HTML code:
<html>
<head>
<title>Space-O | Ajax submit form</title>
</head>
<body>
<h1>jQuery post form data using .ajax() method</h1>
<!-- our form -->
<form id='userForm'>
<div>
<input type='text' name='name' placeholder='Name' />
</div>
<div>
<input type='text' name='email' placeholder='Email' />
</div>
<div>
<input type='text' name='number' placeholder='Mobile Number' />
</div>
<div>
<!-- <input type='submit' value='Submit' /> -->
<button >insert</button>
</div>
</form>
<style type="text/css">
table {
border-collapse: collapse;
}
table, th, td {
border: 1px solid black;
}
</style>
<!-- where the response will be displayed -->
<div id='response'></div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js "></script>
<script>
$(document).ready(function(){
$('#userForm').submit(function(){
// show that something is loading
$('#response').html("<b>Loading response...</b>");
// Call ajax for pass data to other place
$.ajax({
type: 'POST',
url: 'myscript.php',
data: $("#userForm").serialize() // getting filed value in serialize form
})
.done(function(data){ // if getting done then call.
// show the response
$('#response').html(data);
})
.fail(function() { // if fail then getting message
// just in case posting your form failed
alert( "Posting failed." );
});
// to prevent refreshing the whole page page
return false;
});
});
</script>
</body>
</html>
PHP script:
// this is just a test
//send back to the ajax request the request
$con=mysqli_connect('localhost','root','','test');
$name=$_POST['name'];
$email=$_POST['email'];
$fav=$_POST['number'];
$result=mysqli_query($con,"insert into user values('','$name','$email','$fav')");
if ($result) {
echo $msg="success";
} else {
echo $msg="not submit";
}
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