jQuery validation not work in Ajax Form Submit - javascript

I submit my form using jQuery $.post method and need to validation form using jQuery validation plugin like this :
HTML :
<form name="myform" id="myform" action="send.php">User:
<input type="text" value="" name="user" />
<br/>Password:
<input type="password" name="password" value="" />
<input type="hidden" name="xyz" value="123" />
<input type="hidden" name="submit" value="true" />
<?php $token=N oCSRF::generate( 'csrf_token' );?>
<input type="hidden" name="csrf_token" value="<?php echo $token; ?>">
</form>
<button class="btn btn-info" id="ajax-form-1">Run Code</button>
<div id="ajax-form-msg1"></div>
JS:
$('#myform').validate({
rules: {
user: {
required: true
},
password: {
required: true
}
},
submitHandler: function (form) {
$("#ajax-form-1").click(function () {
$("#ajax-form-msg1").html("<img src='loading.gif'/>");
// var formData = $("#myform").serialize(); //or
var formData = $("#myform").serializeArray();
var URL = $("#myform").attr('action');
$.post(URL,
formData,
function (data, textStatus, jqXHR) {
$("#ajax-form-msg1").html('<pre><code class="prettyprint">' + data + '</code></pre>');
}).fail(function (jqXHR, textStatus, errorThrown) {
$("#ajax-form-msg1").html('<pre><code class="prettyprint">AJAX Request Failed<br/> textStatus=' + textStatus + ', errorThrown=' + errorThrown + '</code></pre>');
});
});
}
});
In action, jQuery validation not work with my form. how do fix this ?
DEMO : http://jsfiddle.net/fcuswvf1/2/

You are binding the click handler on the button inside the submitHandler. This way, when a user clicks the button, the click event doesn't fire. You can simply remove the click handling lines from the 'submitHandler, so that the code inside is called directly wheneversubmitHandler` is triggered.

Related

How to check if submitting form with jQuery if successful?

I have following codes as form, and I wanted to check if the form is been successfully submitted.
I was wondering how I can check on the console. I want to check if the form is been successfully submitted so I can display another form.
<form id="signup" data-magellan-target="signup" action="http://app-service-staging.com" class="epic_app-signup" method="POST">
<div class="grid__column " style="width: 100%;">
<input type="text" name="first_name" placeholder="Name" required/>
</div>
<div class="grid__column " style="width: 100%;">
<input type="text" name="password" placeholder="Password" required/>
</div>
<div class="grid__column " style="width: 100%;">
<input type="text" name="confimred-password" placeholder="Confirmed password" />
</div>
<div class="grid__column " style="width: 100%;">
<input type="date" name="startdate" id="startdate" min="2019-12-16">
</div>
</div>
<button type="submit" class="grid__column" style="width: 50%;"></button>
</div>
</div>
</div>
</form>
and the script,
$('.epic_app-signup').on('submit', function(e) {
e.preventDefault();
var formData = $('.epic_app-signup').serializeArray();
var jsonData = {};
formData.forEach(function(item, index) {
jsonData[item.name] = item.value;
});
console.log('data\n', jsonData);
$.ajax({
url: 'http://app-service-staging.com/api/auth/register',
type:'POST',
data: jsonData,
contentType: 'application/json'
}).done(function(data, textStatus, jqXHR) {
if (textStatus === 'success') {
}
});
});
});
you can do this by various ways currently you are not using ajax request if you want to achieve this without ajax let follow these steps
when user click on submit button your form is submitted received form information(you define the path in action attribute where form submitted) after processing successfully redirect toward a new form
second solution use jquery ajax request
//first form
<form action='test.php' id='form_1' method='post'>
<label>Full Name</label>
<input type='text' name='full_name'>
<input type='submit'>
</form>
//second form
<form action='test.php' id='form_2' method='post' style='display:none'>
<label>Father Name</label>
<input type='text' name='father_name'>
<input type='submit'>
</form>
use jquery cdn
<script src='https://code.jquery.com/jquery-git.js'></script>
<script>
$("#form_1").submit(function(e) {
e.preventDefault(); // avoid to execute the actual submit of the form.
var form = $(this);
var url = form.attr('action');
$.ajax({
type: "POST",
url: url,
data: form.serialize(), // serializes the form's elements.
success: function(data)
{
alert("form submitted successfully");
$('#form_1').hide();
$('#form_2').show();
},
error:function(data){
alert("there is an error kindly check it now");
}
});
return false;
});
</script>

