$('#myForm')[0].reset();
}
});
}
controller
public function newShortestPath(){
echo "REACHED";
$var = $this->input->post('conid');
//$this -> load -> model('edge_model');
//$this->edge_model->setConstraints();
}
really need your help guys. thanks in advance. :)
UPDATE
Im no longer getting any error by fixing the errors at calling the base_url() function but at the same time I cant fetch the passed data, is there any other way fetching data from post to controller?
you have to echo the base_url and wrap your var with ''
var base_url = '<?php echo base_url(); ?>';
Update
based on the error you are getting which is:
Fatal error: Call to undefined function base_url()
as you mentioned in your comment.
what you have to do is to go to application/config/autoload.php
and find
$autoload['helper'] = array();
then change it to
$autoload['helper'] = array('url');
or manually load the url helper:
$this->load->helper('url');
read more here
Tips:
Tip 1:
You can setup all url:
var url = <?php echo site_url('welcome/newShortestPath'); ?>;
Tip 2:
Use site_url() instead base_url(). base_url must be used to resources (img, css, js, etc.)
To build URL's use site_url()
I am writing a plugin and the widget contains drop-down lists and needs to retrieve values for each successive drop-down list from the database using AJAX based on the previous drop-down list's selected value.
I am going off a slightly modified version of the code from the WordPress codex found here:http://codex.wordpress.org/AJAX_in_Plugins
For some reason the action function either isn't being called or the there is some error in my code:
// add action hooks
add_action('wp_footer', 'er_qm_ajax_handler' );
add_action('wp_ajax_repopulate_widget_club_type', 'repopulate_widget_club_type');
add_action('wp_ajax_nopriv_repopulate_widget_club_type', 'repopulate_widget_club_type');
// action for the wp_footer hook to insert the javascript
function er_qm_ajax_handler(){
?>
<script type="text/javascript" >
jQuery(document).ready(function($) {
jQuery('#widget_manufacturer').val('3').change(function(){
var manufacturer = $("#widget_manufacturer").val();
var data = {
action: 'repopulate_widget_club_type',
selected_manufacturer: manufacturer
};
// since 2.8 ajaxurl is always defined in the admin header and points to admin-ajax.php
jQuery.post(ajaxurl, data, function(response) {
alert('Got this from the server: ' + response);
});
});
});
</script>
<?php
}
// The action function
function repopulate_widget_club_type(){
die("Got to the action");
}
Not exactly sure what I am doing wrong as I am getting no response from the AJAX post, I know that the jQuery .change() is working as I set up some alert()'s to test it out once I altered the dropdown lists values.
Any thoughts and help will be greatly appreciated, thanks!
You need to define the value of ajaxurl like this:
var ajaxurl = '<?php echo admin_url( 'admin-ajax.php' ); ?>'
It's prefferable to define it in javascript's global scope so before this line:
jQuery(document).ready(function($) {
Hi guys Im trying to make a drop-down list of countries and when the user select a country that redirect to a specific page created dynamically actually i manage to make the redirection work using javascript, but i need to take more parameters to the method inside the controler like the county "id" with out exposing it on the uri, so is that possible using $_post also i should not use button submit.
this is my code
view page
<?php echo form_open('site/country');
$options = array();
$js = 'id="country" onChange="window.location.href= this.form.CTRY.options[this.form.CTRY.selectedIndex].value"';
$options['#'] = "(please select a country)" ;
foreach ($list as $row):
$value= site_url()."/site/country/".url_title($row->name);
$options[$value] = $row->name ;
endforeach;
echo form_dropdown('CTRY', $options,'',$js);
//$test =array ('number' => 10)
//echo form_hidden($test);
echo form_close();?>
this is my method in controller
function country($data)
{
echo 'this is taking you to county= '.$data;
}
Why don't you do something like #Joseph Silber described.
Use jQuery to perform an ajax request when the drop-down list is changed, also managing the redirect?
Something like;
$('#my-drop-down').change(function() {
$.ajax({
url: '/site/country',
type: 'post',
data: $("#my-form").serialize(),
success: function( response.redirect ) {
window.location.href = response.redirect;
},
error: function( response ) {
alert('Oops');
}
});
});
By using serilaize(), you can post all data from the form, or just add the parameters you want to pass.
E.g data: 'id='+$('#id').val()+'&country='+$('#country').val()
How make Zend_Form submission without reload a page - with Ajax?
This is code for create form that reload a page when submitted, what should be change or add that this form will submit with ajax (1.regular solution 2.jquery solution):
Form:
class Application_Form_Login extends Zend_Form
{
public function init()
{
$username=new Zend_Form_Element_Text('username');
$username ->addFilter('StringToLower')
->addValidator('alnum');
$password=new Zend_Form_Element_Text('password');
$password->addFilter('StringToLower')
->addValidator('alnum');
$submit=new Zend_Form_Element_Submit('submit');
$this->addElements(array($username,$password,$submit));
}
}
Controller:
$form = new Application_Form_Login();
$request = $this->getRequest();
if ($request->isPost()) {
if ($form->isValid($request->getPost())) {
if ($this->_process($form->getValues())) {
//code indside
}
}
}
$this->view->form = $form;
View:
<?
echo $this->form;
?>
My proposal that I don't think is proper(does form make filtering and validation?) for View:
<?
echo $this->form;
?>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.4/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('form').submit(function(){
var sendData=$(this).serialize();
$.ajax(
{
url:'',
dataType:'json',
type:'POST',
data:sendData,
success: function(data) {
}
});
return false;
});
});
</script>
Thanks
Well,
for filtering/validation you might want to send the form using Ajax and by knowing at the server-side that it is an Ajax request (you can use a flag for that, like a header, search for knowing if a request is ajax or not) and sending back only the form 'area'. Then when you receive it you can overwrite it.
There is currently no wiser way to do it with Zend_Form I think.