Question

<?php 
if($_SERVER['REQUEST_METHOD']=='POST'){ 
    $username = $_POST['username']; 
    $password = $_POST['password']; 
    $sql = "SELECT * FROM main_member WHERE username='$username' AND password='$password'"; 
    require_once('connect.php'); 
    sql $result = mysqli_query($con,$sql); 
    $check = mysqli_fetch_array($result);  
    if(isset($check));
    { 
        echo "success"; 
    }else{ 
        echo "Gagal"; 
    } 
    mysqli_close($con); 
}
?>

它说解析错误语法错误，在第7行的C：\ xampp \ htdocs \ php \ login.php中意外的$ result T_VARIABLE，请帮助我们，我真的很感激

Answer 1

此行中有错误

sql $result = mysqli_query($con,$sql);

将此更改为

$result = mysqli_query($con,$sql);

更改此行

if(isset($check));

到

if(isset($check))

得到解析/语法错误，无法找出原因

1 个答案: