I need to convert to current PHP

Question

I have an old form that would like to adjust a page with current PHP parameters. Follows code:

<?php include "login_senha.php";

$email = $_POST['email'];

$sql = mysql_query("SELECT * FROM usuarios WHERE email = '$email'");
while($linha = mysql_fetch_array($sql)) {
    $email_db = $linha['email'];
}
$cont = mysql_num_rows($sql);

if($cont == 0){
echo"
<meta http-equiv='refresh' content='0; url=cadastro02.php' />
<script type='text/javascript'>alert('Login nao existe')</script>
";
} else {
if($senha_db !=$senha){
echo"
<meta http-equiv='refresh' content='0; url=cadastro02.php' />
<script type='text/javascript'>alert('Senha nao corresponde')</script>
";
} else {
session_start();
$_SESSION['email_usuario'] = $email;

header("Location: segunda.php");
}
}
$sql = mysql_query("SELECT * FROM dropdown");
while($record = mysql_fetch_array($sql)) {
    echo '<option value="' . $record['item'] , '">' . $record['item'] .'"</option>';

}

mysql_close($db);

?>

If I’m not mistaken today is no more mysql and yes mysqli, correct?

Answer 1

It works yes, there must be some other error in your code...

Take a look at the php manual for every function that doesn’t work.

https://www.php.net/manual/en/mysqli-result.fetch-array.php

I tested this code below using shaman and local base and it worked correctly.

<?php
$link = mysqli_connect("localhost", "root", "", "base_teste");

$mysqli = new mysqli("localhost", "root", "", "base_teste");
/* check connection */
if ($mysqli->connect_errno) {
    echo printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

$query = "SELECT * FROM usuarios";
$result = mysqli_query($link, $query);

while ($linha = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    $email_db = $linha['email'];
    echo '<br> email: ' . $email_db;
}

Answer 2

SO change from "mysql_" to "mysqli_", use CTRL + SPACE that will fill in correctly. It’s all about testing...