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所以我目前正在Xampp中使用php和mysql構建一個用戶認證系統。用戶認證PHP mySQL不工作
我設法讓它識別用戶是否存在他們的電子郵件地址，但其他功能似乎沒有工作。例如，要檢查用戶是否已經激活了他們的賬戶，因爲即使我在數據庫中將其活動狀態更改爲1，他們也沒有回來。或者使用登錄功能，即使電子郵件和密碼都正確，但會說它們不正確。
這裏是我的login.php腳本
<?php
include 'init.php';
function sanitize($data){
return mysql_real_escape_string($data);
}
//check if user exists
function user_exists($email){
$email = sanitize($email);
//$query = mysql_query("SELECT COUNT('ID') FROM 'register' WHERE 'email' = '$email'");
return (mysql_result(mysql_query("SELECT COUNT(ID) FROM register WHERE email = '$email'"),0) == 1)? true : false;
}
//check if user has activated account
function user_activate($email){
$email = sanitize($email);
//$query = mysql_query("SELECT COUNT('ID') FROM 'register' WHERE 'email' = '$email'");
return (mysql_result(mysql_query("SELECT COUNT(ID) FROM register WHERE email = '$email' AND 'active' =1"),0) == 1)? true : false;
}
function user_id_from_email($email){
$email = sanitize($email);
return (mysql_result(mysql_query("SELECT id FROM register WHERE email = '$email'"),0,'id'));
}
function login($email,$password){
$user_id = user_id_from_email($email);
$email = sanitize($email);
$password = md5($password);
return (mysql_result(mysql_query("SELECT COUNT(id) FROM register WHERE email = '$email' AND 'password' ='$password'"),0) == 1)? $user_id : false;
}
if(empty($_POST)=== false){
$email = $_POST['email'];
$password = $_POST['password'];
}
if(empty($email)|| empty($password) === true){
$errors[] = "You must enter a username and a password";
}
else if(user_exists($email) === false){
$errors[] = "Email address is not registered";
}
else if(user_activate($email) === false){
$errors[] = "You haven't activated your account yet";
}
else{
$login = login($email, $password);
if($login === false){
$errors[] = "email/password are incorrect";
} else {
echo "ok";
}
}
print_r($errors);
/*$email = $_POST['email'];
$password = $_POST['password'];
if($email&&$password){
$connect = mysql_connect("localhost","root","") or die ("Couldn't Connect");
mysql_select_db("users") or die("Couldn't find Database");
}
else
die("Please enter a username and a password");
$query = mysql_query("SELECT * FROM register WHERE email = '$email'");
$numrows = mysql_num_rows($query);
echo $numrows;*/
?>
我的數據庫稱爲'users'而且目前只有1臺名爲'register'。與行：id, firstname, lastname, email, password, and active。
不好編程！爲什麼要查詢數據庫4次，查找同一個表，同一個用戶，同一個會話查詢一次並將數據導入變量，然後比較php中的值，而不是單獨查詢所有值的數據庫。 – Spidey