I have php page with a form, I want to open it in a dialog box and submit and show the data on same dialog without reloading the page
php
<form accept-charset="UTF-8">
<button type="submit" class="addButtonClass"></button>
please advice
$('.addButtonClass').submit(function (event) {
event.preventDefault();
....
.. Now your usual ajax post
}
Modify this
$('#add_user1').submit(function(e){
e.preventDefault(); //** prevent normal form submission and page reload
$.ajax({
type: "POST",
cache: false,
url: $link,
data: $('#add_user1').serializeArray(),
success: function (data) {
var json_obj = $.parseJSON(data);
var result = json_obj['result'];
var usergroup = json_obj['usergroup'];
var success = json_obj['success'];
var errorAry = {
}
}
});
return false;
});
Related
In my form i am checking if there is same value in database or not when form submitted. The code below works fine and is giving the rigth result of AJAX post but the problem is when giving alert according to the wrong result, javascript alert and focus works but form still submits after these.
Button for submit:
<input type="submit" name="kaydet" class="btn btn-success form-control"
onClick="return kaynak_kontrol()" value="Kaydet">
AJAX:
<script type="text/javascript">
function kaynak_kontrol(){
Form=document.forms['depo_kayit'];
var depo_sube_no = document.getElementById('depo_sube_no').value;
var depo_firma_no = document.getElementById('depo_firma_no').value;
var depo_kodu = document.getElementById('depo_kodu').value;
var dataString ="depo_sube_no="+depo_sube_no+"&depo_firma_no="+depo_firma_no+"&depo_kodu="+depo_kodu;
$.ajax({
type: "POST",
url: "depo_kodu_kontrol.php",
data: dataString,
success: function(result){
if(result != 0){
alert("Aynı şubede aynı isimde iki depo olamaz!");
document.getElementById('depo_kodu').focus();
return false;
} else {
return true;
Form.submit();
}
}
});
}
</script>
Can you help me why i return false is not working and form still submits?
you need to prevent the form submission manually using jQuery event.preventDefault()
here is a small fix
function kaynak_kontrol(event){ // notice the new parameter !
event.preventDefault();
//the rest of your code
I have 10 form in a page & there data is submitted through ajax, Now i don't want to create ajax script for each form. So here is what i tried
var form_id = $(this).closest("form").attr('id');
$(document).on("submit", "#"+form_id, function(e){
e.preventDefault();
var postData = $("#"form_id).serialize()
var send = true;
var ptel = 1;
$("#"+form_id).find("input").each(function () {
if ($(this).val() === '') { send = false; ptel = 0; }
});
if(ptel == 0) { bootbox.alert('Please Fill All fields'); }
if(send){
$('form_id').trigger("reset");
$.ajax({
type: "POST",
url: "/ajax/X-Profile",
data: postData,
cache: false,
success: function(msg)
{
bootbox.alert('Your Profile has been updated.');
}
});
}
return false;
});
var form_id results in undefined because when page loads no attribute was defined to it
Above codes are just to make you understand,
So my question is how can i make 10 forms submit through single ajax function
You can create a function such as SendForm() and attach it to the forms onsubmit attribute
<form id="yourid" onsubmit="SendForm(this);return false;">
inside the SendForm() function place your script
for instance:
function SendForm(form) {
var postData = form.serialize();
// .......etc
}
To know which form was submitted in PHP, you can place a hidden input inside the form or have a second parameter on SendForm which gets sent through, such as SendForm(node,formtype)
if the form isn't submitting or page reloads, remove the onsubmit attribute and add this to your JS instead
$(document).on("submit","form", function(e){
e.preventDefault();
SendForm($(this));
return false;
});
Have javascript function like below
<script type="text/javascript">
function submitForm(formID) {
data = $('#'+formID).serialize();
//your ajax code
}
</script>
Now use input button with onclick like below, and pass form Id as a parameter
<input type="button" value="Submit" onclick="submitForm({formID})">
This will work without refreshing the page. And on success you can trigger reset() function to form that particular form values.
I have a page where I'm displaying some information. You can select a option and the page will then display a form by loading the form using ajax response:
$("body").on("change", "#patient_id", function(event){
var prescription_id = $(this).val();
event.preventDefault(); // disable normal link function so that it doesn't refresh the page
var curr_data = {
action: 'refferedcall',
prescription_id: prescription_id,
dataType: 'json'
};
$.post(hmgt.ajax, curr_data, function(response) {
$('.prescription_content').html(response);
return true;
});
});
This works fine. But this view is a form. I want to then submit the included form with Ajax as well. But I can't seem to do it. I think it is because if I set up a button handler for the form it doesn't work as the form isn't present when the main page and JQuery script is loaded.
So to be clear, I'm loading this div onto my main page using JQuery and Ajax load. I then want to simply submit this form with Ajax also.
