I have created an application using xampp (apache and mysql). I have the following HTML code:
<!DOCTYPE html>
<html>
<head>
<title>Name</title>
</head>
<body>
<div id="main">
<h1>Details</h1>
<div id="name">
<h2>Name</h2>
<hr/>
<Form Name ="form1" Method ="POST" ACTION = "name.php">
<label>Name: </label>
<input type="text" name="per_name" id="name" required="required" placeholder="please enter name"/><br/><br />
<label>Age: </label>
<input type="text" name="per_age" id="age" required="required" placeholder="please enter age"/><br/><br />
<input type="submit" value=" Submit " name="submit"/><br />
</form>
</div>
</div>
</div>
</body>
</html>
and the following PHP code:
<?php
if(isset($_POST["submit"])){
$servername = "localhost";
$username = "root";
$password = "password";
$dbname = "details";
$connection = new mysqli($servername, $username, $password, $dbname);
if ($connection->connect_error) {
die("Connection failed: " . $connection->connect_error);
}
$sql = "INSERT INTO persons (person_name, person_age)
VALUES ('".$_POST["per_name"]."','".$_POST["per_age"]."')";
if ($connection->query($sql) === TRUE) {
echo "person added";
} else {
echo "person not added";
}
$connection->close();
}
?>
Instead of calling the php file using <Form Name ="form1" Method ="POST" ACTION = "name.php"> how would i create a simple ajax file to call the PHP file? i have tried to do this but can't seem to get anywhere, can anyone help me please? AJAX:
$(document).ready(function(){
$("#submit").click(function(){
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "name.php",
data: dataString,
cache: false,
success: function(result){
alert(result);
});
}
return false;
});
});
