Sending Form Data to MySQL - php

Question

I'm trying to send form data to MySQL. I've found tonnes of code on the net, so I'm copying, and pasting and creating PHP scripts and adding them to my phpAdmin, but I really have no idea.
I don't know what to add to my form as an action to tell the form to send the data to my database. I already have an action code in the form, am I able to have two?
I have created a table in my phpAdmin, but I don't know if it actually will work.
This is my PHP code that I have stuck in my public_html folder, with generic username, database name, and password. I use an IP address instead of Localhost, as this database is on the Internet through Bluehost rather than my own computer.
<?php
// This function will run within each post array including multi-dimensional arrays
function ExtendedAddslash(&$params)
{
foreach ($params as &$var) {
// check if $var is an array. If yes, it will start another ExtendedAddslash() function to loop to each key inside.
is_array($var) ? ExtendedAddslash($var) : $var=addslashes($var);
}
}
// Initialize ExtendedAddslash() function for every $_POST variable
ExtendedAddslash($_POST);
$submission_id = $_POST['submission_id'];
$name = $_POST['name'] ;
$email = $_POST['email'] ;
$homeclub = $_POST['homeclub'] ;
$course1 = $_POST['course1'] ;
$course2 = $_POST['course2'] ;
$winner = $_POST['winner'] ;
$db_host = 'localhost';
$db_username = 'userrname';
$db_password = '';
$db_name = 'dbName';
mysql_connect( $db_host, $db_username, $db_password) or die(mysql_error());
mysql_select_db($db_name);
// search submission ID
$query = "SELECT * FROM `table_name` WHERE `submission_id` = '$submission_id'";
$sqlsearch = mysql_query($query);
$resultcount = mysql_numrows($sqlsearch);
if ($resultcount > 0) {
mysql_query("UPDATE `table_name` SET
`name` = '$name',
`email` = '$email',
`homeclub` = '$homeclub',
`course1` = '$course1',
`course2` = '$course2'
`winner` = '$winner'
WHERE `submission_id` = '$submission_id'")
or die(mysql_error());
} else {
mysql_query("INSERT INTO `table_name` (submission_id, formID, IP,
name, email, homeclub, course1, course2, winner)
VALUES ('$submission_id', '$formID', '$ip',
'$name', '$email', '$homeclub', '$course1', '$course2', '$winner') ")
or die( mysql_error());
}
?>
Any assistance will be greatly appreciated.

Answer 1

I'm sorry but your code is very outdated. You need to validate your value in case something is missing. I used Nulls coaling to do this (-> ?? ""), then you use mysql. You could use mysqli but I prefer and recommend PDO because it's easier. And instead to ask database if there is an insert first, just say hey update if there is an insert with the same id, or insert is (that makes Replace). But in order to do this, make submission_id unique in database
https://joshuaotwell.com/use-mysql-unique-constraint-in-phpmyadmin/
PS: this code is not tested but it should work, otherwise send me your full error report message
<?php
// read variables or assign them a default value
$submission_id = $_POST['submission_id'] ?? 0;
$name = $_POST['name'] ?? "";
$email = $_POST['email'] ?? "";
$homeclub = $_POST['homeclub'] ?? "";
$course1 = $_POST['course1'] ?? ""; // make array out of this
$course2 = $_POST['course2'] ?? "";
$winner = $_POST['winner'] ?? "";
// mysql is deprecated use pdo instead
$pdo = new PDO('mysql:host='.$db_host.';dbname='.$db_name, $db_username, $db_password)
// if exists -> replace, if not exist insert (replace makes that in one query)
// submission_id must be flagged as unique in mysql otherwise it will insert a new row every time
// I use prepared statements so first tell database what to do then the value
// prevents hacking and increase secureity
$query = 'REPLACE table_name SET
name = :name
, email = :email
, homeclub = :homeclub
, course1 = :course1
, course2 = :course2
, winner = :winner
, submission_id = :submission_id';
$statement = $pdo->prepare($query);
$statement->execute(array(
':name' = $name
, ':email' = $email
, ':homeclub' = $homeclub
, ':course1' = $course1
, ':course2' = $course2
, ':winner' = $winner
, ':submission_id' = $submission_id
));