Try to pass 2 variables to php using ajax jquery - php

Question

I've been looking to find a way to update some data into my database when execute onclick function.
The idea is to update the value of a button contains the word "Active" and when execute onclick like to update into mysql to "Inactive"
This is the button:
$register_status = '<button class="btn btn-success btn_status btn-block" data-status="Active">Active</button>';
How to give the data variable so that we can fetch it in PHP using $_POST? The above representation of the data variable shows an error.
scripts.js
$(document).ready(function(){
$('body').click('.btn_status', function(e){
var button = jQuery(e.target);
if(button.data('status') == 'Active'){
var data = {id:id, register_status:Inactive};
$.ajax({
type: "POST",
dataType: "json",
url: "ajax/updateStatus.php",
data: data,
success: function(data) {
alert("update!");
}
});
}else if(button.data('status') == 'Inactive'){
alert("nothing");
}
});
})
ajax/updateStatus.php
include("db_connection.php");
if(isset($_POST))
{
$id = $_POST['id'];
$register_status = $_POST['register_status'];
$query = "UPDATE `users` SET `register_status` = '$register_status' WHERE `id` = '$id'";
if (!$result = mysqli_query($db,$query)) {
exit(mysqli_error());
}
}
UPDATE:
This is another function to delete. This function is run correctly.
function DeleteUser(id) {
$.post("ajax/deleteUser.php", {
id: id
},
function (data, status) {
readRecords();
}
);
}
deleteUser.php
<?php
if(isset($_POST['id']) && isset($_POST['id']) != "")
{
include("db_connection.php");
$user_id = $_POST['id'];
$query = "DELETE FROM users WHERE id = '$user_id'";
if (!$result = mysqli_query($db,$query)) {
exit(mysqli_error());
}
}
?>
and when do a click I can remove:
<li onclick="DeleteUser('.$row['id'].')">Delete</li>
However, this doesn't seem to work. If there's a better way to do what I mentioned above. thanks

Answer 1

Based on the example, you will want to make some changes:
JavaScript
$(function(){
function updateStatus(id, st){
$.post("ajax/updateStatus.php", {"id": id, "register_status": st}, function(data, status){
console.log("Update Success", data, status);
readRecords();
});
}
$('body').on('click', '.btn_status', function(e){
var button = $(this);
updateStatus(1001, (button.data("status") === "Active" ? "Inactive" : "Active"));
});
});
First, using the click() as you had, was not exactly the best way to do it. Using .on() will be better as it can handle any new or dynamically created items.
This script will require id to be defined someplace. The AJAX needs to send the id info to the PHP. The code you posted does not show this, so I am not sure what it's value should be.
PHP (updateStatus.php)
include("db_connection.php");
if(isset($_POST['id'])){
$id = (int)$_POST['id'];
$register_status = ($_POST['register_status'] == "Inactive" ? "Inactive" : "Active");
$query = "UPDATE `users` SET `register_status` = '$register_status' WHERE `id` = '$id'";
if (!$result = mysqli_query($db,$query)) {
exit(mysqli_error());
}
}
These updates can help prevent SQL Injection. I would advise using Prepared Statements to further prevent it.
Update
If you're button is going to be something like this:
<button onclick="ChangeStatus('.$row['id'].') class="btn btn-success btn_status btn-block" data-status="Active">Active</button>
Notice that this is only passing in just the ID from PHP. If these items will always be "Active", then yes, you can do this with something like:
function ChangeStatus(id){
$.post("ajax/updateStatus.php", {id: id}, function(data, status){ console.log(status, data)});
}
With PHP like:
<?php
if(isset($_POST['id']) && $_POST['id'] != ""){
include("db_connection.php");
$user_id = (INT)$_POST['id'];
$query = "UPDATE `users` SET `register_status` = 'Inactive' WHERE `id` = '$user_id'";
if (!$result = mysqli_query($db,$query)) {
exit(mysqli_error());
}
}
?>
Hope this helps.