Data can't submit through ajax, php and jquery in to database
Question
I try to insert data through the ajax, PHP and jquery but the code is not working properly. anyone can help me what is going wrong. even I used serialize() method also but no result.
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
<form id="myform">
<label>Username</label> <input name="name" type="text">
<label>Password</label> <input name="password" type="text">
<button type="submit" name="submit" id="submit">Save</button>
</form>
<script>
$(document).ready(function() {
$("#submit").click(function(event) {
event.preventDefault();
$.ajax({
url: "new.php",
type: 'POST',
data: $("#myform").serialize(),
success: function(data) {
alert(data);
$('form').trigger("reset");
}
});
});
});
</script>
</body>
</html>
And this is the new.php it means the PHP code in another file.
<?php
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', 'root');
define('DB_DATABASE', 'test');
$db = mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
if(isset($_POST['submit'])) {
$username=$_POST['name'];
$password=$_POST['password'];
$sql="INSERT INTO ourteam(name, position) VALUES ('$username','$password')";
$result=mysqli_query($db,$sql);
?>
1 answers
solution1
0 2018-12-23 12:56:59
Your issue is that this method expects a response from the server in-order to be able to store data as the server response.
success: function(data) {}
You can return this data by outputting it using echo, which we will also json_encode() for the JS to be able to understand the output.
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', 'root');
define('DB_DATABASE', 'test');
header('Content-Type: application/json'); # Declare what content type we are returning
$db = mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
if(empty(array_diff(array('name', 'password'), array_keys($_POST)))) {
$stmt = $db->prepare('INSERT INTO ourteam(name, position) VALUES (?, ?)');
$stmt->bind_param('ss', $_POST['name'], $_POST['password']);
die(json_encode($stmt->execute())); # die() will halt the program here so no other instructions are executed
}
I updated your code to use prepared statements to prevent SQL Injections .
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