解析錯誤：語法錯誤，意外的'$密碼'（T​​_VARIABLE）第4行

Question

我試圖創建一個簡單的登錄在HTML使用phpmyadmin數據庫和xampp服務器，但然後這個錯誤顯示*解析錯誤：語法錯誤，意外的'$密碼'（T​​_VARIABLE）*

這是我的代碼。

<?php 

$username = $_POST['user'] 
//*here's the Erro// $password = $_POST['pass'] 

$username = stripcslashes($username); 
$password = stripcslashes($password); 
$username = mysql_real_escape_string($username); 
$password = mysql_real_escape_string($password); 

mysql_connect("localhost", "root", ""); 
mysql_select_db("Login"); 

$result = mysql_query("select * from users where username = '$username' and password = '$password'") or die("Failed to query Database".mysql_error()); 

$row = mysql_fecth_array($result); 

if ($row['username'] == $username && $row['password'] == $password){ 
    echo "Login Succesful! Welcome" =.$row['username']; 
}else { 
    echo "Login Failed" 
} 

?> 

Answer 1

你有多個錯誤：

<?php 

$username = $_POST['user']; //<-------put ';' 
//*here's the Erro// $password = $_POST['pass'] 


$username = stripcslashes($username); 
$password = stripcslashes($password); 
$username = mysql_real_escape_string($username); 
$password = mysql_real_escape_string($password); 



mysql_connect("localhost", "root", ""); 
mysql_select_db("Login"); 


$result = mysql_query("select * from users where username = '$username' and password = '$password'") or die("Failed to query Database".mysql_error()); 

$row = mysql_fecth_array($result); 

if ($row['username'] == $username && $row['password'] == $password){ 
echo "Login Succesful! Welcome =".$row['username']; //<-------change this line 
}else { 
echo "Login Failed";//<-------put ';' 

} 

?> 

Answer 2

您忘記了分號。

$username = $_POST['user']; 

你也忘了這裏。

echo "Login Failed";