無法弄清楚爲什麼我的查詢不會通過

Question

這一整天都在工作，訪問了很多論壇，並接受了很多建議，但似乎仍然有些問題。

我試圖通過SQL提交查詢，一旦所有正確的用戶信息被輸入，然後重定向用戶返回index.html作爲一個'登錄'用戶一旦完成。

下面是代碼：

error_reporting(E_ALL); ini_set('display_errors', 1); mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); 
//setup database connection 
$dbhost = 'localhost'; 
$dbuser = 'root'; 
$dbpass = ''; 
$mysql_database = "21st"; 

$conn = mysqli_connect($dbhost, $dbuser, $dbpass) or die("Could not connect database"); 
     mysqli_select_db($conn, $mysql_database) or die("Could not select database"); 

//get user info  
$username = $_POST['username']; 
$email = $_POST['email']; 
$password = $_POST['password']; 
$password2 = $_POST['password2']; 
//if passwords are equal and longer than 5 characters, and username is longer than 4 characters 
if ($password != $password2) || (strlen($username)) < 4 || (strlen($password < 5)) { 
     header('Location: index.html'); 
} else { 
     //encrypt passwords 
     $password = md5($password); 
     $password2 = md5($password2); 
     //sanitize email 
     $sanitized_email = filter_var($email, FILTER_SANITIZE_EMAIL); 
     //insert query into database 
     $sql = "INSERT INTO members (username, email, password) VALUES ('$username','$email','$password')"; 
     mysqli_query($conn, $sql); 
     //redirect user to index.html once all is completed 
     header('Location: index.html'); 
     exit(); 
    } 
} 

Answer 1

您的編碼錯誤：

if ($password != $password2) || (strlen($username)) < 4 || (strlen($password < 5)) { 
     header('Location: index.html'); 
} 

應該

(strlen($username) < 4) 
(strlen($password) < 5) 

編輯

你if else語法不正確（docs）

正確的代碼應該是這樣的：

if (($password != $password2) || (strlen($username) < 4) || (strlen($password) < 5)) { 
    header('Location: index.html'); 
} else { 
     // Handle your false 
}