I am submitting a number of forms on my page via php using Ajax. The code works great in forms preloaded with the page. However, I need to submit some dynamic forms that don't load with the page, they are called via other javascript functions.
Please, I need someone to help me review the code for use for forms that don't load with the page. Also the 'failure' condition is not working.
The code is below:
<script type="text/javascript">
feedbar = document.getElementById("feedbar");
jQuery(document).ready(function() {
$('#addressform').on('submit', function (e) {
$.ajax({
type: 'post',
url: 'data/process.php',
data: $('#addressform').serialize(),
success: function () {
feedbar.innerHTML='<div class="text-success">New Addressed Saved Successfully</div>';
},
failure: function () {
feedbar.innerHTML='<div class="text-danger">Error Saving New Address</div>';
}
});
e.preventDefault();
});
});
Thanks.
You need to bind event by existing html (e.g body).
Event handlers are bound only to the currently selected elements; they must exist on the page at the time your code makes the call to .on()
see api: https://api.jquery.com/on/
Try like this:
$("body").on('submit', '#addressform',function (e) {
$.ajax({
type: 'post',
url: 'data/process.php',
data: $('#addressform').serialize(),
success: function () {
feedbar.innerHTML='<div class="text-success">New Addressed Saved Successfully</div>';
},
failure: function () {
feedbar.innerHTML='<div class="text-danger">Error Saving New Address</div>';
}
});
e.preventDefault();
});
});
you can delegate to document:
$(document).on('submit', '#addressform', function (e) {
$.ajax({
type: 'post',
url: 'data/process.php',
data: $(this).serialize(), // <----serialize with "this"
success: function () {
feedbar.innerHTML='<div class="text-success">New Addressed Saved Successfully</div>';
},
error: function () { //<----use error function instead
feedbar.innerHTML='<div class="text-danger">Error Saving New Address</div>';
}
});
e.preventDefault();
});
});
As you have posted this line as below:
I need to submit some dynamic forms that don't load with the page
What i understand with this line is you want a common submit function for all forms which are generated dynamically, then you can do this:
$(document).on('submit', 'form', function (e) {
$.ajax({
type: 'post',
url: 'data/process.php',
data: $(this).serialize(), // <----"this" is current form context
success: function () {
//some stuff
},
error: function () { //<----use error function instead
//some stuff
}
});
e.preventDefault();
});
});
For your last comment:
You can try to get the text in ajax response like this:
success: function (data) {
feedbar.innerHTML='<div class="text-success">'+ data +'</div>';
},
error: function (xhr) { //<----use error function instead
feedbar.innerHTML='<div class="text-danger">' + xhr.responseText + '</div>';
}
if Success:
here in success function you get the response in data which is the arguement in success function, this holds the response which it requested to the serverside.
if Error:
Same way if something goes wrong at the serverside or any kind of execption has been occured then xhr which is the arguement of error function holds the responseText.
And finally i suggest you that you can place your response in feedbar selector using jQuery this way:
var $feedbar = $('#feedbar');
so in success function:
$feedbar.html('<div class="text-success">'+ data +'</div>');
so in error function:
$feedbar.html('<div class="text-success">'+ xhr.responseText +'</div>');
Related
I have an Asp.Net MVC web application. I want to run some code on the successful response of the API method which is called on form submit.
I have the below Code.
#using (Html.BeginForm("APIMethod", "Configuration", FormMethod.Post, new { #class = "form-horizontal", id = "formID" }))
{
}
$('#formID').submit(function (e) {
$.validator.unobtrusive.parse("form");
e.preventDefault();
if ($(this).valid()) {
FunctionToBeCalled(); //JS function
}
}
But FunctionToBeCalled() function gets called before the APIMethod(), but I want to run the FunctionToBeCalled() function after the response of APIMethod().
So I made the below changes by referring this link. But now the APIMethod is getting called twice.
$('#formID').submit(function (e) {
$.validator.unobtrusive.parse("form");
e.preventDefault();
if ($(this).valid()) {
//Some custom javasctipt valiadations
$.ajax({
url: $('#formID').attr('action'),
type: 'POST',
data: $('#formID').serialize(),
success: function () {
console.log('form submitted.');
FunctionToBeCalled(); //JS function
}
});
}
}
function FunctionToBeCalled(){alert('hello');}
So I am not able to solve the issue.