<div class="prescription_content">
<div class="title">Submit News</div>
<form role="form" id="ref_form" name="ref_form_p">
<div class="form-group">
<label for="pat_ref_hosp">Hospital to Refer:</label>
<input type="text" class="form-control" id="pat_ref_hosp" name="pat_ref_hosp" value="<?php if(!empty($result->reffer_hospital)){ echo $result->reffer_hospital; }?>">
</div>
<input type="hidden" class="form-control" id="pres_note" name="pres_note" value="<?php echo $result->priscription_id ;?>">
<button type="button" id="<?php echo $result->priscription_id ;?>" class="btn btn-success reffering_status">Refer Now</button>
</form>
</div>
TIA
Then I submitted form again using ajax through below button click event:
$("body").on("click", ".reffering_status", function(event){
event.preventDefault(); // disable normal link function so that it doesn't refresh the page
var prescription_id = $("#pres_note").val();
var pat_ref_hosp = $("#pat_ref_hosp").val();
var curr_data = {
action: 'reffering_status',
dataType: 'json',
prescription_id: prescription_id,
pat_ref_hosp : pat_ref_hosp,
};
console.log(curr_data);
)};
Here is log displaying
Object {action: "reffering_status", dataType: "json", prescription_id: "1", pat_ref_hosp: ""}
pat_ref_hosp is empty
I don't know how to display ajax in jsfiddle
https://jsfiddle.net/3ggq3Ldm/
Yes the way you are doing it will not work because the contents of the DIV you are loading-in is not loaded into the DOM when your initial
$("body").on("click", ".reffering_status", function(event){});
call is made.
If I am understanding you correctly, this is the behaviour you want to achieve:
$("#patient_id").on("change", function(event) {
var prescription_id = $(this).val();
event.preventDefault(); // disable normal link function so that it doesn't refresh the page
var curr_data = {
action: 'refferedcall',
prescription_id: prescription_id,
dataType: 'json'
};
$.post(hmgt.ajax, curr_data, function(response) {
$(".prescription_content").html(response);
$(".reffering_status").on("click", function(event){
event.preventDefault(); // disable normal link function so that it doesn't refresh the page
var prescription_id = $("#pres_note").val();
var pat_ref_hosp = $("#pat_ref_hosp").val();
var curr_data = {
action: 'reffering_status',
dataType: 'json',
prescription_id: prescription_id,
pat_ref_hosp : pat_ref_hosp
};
console.log(curr_data);
)};
return true;
});
});
You simply need to run the code that attaches your click listener AFTER the DOM has already been updated with the new information.
Please let me know if this code does what you were intending it to.
My users will see a google repcatcha2 (nocaptcha) in a web page. When they solve the captcha (put the tick in the box) the form should be automatically submit.
Is there any way to do that?
Sure you can do it.
In this post I've explained how to insert reCaptcha into a site and to code javascript to verify user and site.
Add a name to your form with reCaptcha: <form method="post" name="myform">
Add document.myform.submit(); code for submitting of the form upon the site verification success event:
<script type='text/javascript'>
var onReturnCallback = function(response) {
var url='proxy.php?url=' + 'https://www.google.com/recaptcha/api/siteverify';
$.ajax({ 'url' : url,
dataType: 'json',
data: { response: response},
success: function( data ) {
var res = data.success.toString();
if (res)
{ document.myform.submit(); }
} // end success
}); // end $.ajax
}; // end onReturnCallback
</script>
within the form element I send data with a href. Using the JavaScript (as defined below), it works perfectly.
I want to send the forms now with Ajax, unfortunately it does not work. Do you have a solution for me. jQuery is included on the page.
Thank You!
function sendData(sFm, sID, sFn, sCl) {
var form = document.getElementById(sFm);
form.sid.value = sID;
form.fnc.value = sFn;
form.cl.value = sCl;
form.submit();
}
Send Data
My new Code:
function sendData(sFm, sID, sFn, sCl) {
var form = $("#"+sFm);
form.submit( function() {
var sid = $('input[name="sid"]').val(sID);
var fnc = $('input[name="fnc"]').val(sFn);
var cl = $('input[name="cl"]').val(sCl);
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
data: {sid: sid, fnc: fnc, cl: cl}
}).done(function( e ) {
});
});
form.submit();
}
$("form").submit(function() { // for all forms, if you want specially one then use id or class selector
var url = $(this).attr('action'); // the script where you handle the form input.
var method = $(this).attr('method');
$.ajax({
type: method,
url: url,
data: $(this).serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // handle response here
}
});
return false; // avoid to execute the actual submit of the form.
});
:) , Thanks
Just because we don't want to submit all form by ajax so we can set a data-attribute to form tag to indicate, will it should be submit by ajax or normally ,, so ,,
$("form").submit(function() { ...
if($(this).attr('data-ajax') != "true") return true; // default hanlder
...
so If form is written like this :-
<form action="/dosomething" method="post" data-ajax="true"> </form> // will be sumit by ajax
<form action="/dosomething" method="post" data-ajax="false"> </form>
// will be sumit by default method