PHP - Function Runs 100 Queries - Need This Reduced!

Hello all, 
I have a php file connecting to phpmyadmin on xampp. I set the localhost password to ‘password’ I run the script with ajax and it does work because I tested it with a filecreate. I get NO errors and all I want to do is check and see if the email exists and if not then create a row in the database.

<?php
// Connect to Phpmyadmin database localhost.
$hostname = "localhost";//"";127.0.0.1
$username = "root";
$password = "password";//"password";
$dbname = "test";

$conn = new mysqli($hostname, $username, $password, $dbname);
if($conn -> connect_error) {
	die("Connection failed: " . $conn -> connect_error);
}
// USED to CHANGE the db. - (mysqli_select_db($con, "test") or die());

/* Check to make all fields entered -make sure submitted- #2*/
if(isset($_POST["register"])) {
	if(!$_POST["email"] | !$_POST["password"] | !$_POST["password2"]) {
		die("You did not fill in all the fields");
	}

/* do escape strings for SQL injection */
$emailsafe = mysqli_real_escape_string($conn, $_POST["email"]);
$passsafe = mysqli_real_escape_string($conn, $_POST["password"]);
$pass2safe = mysqli_real_escape_string($conn, $_POST["password2"]);

/* Check and see if email EXISTS in db. */

/* Insert email in db if does not exist */
$sql = "SELECT email FROM users WHERE email = '$emailsafe'";//limit 1
$result = $conn->query($sql) or die("invalid email check " . mysqli_error($conn));
$numrows = mysqli_num_rows($result);
if($numrows != 0) {
	die("Sorry the email " . $_POST["email"] . "is already in use");
}
// check if passwords both match
if($_POST["password"] != $_POST["password2"]) {
	die("Passwords did NOT match");
}
// if passwords match encrypt
$hashedpass = password_hash($passsafe, PASSWORD_DEFAULT);
// With everything escaped and hashed INSERT into db.
$sqlINSERT = "INSERT INTO users(email, password) " .
	"VALUES('" . $emailsafe ."','". $hashedpass ."')";
if($query = $conn->query($sqlINSERT)=== TRUE) {
	echo '<script type="text/javascript">alert("Successfully INSERTed");</script>';
} else {
	die("Unsuccessful");
}
$conn->close();
?>

I ran my php with ajax and it creates the file successfully so i know it runs. I wanted to setup PHP debugging with atom or visual studio but i have not been able to successfully do that... I tried with xdebug so i just use the network tab inchrome to find if errors exist.
I wish i was able to run debug line by line I use xampp for some reason my network tab makes me use mysqli I thought the newest php just used mysql I downloaded the php7 and put it on my hard drive but there was no php.ini to setup xdebug. I found a php.ini in xampp but that did not correctly do it. I also downloaded the NEWEST xampp thinking that was the problem.. packages on atom followed many different guides online but nothing was successful.

Hello,

I had updated from PHP 5.x to PHP 7.2. The issue is that my cron files are not running now. The sites based on php 5 code runs without any issues.

The cron files when run from cmd are producing errors like to change to mysqli.

Any solutions?

Thanks.

<?PHP

class database {
	// Variables
	public $test;
	
	// Constructor
	public function __construct()
	{
		$test = "4";
	}
	
	
	// Functions
	//
	public function test()
	{
		var_dump($this->test);
	}
	
}

$test = new database;

$test->test();

?>

I have the following php code that errors as indicated:

$query = $con->query('SELECT FILENAME, country, area, city FROM download WHERE FILENAME is not null');

Fatal error: Uncaught PDOException: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'country' in 'field list' in /home/larry/web/test/public_html/report1.php:47 Stack trace: #0 /home/larry/web/test/public_html/report1.php(47): PDO->query('SELECT FILENAME...') #1 {main} thrown in /home/larry/web/test/public_html/report1.php on line 47

The Select statement doesn't error when run in mysql shell or phpmyadmin. Here's the result of show create table download:

localhost/test/download/        http://localhost/phpmyadmin/tbl_sql.php?db=test&table=download&token=5739c407033be3e118287bc7a9041c2c
 Current selection does not contain a unique column. Grid edit, checkbox, Edit, Copy and Delete features are not available. 

Your SQL query has been executed successfully.

show create table download

download    CREATE TABLE `download` (
  `ID` int(5) NOT NULL AUTO_INCREMENT,
  `LOG_TIME` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `IP_ADDRESS` int(64) unsigned NOT NULL,
  `FILENAME` varchar(50) COLLATE utf8_general_mysql500_ci DEFAULT NULL,
  `country` varchar(50) COLLATE utf8_general_mysql500_ci DEFAULT NULL,
  `area` varchar(50) COLLATE utf8_general_mysql500_ci DEFAULT NULL,
  `city` varchar(50) COLLATE utf8_general_mysql500_ci DEFAULT NULL,
  PRIMARY KEY (`ID`),
  UNIQUE KEY `ID` (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=1266 DEFAULT CHARSET=utf8 COLLATE=utf8_general_mysql500_ci    

Does anyone have an idea why this is happening?

Good day everyone. I go by the nickname Sbosh and I am a newbie in PHP.

I am currently learning php using video tutorials. I need help with regards to php queries.

I have the following code:

"SELECT 'food', 'calories' FROM 'diet' ORDER BY 'id'"

which is supposed to display content from a table inside phpmyadmin which i created manually using phpmyadmin. But It gives an error saying i must check the MariaDB version on how to right this query, something like that.

Please assist on the proper way of writing this code. Again I am a newbie in PHP and just starting to learn, so your help will be highly appreciated.