If you want to execute some work on success, fail, etc. situation of form submission, then you would need to use Ajax call in your view. As you use ASP.NET MVC, you can try the following approach.
View:
$('form').submit(function (event) {
event.preventDefault();
var formdata = $('#demoForm').serialize();
//If you are uploading files, then you need to use "FormData" instead of "serialize()" method.
//var formdata = new FormData($('#demoForm').get(0));
$.ajax({
type: "POST",
url: "/DemoController/Save",
cache: false,
dataType: "json",
data: formdata,
/* If you are uploading files, then processData and contentType must be set to
false in order for FormData to work (otherwise comment out both of them) */
processData: false, //For posting uploaded files
contentType: false, //For posting uploaded files
//
//Callback Functions (for more information http://api.jquery.com/jquery.ajax/)
beforeSend: function () {
//e.g. show "Loading" indicator
},
error: function (response) {
$("#error_message").html(data);
},
success: function (data, textStatus, XMLHttpRequest) {
$('#result').html(data); //e.g. display message in a div
},
complete: function () {
//e.g. hide "Loading" indicator
},
});
});
Controller:
public JsonResult Save(DemoViewModel model)
{
//...code omitted for brevity
return Json(new { success = true, data = model, message = "Data saved successfully."
}
Update: If SubmitButton calls a JavaScript method or uses AJAX call, the validation should be made in this method instead of button click as shown below. Otherwise, the request is still sent to the Controller without validation.
function save(event) {
//Validate the form before sending the request to the Controller
if (!$("#formID").valid()) {
return false;
}
...
}
Update your function as follows.
$('#formID').submit(function (e) {
e.preventDefault();
try{
$.validator.unobtrusive.parse("form");
if ($(this).valid()) {
$.ajax({
url: $('#formID').attr('action'),
type: 'POST',
data: $('#formID').serialize(),
success: function () {
console.log('form submitted.');
FunctionToBeCalled(); //JS function
}
});
}
}
catch(e){
console.log(e);
}
});
Check the browser console for fetching error. The above code will prevent of submitting the form.
I think line $.validator.unobtrusive.parse("form") were throwing error.
For that use you need to add the following jQuery libraries.
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.9/jquery.validate.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery-validation-unobtrusive/3.2.11/jquery.validate.unobtrusive.min.js"></script>
I think you should remove razor form tag if you want to post your form using ajax call and add post api URL directly to ajax request instead of getting it from your razor form tag using id:
Here is the revised version of your code :
<form method="post" id="formID">
<!-- Your form fields here -->
<button id="submit">Submit</button>
</form>
Submit your form on button click like:
$('#submit').on('click', function (evt) {
evt.preventDefault();
$.ajax({
url: "/Configuration/APIMethod",
type: 'POST',
dataType : 'json',
data: $('#formID').serialize(),
success: function () {
console.log('form submitted.');
FunctionToBeCalled(); //JS function
}
});
});
function FunctionToBeCalled(){alert('hello');}
You need to use Ajax.BeginForm, this article should help [https://www.c-sharpcorner.com/article/asp-net-mvc-5-ajax-beginform-ajaxoptions-onsuccess-onfailure/ ]
The major thing here is that I didn't use a submit button, I used a link instead and handled the rest in the js file. This way, the form would nver be submitted if the js file is not on the page, and with this js file, it initiates a form submission by itself rather than th form submitting when the submit button is clicked
You can adapt this to your solution as see how it respond. I have somthing like this in production and it works fine.
(function() {
$(function() {
var _$pageSection = $('#ProccessProductId');
var _$formname = _$pageSection.find('form[name=productForm]');
_$formname.find('.buy-product').on('click', function(e) {
e.preventDefault();
if (!_$formname.valid()) {
return;
}
var formData = _$formname.serializeFormToObject();
//set busy animation
$.ajax({
url: 'https://..../', //_$formname.attr('action')
type: 'POST',
data: formData,
success: function(content) {
AnotherProcess(content.Id)
},
error: function(e) {
//notify user of error
}
}).always(function() {
// clear busy animation
});
});
function AnotherProcess(id) {
//Perform your operation
}
}
}
<div class="row" id="ProccessProductId">
#using (Html.BeginForm("APIMethod", "Configuration", FormMethod.Post, new { #class = "form-horizontal", name="productForm" id = "formID" })) {
<li class="buy-product">Save & Proceed</li>
}
</div>
From C# I am returning this way
return PartialView("~/View.cshtml", model);
In view, I am updating the data this way. After binding this way I am getting request multiple times.
