Getting parse / syntax error and can not figure out why

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); 
}
?>

it says parse error syntax error, unexpected $result T_VARIABLE in C:\xampp\htdocs\php\login.php on line 7, please help me guys i really appreciate it

i dont know, i just copied it from someone's project – Adam Bayu Jun 24 '16 at 07:18 — Adam Bayu, Jun 24 '16 at 07:18

Shibon · Accepted Answer · 2016-06-24T07:36:35.740

0

you have error in this line

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

change this to

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

change this line

if(isset($check));

to

if(isset($check))

edited Jun 24 '16 at 07:36

answered Jun 24 '16 at 03:17

Shibon

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wow thanks man, it worked but another error appears and it say Parse error: syntax error, unexpected 'else' (T_ELSE) in C:\xampp\htdocs\php\login.php on line 12,, can you help me again? please >_ – Adam Bayu Jun 24 '16 at 07:20
if(isset($check)) no need of semi colon – Shibon Jun 24 '16 at 07:37
omg it works,, thanks man i'm so happy right now*tears – Adam Bayu Jun 24 '16 at 07:49

Getting parse / syntax error and can not figure out why

1 Answers1