Getting value from an input field in html off a button click

I have a button and input field, and want to send the value of that field into an ajax call. I'm having a brain freeze at the moment and could use some help. Here's what I have so far.
function submit() {
$.ajax({
type: 'POST',
url: 'http://localhost:8000/getCoords',
dataType: 'json',
data: {name: 'Abduh'},
success: (success) => {
console.log(success);
},
error: (error) => {
console.log(error);
}
});
}
<input type=text id=search placeholder=Search />
<br/>
<button id=submit onclick="submit()">Submit</button>
you should sue quote around properties values
<input type='text' id='search' placeholder=Search />
<br/>
<button id='submit' onclick="submit();">Submit</button>
$("#submit").on("click",function(){
var url= 'http://localhost:8000/getCoords';
$.post( url, { name: $("#search").val() }).done(function(data) {
alert( "Data Loaded: " + data );
});
});
<input type="text" id="search" placeholder="Search" />
<button id="submit">Submit</button>

How to post mutliple forms from a webpage

I want to post 2 forms using javscript, but I can't seem to figure it out. Can someone help me?
It seems like I need to submit the first form Async according to this but I don't follow their code: Submit two forms with one button
HTML
<form name="form1" action="https://test.salesforce.com/servlet/servlet.WebToLead?encoding=UTF-8" method="POST" target = "me1">
<input type="hidden" name="oid" value="00Df00000000001" />
</form>
<form name="form2" action="https://test.salesforce.com/servlet/servlet.WebToLead?encoding=UTF-8" method="POST" >
<input type="hidden" name="oid" value="00Df00000000001" />
</form>
<input name="submit" value="Submit Form" onclick="submitForms ()" type="button">
JS
function submitForms(){
document.forms["form1"].submit(); /* should be Async?*/
document.forms["form2"].submit();
}
var str1 = $( "form1" ).serialize();
var str2 = $( "form2" ).serialize();
ajaxReq(url,str1);
ajaxReq(url,str2);
function ajaxReq(url,data){
var request = $.ajax({
url: url,
type: "POST",
data: data,
dataType: "html"
});
request.done(function(html) {
console.log('SUCCESS')
});
request.fail(function( jqXHR, textStatus ) {
console.log("AJAX REQUEST FAILED" + textStatus);
});
}

Sending values to the action script continuously. How to do that?

I want to repeatedly send values of username and password to the php script. How do I do this ? Like to send the values to the action script, we use submit button but how can I send the values automatically to the script and that too continuously ?
<form method="post" action="processor.php">
<input type="username" value="suhail" />
<input type="password" value="secret_code" />
<input type="submit" />
</form>
Using the jQuery form plugin you can do the following:
setInterval(function() {
$('form').ajaxSubmit();
}, 1000);
Another solution is to target the form to an iframe so if you submit the form, it doesn't reload the page:
HTML:
<form id="myform" method="post" action="processor.php" target="frm">
<input type="username" value="suhail" />
<input type="password" value="secret_code" />
<input type="submit" />
</form>
<iframe name="frm" id="frm"></iframe>
JS:
var form = document.getElementById('myform');
setInterval(function() {
form.submit();
}, 1000);
try something like this
JAVASCRIPT
<script language=javascript>
var int=self.setInterval(function(){send_data()},1000);
function send_data()
{
document.getElementById('my_form').submit()
}
</script>
HTML
<form method="post" id="my_form" action="processor.php">
<input type="username" value="suhail" />
<input type="password" value="secret_code" />
</form>
<form id="myform" method="post" action="processor.php">
<input type="username" value="suhail" />
<input type="password" value="secret_code" />
<input type="submit" />
</form>
<script type="text/javascript">
var count=100,i=0;
for(i=0;i<count;i++) {
document.getElementById('myform').submit();
}
</script>
This will submit the form 100 times
Use Ajax, it's really easy with jQuery. To send the form data to the processor.php script:
var sendForm = function () {
$.ajax({
type: 'post',
url: 'processor.php',
dataType: 'JSON',
data: {
username: $('#username').val(),
password: $('#password').val()
},
success: function (data) {
// do something with the answer from server?
},
error: function (data) {
// handle error
}
});
}
So, sendForm is a function that sends the form data to the server. Now, wee need to set a timer that will invoke it repeatedly:
window.setInterval(sendForm, 1000); // sends form data every 1000 ms
You may you $.post or $.get or $.ajax request repeatedly to send continuous request.
$(document).ready(function(){
setInterval(function() {
var username = $("#username").val();
var password = $("#password").val();
var dataString = 'username='+username+"&password="+password;
$.post('login.php',dataString,function(response){
//your code what you want to do of response
alert(response);
});
}, 1000);
});
and html code is like following
<form method="post" action="processor.php">
<input type="username" value="suhail" id="username"/>
<input type="password" value="secret_code" id="password"/>
<input type="submit" />
</form>
This is a full HTML file doing what you want, read the comments.
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<form method="post" action="processor.php">
<input type="username" id="username" value="suhail" />
<input type="password" id="password" value="secret_code" />
<input type="submit" />
</form>
<script>
function send_request(username, password) {
var dataString = 'username='+username+"&password="+password;
$.post('login.php',dataString,function(response){
// You can check if the login is success/fail here
console.log(response);
// Send the request again, this will create an infinity loop
send_request(username, password);
});
}
// Start sending request
send_request($('#username').val(), $('#password').val());
</script>
Try this,
JS:
$(document).ready(function(){
var int=self.setInterval(function(){statuscheck()},1000);
function statuscheck()
{
var username = $("#username").val();
var password = $("#password").val();
$.ajax({
type:"post",
url:"processor.php",
dataType: "html",
cache:false,
data:"&username="+username+"&password="+password,
success:function(response){
alert(response);
}
});
}
});
HTML:
<form method="post" action="processor.php">
<input type="username" value="suhail" id="username"/>
<input type="password" value="secret_code" id="password"/>
<input type="submit" />
</form>

