I have a button in my form .i am passing some parameters to another form on button click.
My code is
<?php echo CHtml::link('newForm',"",
array('class'=>'btn btn-success',
'onclick'=>'test()'
));?>
My js code is like this:
<script type="text/javascript">
function test(){
$.ajax({
type: 'POST',
url: '<?php echo $this->createUrl('company/registration'); ?>',
data:{ids:values},
success:function(data){
}
error: function(data) {
alert("Error occured.please try again");
}
});
</script>
My controller code is:
public function actionRegistration()
{
$arrList=array();
$arrList=$_POST;
$model=new Registration;
$this->render('_newRegistration',array('model'=>$model,));
}
Here my view doesnot open.What is problem with my code???
If i understood you correctly, this few things does matter
1.Do you want open another file in modal popup (bootstrap)?
2.And in your controller you should use renderPartial instead of render.
3.and in your ajax success function you should render the data to the modal popup body
so your code in controller something like .
public function actionRegistration()
{
$arrList=array();
$arrList=$_POST;
$model=new Registration;
$this->renderPartial('_newRegistration',array('model'=>$model));
}
and in your ajax success function (if you are using modal popup),then your code something like
$.ajax({
type: 'POST',
url: '<?php echo $this->createUrl('company/registration'); ?>',
data:{ids:values},
success:function(data){
$("#your_modal_popup_id").html(data);//bootstrap modal popup
$("#your_modal_popup_id").modal('show');//bootstrap modal popup
}
error: function(data) {
alert("Error occured.please try again");
}
});
hope this help you
Related
I am using admin ajax but it is not working. Kindly, help me to find out the problem. Here is jquery code
jQuery(document).ready(function($) {
jQuery('#newPostsForm').submit(ajaxSubmit);
function ajaxSubmit(){
var newPostsForm = jQuery(this).serialize();
jQuery.ajax({
type:"POST",
url: "<?php echo admin_url('admin-ajax.php'); ?>",
data: newPostsForm,
success:function(data){
jQuery("#feedback").html(data);
}
});
return false;
}
}):
If I alert the var "newPostsForm" , it shown the posted values.but it is now proceeding to ajax. Here is the from I am using
<form type="post" action="" id="newPostsForm">
<input type="hidden" name="action" value="addPosts"/>
<!-- input fields -->
</form>
An here is the WordPress function I am using. this function is another file. HTML and javascript are in same file
function addPosts(){
echo "<pre>";
print_r($_POST);
die();
}
add_action('wp_ajax_addPosts', 'addPosts');
add_action('wp_ajax_nopriv_addPosts', 'addPosts'); // not really needed
Check to see if the script is getting processed by PHP before it is sent to the client. Change the code to something similar to this:
jQuery(document).ready(function($) {
jQuery('#newPostsForm').submit(ajaxSubmit);
function ajaxSubmit() {
var newPostsForm = jQuery(this).serialize();
var url = "<?php echo admin_url('admin-ajax.php'); ?>";
alert("Submitting to URL: " + url);
jQuery.ajax({
type:"POST",
url: url,
data: newPostsForm,
success:function(data){
jQuery("#feedback").html(data);
},
error: function (xhr, status, err) {
alert("Got status " + status + " and error: " + err);
}
});
return false;
}
});
If you get an actual URL like https://mysite.example.org then check that the URL goes to a valid location. If you get <?php echo admin_url('admin-ajax.php'); ?> then your code is not getting processed by PHP, and the AJAX call will fail because you are not using a valid URL.
The problem seems that the AJAX URL is not accessible in JS code. If the JS code written into a PHP page then only the code will work. Because the PHP code cant be executed into the JS files.
NOW the solution is to localized the JS file. Please follow the code.
wp_localize_script( 'handle', 'settings', array('ajaxurl' => admin_url( 'admin-ajax.php' )));
Write the above code just under where you have enqueued your js file.
NOW in JS file :
jQuery.ajax({
type:"POST",
**url: settings.ajaxurl,**
data: newPostsForm,
success:function(data){
jQuery("#feedback").html(data);
}
});
Hope it will work at your choice.
i want to refresh a particular div on ajax success, im using the below code but the whole page getting refreshed.
<script type="text/javascript">
$('#post_submit').click(function() {
var form_data = {
csrfsecurity: $("input[name=csrfsecurity]").val(),
post_text: $('#post_text').val()
};
$.ajax({
url: "<?php echo site_url('/post_status'); ?>",
type: 'POST',
data: form_data,
success: function(response){
$(".home_user_feeds").html("markUpCreatedUsingResponseFromServer");
}
return false;
});
return false;
});
</script>
you have an extra return false which is inside the $.ajax block which most probably causes an error so your form isn't submitted via ajax. If you remove that, you shouldn't have any issues.
