MySQLi登錄腳本彈出空白頁

Question

這絕對沒有。查詢運行，但在此之後，沒有任何反應。這裏是login.php的代碼

我沒有收到任何錯誤，只是一個空白頁。這是純粹的骨頭腳本。謝謝您的幫助。 MySQLi新手，剛剛停止使用舊的東西。

我得到的錯誤是： 注意：使用未定義的常量登錄 - 在文件路徑/ login2.php假定「登錄」第15" 行

<?php 
// This works 
$con = new mysqli('localhost', 'username', 'password', 'kristen_listing'); 

if($con->connect_errno > 0){ 
die('Unable to connect to database [' . $con->connect_error . ']'); 
} 


// checking the user 

// in the index.php file I have submit named "login" 
if(isset($_POST[‘login’])){ 

$username = $_POST[’username’]; 

$password = $_POST[‘password’]; 

$sel_user = <<<SQL 
SELECT * 
FROM `listing` 
WHERE `id` = 1 
AND `username` = '$username' 
AND `password` = '$password' 
SQL; 

$run_user = mysqli_query($con, $sel_user); 

$check_user = mysqli_num_rows($run_user); 

if($check_user>0){ 

$_SESSION[‘username’]=$username; 

print "Logged in"; 

} 

else { 

print "Invalid username or password"; 

} 

} 




?> 

Answer 1

嘗試啓用錯誤處理這個：

error_reporting(E_ALL);

注意，不要使用在實際環境中

Answer 2

有可能你的一個實際查詢有錯誤。添加額外的日誌和錯誤檢查，看看它可能已停止：

<?php 
echo "Start of PHP"; 
// This works 
$con = new mysqli('localhost', 'username', 'password', 'kristen_listing'); 

if($con->connect_errno > 0){ 
die('Unable to connect to database [' . $con->connect_error . ']'); 
} 


// checking the user 

// in the index.php file I have submit named "login" 
if(isset($_POST[‘login’])){ 

$username = $_POST[’username’]; 

$password = $_POST[‘password’]; 

$sel_user = <<<SQL 
SELECT * 
FROM `listing` 
WHERE `id` = 1 
AND `username` = '$username' 
AND `password` = '$password' 
SQL; 

$run_user = mysqli_query($con, $sel_user) or die(mysqli_error($con)); 

$check_user = mysqli_num_rows($run_user); 
echo "Number of Returned Users".$check_user; 

echo "Before If Statement"; 

if($check_user>0){ 

$_SESSION[‘username’]=$username; 

print "Logged in"; 

} else { 
echo "Incorrect Username/Password"; 
} 
echo "end"; 

我添加了一個「或死」的mysqli_query也加入呼應「$ check_user」，呼應如果語句之前和之後的呼應如果聲明。

我還添加了處理不正確的用戶名/密碼。

作爲一個附註，我想讓你知道你有SQL注入的風險。這裏有一個很好的問題從堆棧溢出：How can I prevent SQL-injection in PHP?

Answer 3

我得到它在Papchonk和minifisch的幫助下工作。由於某種原因變量沒有從表單中傳遞，所以我編輯了一下。這是準系統。它應該返回1個用戶。這是代碼。

<?php 
// error_reporting(E_ALL); 
$username=$_POST['username']; 
$password=$_POST['password']; 
$login=$_POST['login']; 

// This works 
$con = new mysqli('localhost', 'username', 'password', 'kristen_listing'); 

if($con->connect_errno > 0){ 
die('Unable to connect to database [' . $con->connect_error . ']'); 
} 


// checking the user 


if(isset($login)){ 


$sel_user = <<<SQL 
SELECT * 
FROM `listing` 
WHERE `id` = 1 
AND `username` = '$username' 
AND `password` = '$password' 
SQL; 

$run_user = mysqli_query($con, $sel_user) or die(mysqli_error($con)); 

$check_user = mysqli_num_rows($run_user); 
echo "Number of Returned Users".$check_user; 

if($check_user>0){ 

$_SESSION[‘username’]=$username; 

print "Logged in"; 

} else { 
echo "Incorrect Username/Password"; 
} 
echo "end"; 
} 
?>