Ajax is giving me an error when I send data

I am trying to submit a form by Ajax but I am unable to . I have multiple forms and I am using (this) to submit the data. I am getting the error From error:0 error.The alert messages are showing me that I have the value.
<script type="text/javascript">
$(document).ready(function() {
$(".submitform").click(function (){
alert ($(this).parent().serialize());
$.ajax({
type: "POST",
url: "reply_business.php",
timeout:5000,
data: $(this).parent().serialize(),
beforeSend: function(xhr){
$('#load').show();
},
success: function(response){
$(this).parent().find('.sentreply').append(response);
$('.sentreply div:last').fadeOut(10).fadeIn(2000);
//uncomment for debugging purposes
//alert(response);
},
error: function(jqXHR) {
alert ('From error:' + jqXHR.status + ' ' +jqXHR.statusText);
},
complete: function(jqXHR, textStatus){
//uncomment for debugging purposes
//alert ('From complete:' + jqXHR.status + ' ' +jqXHR.statusText + ' ' + textStatus);
$('#load').hide();
}
});
});
});
</script>
I am creating the form below by the PHP code
foreach ($array['business_ids'] as $business)
{
?>
<form >
<input type="hidden" name="b_id" value="<?php echo $business ; ?>" />
<input type="hidden" name="c_id" value="<?php echo $sqlr['conversation_id']; ?>" />
<input type="hidden" name="q_id" value="<?php echo $sqlr['query_id']; ?>" />
<input type="hidden" name="u_id" value="<?php echo $sqlr['u_id']; ?>" />
<textarea name="reply">Type the reply here.</textarea>
<input type="submit" class="submitform" value="Submit">
</form>
<?php
}
I do not understand why Ajax isn't able to send the data.
Without seeing the markup or the network traffic, we can only guess. Perhaps $(this).parent() isn't the form?
It's typically safer to attach $(form).submit() than $(button).click() for this reason and because $(button).click() doesn't capture form submit by hitting the enter key.
Edit Here's an example:
<form id="theform">
<input type="text" id="thetext" name="thetext" />
<input type="submit" value="send" />
</form>
<form id="anotherform">
<input type="text" id="anothertext" name="anothertext" />
<input type="submit" value="save 2" />
</form>
<script>
$(document).ready(function () {
$("#theform").submit(function (e) {
var data = {
thetext: $("#thetext").val()
};
$.ajax("/the/server/url", {
type: "POST",
contentType: "application/json; charset=utf-8",
data: JSON.stringify(data),
success: function (r) {
alert("done");
}
});
// Technically you need only one or the other
e.preventDefault();
return false;
});
});
</script>
You seem to have submitted the form after starting the Ajax request - and when the page is unloaded, the request is cancelled. Since your form has no action, the submit will just reload the current page so you might not notice it.
To prevent this, you will need to preventDefault() the catched event. Also, you should not handle just click events on submit-buttons, but rather the submit-event of the <form> itself.

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