Answers for "ajax submit form data"

jQuery AJAX form submit

// jQuery ajax form submit example, runs when form is submitted
$("#myFormID").submit(function(e) {
    e.preventDefault(); // prevent actual form submit
    var form = $(this);
    var url = form.attr('action'); //get submit url [replace url here if desired]
    $.ajax({
         type: "POST",
         url: url,
         data: form.serialize(), // serializes form input
         success: function(data){
             console.log(data);
         }
    });
});

ajax submit form data

$(function() {
  $('form.my_form').submit(function(event) {
    event.preventDefault(); // Prevent the form from submitting via the browser
    var form = $(this);
    $.ajax({
      type: form.attr('method'),
      url: form.attr('action'),
      data: form.serialize()
    }).done(function(data) {
      // Optionally alert the user of success here...
    }).fail(function(data) {
      // Optionally alert the user of an error here...
    });
  });
});

javascript ajax post form data

$(document).ready(function() {

    // process the form
    $('form').submit(function(event) {

        // get the form data
        // there are many ways to get this data using jQuery (you can use the class or id also)
        var formData = {
            'name'              : $('input[name=name]').val(),
            'email'             : $('input[name=email]').val(),
            'superheroAlias'    : $('input[name=superheroAlias]').val()
        };

        // process the form
        $.ajax({
            type        : 'POST', // define the type of HTTP verb we want to use (POST for our form)
            url         : 'process.php', // the url where we want to POST
            data        : formData, // our data object
            dataType    : 'json', // what type of data do we expect back from the server
                        encode          : true
        })
            // using the done promise callback
            .done(function(data) {

                // log data to the console so we can see
                console.log(data);

                // here we will handle errors and validation messages
            });

        // stop the form from submitting the normal way and refreshing the page
        event.preventDefault();
    });

});

ajax submit form data

<form id="contactForm1" action="/your_url" method="post">
    <!-- Form input fields here (do not forget your name attributes). -->
</form>

<script type="text/javascript">
    var frm = $('#contactForm1');

    frm.submit(function (e) {

        e.preventDefault();

        $.ajax({
            type: frm.attr('method'),
            url: frm.attr('action'),
            data: frm.serialize(),
            success: function (data) {
                console.log('Submission was successful.');
                console.log(data);
            },
            error: function (data) {
                console.log('An error occurred.');
                console.log(data);
            },
        });
    });
</script>

jquery ajax form submit example

$.ajax({
			    type: "POST",
			    url: "upload.php",
			    data: { name:name, mobile:mobile, address:address, city:city },		    
			    dataType: "json",
			    success: function(result){
			        			    }
			});

Submit AJAX Forms with JQuery to database

// Database Structure 
CREATE TABLE `contacts` (
 `id` int(11) NOT NULL AUTO_INCREMENT,
 `name` text NOT NULL,
 `email` text NOT NULL,
 `reason` text NOT NULL,
 `message` text NOT NULL,
 PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=5 DEFAULT CHARSET=latin1

<?php
if(isset($_POST['submit_contact']))
{
 $name=$_POST['name'];
 $email=$_POST['email'];
 $reason=$_POST['reason'];
 $message=$_POST['message'];
	
 $host="localhost";
 $username="root";
 $password="";
 $databasename="sample";
 $connect=mysql_connect($host,$username,$password);
 $db=mysql_select_db($databasename);
 
 mysql_query("insert into contacts values('','$name','$email','$reason','$message')");
 echo "submitted";
 exit();
}
?>