$("#LoanCommitteDateSubmitedselected").change(function ()
{
$.ajax({
success: function (data)
{
$("#sectionName").html(data);
}
});
});
After doing this in Ajax form submit I am getting request multiple times in below method.
$('#sectionName').submit(function (e)
{
e.preventDefault();
$.ajax({
success: function (response)
{
}
});
});
Can Anyone advise me how to resolve this issue?
That might be a problem with multiple event registration.
You can try unbinding and then binding your submit function as-
$('#sectionName').off('submit');
$('#sectionName').on('submit',function (e)
{
e.preventDefault();
$.ajax({
success: function (response)
{
}
});
});
When I am using the .focusout() function in jQuery it seems to fire twice when I trigger the event for the second time, here is an example of my code:
$(document).ready(function() {
var baseUrl = "http://annuityadvicecentre.dev/";
if($('html').hasClass('ver--prize-soft')) {
$('#home_telephone').focusout(function () {
var softResponse = {};
var inputVal = $(this).val();
$.ajaxSetup({
headers: { 'X-CSRF-Token' : $('meta[name=_token]').attr('content') }
});
$.ajax({
type: 'POST',
url: baseUrl + "lookup",
data: {
number: inputVal,
phone_type: "mobile",
},
error: function() {
console.log('POST Error: Mobile');
},
}).done(function(data) {
// you may safely use results here
softResponse.isMobile = data;
});
$.ajax({
type: 'POST',
url: baseUrl + "lookup",
data: {
number: inputVal,
phone_type: "landline",
},
error: function() {
console.log('POST Error: Landline');
},
}).done(function(data) {
// you may safely use results here
softResponse.isLandline = data;
});
$(document).ajaxStop(function () {
console.log('All AJAX Requests Have Stopped');
});
});
}
});
Sorry for the messy example as I have just been bootstrapping this up however you can see I am wrapping this focusout function:
$('#home_telephone').focusout(function () {...
Around my AJAX calls, now for some reason when I test this out on the page and un-focus on the #home_telephone element the .ajaxStop() function only runs once which is the functionality I want however if I then click on the element and un-focus again the .ajaxStop() function runs twice. Any idea why this might be happening? Thanks
Try to add e.stoppropogation() within function like:
$('#home_telephone').focusout(function (e) {
e.stopPropagation()();
//your code
});
You're adding a new ajaxStop listener every time the element is unfocused. Just move the:
$(document).ajaxStop(function () {
console.log('All AJAX Requests Have Stopped');
});
call outside of the focusout callback function.
I am a beginner in JavaScript means and programming, and I encountered a problem for a personal project. I made an anime fight website getting some information from MySQL Database each anime has ten videos and photos, through a random button it randomly takes one link for a video and photo. The problem is that it only work only one time if I random again nothing happens. I know that in order to make that work I have to rewrite the code again after the success of the first random for getting a second random but this will create an infinite loop. Can somebody help me solve this issue.
This is the code used:
<script>
$(document).ready(function () {
$('.imgResponsive').click(function(){
$('#hiddenPage').hide();
$('#hiddenPage').html('<center><img src="img/loading.gif"></center>');
$('#hiddenPage').show();
$.ajax({
type: 'POST',
url: 'php/handler.php',
data: {
anime: $(this).prev().val()
},
success: function(response){
$('#hiddenPage').html(response);
$('#random').click(function(){
$('#hiddenPage').hide();
$('#hiddenPage').html('<center><img src="img/loading.gif"></center>');
$('#hiddenPage').show();
$.ajax({
type: 'POST',
url: 'php/handler.php',
data: {
anime: $(this).prev().val()
},
success: function(response){
$('#hiddenPage').html(response);
}
});
})
}
});
})
});
</script>
I understand that $.ajax request overwrites initial .imgResponsive element, am I right? Along with overwritten .imgResponsive you permanently lose click event attached to this element.
