formdata jquery ajax php
//set your ajax url here
//$url =
$(document).on('submit', '#form', function(e){
e.preventDefault();
$.ajax({
type: 'POST',
url: $url,
data: new FormData(this),
dataType: 'json',
contentType: false,
processData:false,//this is a must
success: function(response){
//statements on success
}
});
});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();
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: frm.serialize(),
success: function (data) {
console.log('Submission was successful.');
console.log(data);
},
error: function (data) {
console.log('An error occurred.');
console.log(data);
},
});
});
</script>Submit AJAX Forms with JQuery to database
// Database Structure
CREATE TABLE `contacts` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` text NOT NULL,
`email` text NOT NULL,
`reason` text NOT NULL,
`message` text NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=5 DEFAULT CHARSET=latin1
<?php
if(isset($_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();
}
?>Copyright © 2021 Codeinu
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