How to update and output a database without reloading the page - php

Question

I have data which is a name. And I am having an input type=text where the value is equals to the name, together with a button type=submit. So what i'm trying to do is that, i want to change and update the name in my database and also output the new name without reloading the page at once because I want to run a JS function (Not Alert but a Toast Notification) which is I cannot do. So to shorten, ( EDIT -> UPDATE -> SHOW = without reloading the page).
EDIT:: I'm sorry I forgot to mention. I know jQuery and AJAX is the solution but I am having trouble understanding AJAX not unlike jQuery.
profile.php (FORM CODE)
<form action="profile.php" method="POST">
<input type="text" id="changename" name="changename" class="form-control" placeholder="New Name" required>
<button id="btnchange" type="submit"></button>
</form>
profile.php (PHP CODE)
<?php
if(isset($_POST['changename'])) {
require 'connect.php';
$newname = $_POST['changename'];
$user_id = $_SESSION['temp_user'];
$sql = "UPDATE login SET login_name='$newname' WHERE user_id='$user_id'";
if(mysqli_query($conn, $sql)){
header('location: profile.php');
// MY JS FUNCTION HERE //
}
mysqli_close($conn);
}
?>

Answer 1

How to update and output a databse without reloading the page? what u looking for is AJAX.
I know you have selected the answer, but there's an extra information that you need that might help you in the long run
The are some good ajax tutorials you can follow in the web
Ajax Introduction
What is Ajax ?
And many more you can find on the web.
This is what u need to do, first your form method is GET and on your php script you are requesting an $_POST therefore this will generate an undifined notice error,changename : notice : undifined variable changename so the solution is to use one method in a single form if your form is $_GET you use $variable = $_GET['fieldname'] if form method is POST on your server side use $variable = $_POST['fieldname']; not the way you did. So lets change your form method to POST Then this is what u should do.
edit.php
Update
<script type="text/javascript">
$('document').ready(function(){
$('form').on('submit',function(event){
event.preventDefault();//prevent the form from reloading the page when submit
var formData = $('form').serialize(); // reading form input
$.ajax({
type :'POST',
data : formData,
url : 'profile.php',
dataType : 'json',
encode : true,
success : function(response){
//response = JSON.parse(response);
if(response == "ok"){
$('#success').html('profile updated success');
setTimeout(' window.location.href = "profile.php"; ', 6000); // redirecting
}else{
//could not update
$('#error').html(response);
}
},
error : function(e){
console.log(e); // debugging puporse only
}
});
});
});
</script>
That's what you need on the frontend side.. Please note if you have more than one form in a single page then do not use $('form').on('submit',function(event){}); you then give each form a unique ID then on replace the form on jquery/ajax with the iD of the form u wanna submit.
Then your server side.
I have noticed that you have header('location: profile.php'); that's
for redirecting, since we are sending ajax request to the server and
back to the client, you don't redirect on the server side, your
redirect using the client side after you have received the response
you expected from the server. I have commented that section where I do
redirecting with client side, on the script above.
profile.php
<?php
$message = '';
if(isset($_POST['submit'])) {
require 'connect.php';
//validate your input field
$newname = $_POST['changename'];
$user_id = $_SESSION['temp_user'];
$sql = "UPDATE login SET login_name='$newname' WHERE user_id='$user_id'";
if(mysqli_query($conn, $sql)){
$message = 'ok'; // success
}else{
$message = 'system error could not update please try again later';
}
echo json_encode($message);//sending response back to the client
mysqli_close($conn);
}
?>
That's all you need, note you need to validate all your inputs fields in both the client and server side before using them.
I would also advice you to learn more about prepared statements, using mysqli or PDO they are very easy to use and very safe.
running a query like this : $sql = "UPDATE login SET login_name='$newname' WHERE user_id='$user_id'"; is very unsafe, your inputs are not validated and not filtered and sanitized.
so with prepared statements its easy like :
profile.php prepared
<?php
$message = '';
if(isset($_POST['submit'])) {
require 'connect.php';
//validate your input field
$newname = $_POST['changename'];
$user_id = $_SESSION['temp_user'];
//prepare and bind
$sql = $conn->prepare("UPDATE login SET login_name= ? WHERE user_id= ? ");
$sql->bind_param("si",$newname,$user_id);
if($sql->execute()){
$message = 'ok';
}else{
$message = 'Error Could not update please try again later';
}
echo json_encode($message);
$conn->close();
}
?>
$sql->bind_param("si",$newname,$user_id); This function binds the parameters to the SQL query and tells the database what the parameters.
are.By telling mysql what type of data to expect, we minimize the risk of SQL injections.

Answer 2

First thing to do in your form is to change your form method to POST (and not GET) as you try to retrieve "changename" by POST method.
So change this:
<form action="profile.php" method="GET">
by this:
<form action="profile.php" method="POST">