Ajax datatable Update with PHP MySql - php

Question

I want to update data using Ajax and below is the sample code. The SweetAlert get displaced that it has been updated but it doesn't take effect in the database non thus it gives an error.
Ajax Code
This is the Ajax Script for the form submission when the submit button is clicked.
<script>
$('#submit').on('click', function(event){
event.preventDefault();
var firstname = $('#firstname').val();
var othername = $('#othername').val();
var gender = $('#gender').val();
var id_type = $('#id_type').val();
var id_number = $('#id_number').val();
var issue_date = $('#issue_date').val();
var business_place = $('#business_place').val();
var food_type = $('#food_type').val();
var screened = $('#screened').val();
var sub_metro = $('#sub_metro').val();
var telephone = $('#telephone').val();
var get_date = $('#get_date').val();
var chit_number = $('#chit_number').val();
var remarks = $('#remarks').val();
var user_id = $('#user_id').val();
var vendor_id = $('#vendor_id').val();
if (firstname!="" &&othername!="" && food_type!=""){
$.ajax({
url: "action/vendor_update.php",
type: "POST",
data: {
firstname:firstname, othername:othername, gender:gender,
id_type:id_type, id_number:id_number, issue_date:issue_date,
business_place:business_place, food_type:food_type, screened:screened,
sub_metro:sub_metro, telephone:telephone, get_date:get_date,
chit_number:chit_number, remarks:remarks, user_id:user_id, vendor_id:vendor_id,
},
cache: false,
success: function(data){
if(data.statusCode=200){
$('#dataForm').find('input:text').val('');
alert('Updated');
})
}
else if (data.statusCode=201)
{
alert('Cannot Update');
}
}
});
}else{
alert('All fields are mandatory');
}
});
PHP Code ---> action/vendor_update.php
This code is for the php server side for data insertion into the database.
<?
session_start();
// include('../includes/session.php');
include('./includes/connection.php');
$query = $dbh->$prepare("UPDATE vendors SET Firstname=:firstname, Othername=:othername, Telephone=:telephone, Gender=:gender, IdType=:id_type, IdNumber=:id_number, IdDate=:issue_date, BusinessPlace=:business_place, FoodType=:food_type, Notes=:remarks, ScreenStatus=:screened, ChitNumber=:chit_number, UpdatedBy=:user_id WHERE VendorId=:vendor_id");
$query->execute([
$firstname = $_POST['firstname'];
$othername = $_POST['othername'];
$telephone = $_POST['telephone'];
$gender = $_POST['gender'];
$id_type = $_POST['id_type'];
$id_number = $_POST['id_number'];
$issue_date = $_POST['issue_date'];
$business_place = $_POST['business_place'];
$food_type = $_POST['food_type'];
$remarks = $_POST['remarks'];
$screened = $_POST['screened'];
$chit_number = $_POST['chit_number'];
$user_id = $_POST['user_id'];
$vendor_id = $_POST['vendor_id'];
]);
// $query->execute($_POST);
if ($query->execute()){
echo json_encode(array("statusCode"=>200));
} else {
echo json_encode(array("statusCode"=>500));
}
?>

Answer 1

Here's the cleaned up PHP code:
<?php
// Tip:
// * Use named placeholders
// * Define the query string inside prepare() so you can't "miss" and run
// the wrong query by accident
// * Use single quotes so you can't interpolate variables by accident
// and create ugly SQL injection bugs
$query = $dbh->prepare('UPDATE vendors SET Firstname=:firstname, Othername=:othername, Telephone=:telephone, Gender=:gender, IdType=:id_type...');
// Use ONE of:
// A) if you have slight differences
$query->execute([
'firstname' => $_POST['firstname'],
'othername' => $_POST['othername'],
'telephone' => $_POST['telephone'],
...
]);
// OR
// B) if you're confident the placeholders match 100% and no clean-up
// such as trim() is necessary.
$query->execute($_POST);
if ($query->execute()) {
echo json_encode(array("statusCode"=>200));
} else {
echo json_encode(array("statusCode"=>500));
}
?>
Note: It's worth noting that code like this does not need to exist, that any decent ORM will make this trivial to do. It's worth exploring what options you have there as this could be so much easier.