Php不會在數據庫中創建用戶

Question

我編寫了一個代碼，可以防止sql注入，但現在甚至不會創建用戶。下面是我的代碼：

<?php 
$user = $_GET['username']; 
$pass = $_GET['password']; 

$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "test123"; 

$conn = new mysqli($servername, $username, $password, $dbname); 
function selectInfo($user, $pass){ 
    global $conn; 
    $stmt = $conn->prepare("INSERT INTO users (username, password) VALUES (?, ?)"); 
    $stmt->bind_param("ss", $user, $pass); 
    $stmt->execute(); 
    $stmt->close(); 
    } 
?> 

執行時，我沒有得到任何錯誤，但它並沒有創造我需要的用戶。 對不起，這些代碼可能不正確。我在這個新的。

Answer 1

的真正原因是你是不是調用該函數

做這一點

<?php 
$user = $_GET['username']; 
$pass = $_GET['password']; 

$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "test123"; 

$conn = new mysqli($servername, $username, $password, $dbname); 
function selectInfo($user, $pass){ 
    global $conn; 
    $stmt = $conn->prepare("INSERT INTO users (username, password) VALUES (?, ?)"); 
    $stmt->bind_param("ss", $user, $pass); 
    $stmt->execute(); 
    $stmt->close(); 
    } 
selectInfo($user, $pass); 
?> 

或

<?php 
$user = $_GET['username']; 
$pass = $_GET['password']; 

$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "test123"; 

$conn = new mysqli($servername, $username, $password, $dbname); 

    $stmt = $conn->prepare("INSERT INTO users (username, password) VALUES (?, ?)"); 
    $stmt->bind_param("ss", $user, $pass); 
    $stmt->execute(); 
    $stmt->close(); 


?>