Use the submit event of the form and remove the return false from the ajax callback:
$('#myFormId').on('submit', function() {
var form_data = {
csrfsecurity: $("input[name=csrfsecurity]").val(),
post_text: $('#post_text').val()
};
$.ajax({
url: "<?php echo site_url('/post_status'); ?>",
type: 'POST',
data: form_data,
success: function(response){
$(".home_user_feeds").html("markUpCreatedUsingResponseFromServer");
}
});
return false;
});
Remove the return false from inside your $.ajax function. Its a syntax error. The $.ajax function only expects a json object as an argument. "return false" cannot be part of a json object. You should keep the JavaScript console open during testing at all times - Press Ctrl-Shift-J in Chrome and select console to see any JS errors.
Also suggest you use <input type=button> instead of <input type=submit> or <button></button>
I am trying to show the value of ajax response inside a div and for that I have the following code in my view file.
<script type="text/javascript" src="MY LINK TO JQUERY"></script>
<script type="text/javascript">
$(function(){ // added
$('a.vote').click(function(){
var a_href = $(this).attr('href');
$.ajax({
type: "POST",
url: "<?php echo base_url(); ?>contents/hello",
data: "id="+a_href,
success: function(server_response){
if(server_response == 'success'){
$("#result").html(server_response);
}
else{
alert('Not OKay');
}
}
}); //$.ajax ends here
return false
});//.click function ends here
}); // function ends here
</script>
<a href="1" title="vote" class="vote" >Up Vote</a>
<br>
<div class="result"></div>
my Controller (to which the ajax is sending the value):
function hello() {
$id=$this->input->post('id');
echo $id;
}
Now what I am trying achieve is get the server_response value (the value that is being sent from the controller) in side <div class="result"></div> in my view file.
I have tried the following code but its not showing the value inside the div.
Could you please tell me where the problem is?
The problem is that you have mixed arguments of Ajax success handler. First goes data which your script gives back, then goes textStatus. Theoretically it can be "timeout", "error", "notmodified", "success" or "parsererror". However, in success textStatus will always be successful. But if you need to add alert on error you can add error handler. And yes, change selector in $("#result") to class. So corrected code may look like this:
$.ajax({
type: "POST",
url: "<?php echo base_url(); ?>contents/hello",
data: "id=" + a_href,
success: function(data, textStatus) {
$(".result").html(data);
},
error: function() {
alert('Not OKay');
}
});
success: function(server_response) {
$(".result").html(server_response);
}
<div class="result"></div> // result is the class
The selector should be .result not #result
Try changing <div class="result"></div> to <div id="result"></div>, because that's what you're referencing in your ajax success function
$("#result").html(server_response);
I am using CJuidialog widget to wrap a view file, and I don't want the default 'save' button because I would like to use a javascript to make an ajax call to the server for data validation, then save it. I tried below:
<?php
if($model->isNewRecord)
echo CHtml::submitButton('Create');
else
echo '<button onClick="javascript: _updatedata('.$model->id.');">Save</button>';
?>
when the save button is clicked. it still will go to the actionUpdate to save the form data, however I have created an action to just save my data.
function updatedata(id)
{
var url = '<?php echo Yii::app()->request->baseUrl ?>' + '/index.php?r=user/profileupdate&id='+id;
......
$.ajax({
url: url,
type: 'POST',
dataType: "html",
data:
{
...
},
success: function(data, textStatus, XMLHttpRequest) {
if (data != null && data == "success")
{
//$('#xccdfgrid').trigger('reloadGrid');
$('#userprofile').dialog('close');
}
else
alert(data);
},
......
If you got your question correct then what you are saying that instead of default button you need something that you can perform Ajax Things rite? if yes then you need to have Ajax Submit Button. Yii gives you liberty to use ajax more easily then you are trying to do. Please see CJuiDialog and AjaxSubmitButton
I have made a custom module in magento. I am using ajax in it(prototype.js).i can find the post variable on ajax page. But I am unable to find the file array on ajax page.
I am using following code for this.Please let me know where i am wrong?
//Ajax code on phtml page
new Ajax.Request(
reloadurl,
{
method: 'post',
parameters: $('use-credit-Form').serialize(),
onComplete: function(data)
{
alert(data.responseText);
}
});
//Php code on ajaxpage
public function ajaxAction()
{
$fileData = $_FILES;
echo '<pre>';
print_r($fileData);die;
}
It always print blank. but when I added this line
"VarienForm.prototype.submit.bind(usecreditForm)();"
I can get the value of file array. but draw back now page starts refreshing.
Please give me some suggestion.
Try this:
Event.observe('use-credit-Form', 'submit', function (event) {
$('use-credit-Form').request({
onFailure: function () {
alert('fail.');
},
onSuccess: function (data) {
alert(data.responseText);
}
});
Event.stop(event); // stop the form from submitting
});
Credit: submit a form via Ajax using prototype and update a result div