In that case you need to attach event to element container, eg.
$(document).on('click', '.imgResponsive', function() {....
...
}
instead of
$('.imgResponsive').click(function(){ ....
You have to register EventListener in order to proceed.
Try this:
<script>
var showImage = function(e){
e.preventDefault();
$('#hiddenPage').hide();
$('#hiddenPage').html('<center><img src="img/loading.gif"></center>');
$('#hiddenPage').show();
$.ajax({
type: 'POST',
url: 'php/handler.php',
data: {
anime: $(this).prev().val()
},
success: function(response){
$('#hiddenPage').html(response);
}
});
};
$(document).ready(function () {
$('document').on('click', '.imgResponsive', showImage);
$('document').on('click','#random', showImage);
});
</script>
The way your code is written, you're handling only a single AJAX response (aside from the first one) with no way to handle more AJAX requests triggered by clicking the #random button. You need to write functions for handling button clicks and the AJAX responses instead of using anonymous functions; that way it's modular enough that you can listen for and handle more button clicks in the future.
Something like this:
<script>
$(document).ready(function () {
$('.imgResponsive').click(function(){
$('#hiddenPage').hide();
$('#hiddenPage').html('<center><img src="img/loading.gif"></center>');
$('#hiddenPage').show();
$.ajax({
type: 'POST',
url: 'php/handler.php',
data: {
anime: $(this).prev().val()
},
success: handleResponse
});
$('#random').click(handleButtonClick);
});
function handleButtonClick(e){
$('#hiddenPage').hide();
$('#hiddenPage').html('<center><img src="img/loading.gif"></center>');
$('#hiddenPage').show();
$.ajax({
type: 'POST',
url: 'php/handler.php',
data: {
anime: $(this).prev().val()
},
success: handleResponse
});
}
function handleResponse(response){
$('#hiddenPage').html(response);
}
});
</script>
Edit: Another thing that might be happening is that your #random element is being overwritten every time you do $('#hiddenPage').html(response);. In that case you would need to attach a new event handler to the new #random element every time you handle an AJAX response:
function handleResponse(response){
$('#hiddenPage').html(response);
$('#random').click(handleButtonClick);
}
I want to know that my jquery-ui datepicker is not working in document.ready after an ajax function call. when I put on ajax complete its work successfully please help what should I do. what's the reason for not working
$("#ScheduledArrivalDate").datepicker({
beforeShow: function () {
setTimeout(function () {
$('.ui-datepicker').css('z-index', 2000);
}, 0);
}
});
function getPage(page) {
$.ajax({
type: "POST",
url: page,
data: $("#frm").serialize(),
xhrFields: {
withCredentials: true
},
success: function (html) {
$('#List').empty();
$('#List').append($.parseHTML(html));
},
error: function () {
alert("error");
},
complete: function () {
alert("complete");
}
});
}
$.document.ready() only initiates after a page is loaded without ajax. When you replace/append html in an ajax call and you have a datefield in the new inserted html, you need to initialise it again (at least for the new inserted html block).
You could do this by calling $.datepicker in your success or complete function, like you already did, or by adding $.document.ajaxEnd() to your javascript file, what is initialized after every end of an ajax event (also on error).
Be aware not do double initiate the datepicker, especially when using ajaxEnd. This could lead to unexpected behaviour.
the code inside $(document).ready() will run only after page loads. While you are dynamically adding datepicker if I am not wrong. So do one thig. Take options in a variable like below:
var options = {
beforeShow: function () {
setTimeout(function () {
$('.ui-datepicker').css('z-index', 2000);
}, 0);
}
}
then:
$(document).ready(function(){
$("#ScheduledArrivalDate").datepicker(options);
});
and in ajax call:
function getPage(page) {
$.ajax({
type: "POST",
url: page,
data: $("#frm").serialize(),
xhrFields: {
withCredentials: true
},
success: function (html) {
$('#List').empty();
$('#List').append($.parseHTML(html));
$('#IdOfnewlyAddedDatePicker').datepicker(options);
},
error: function () {
alert("error");
},
complete: function () {
alert("complete");
}
});
}
Let me know if this not work or you are injecting html other than this.