Parse error: syntax error, unexpected '{' in C:\xampp\htdocs\multiusers\users\user_Validate.php on line 13

Question

It seem like the error in the if else statement but I can't trace that error.

Can anyone please help me to solve this error.

<?php
    session_start();
    include("dbconfig.php");
    $table ="";
    $rurl="";
    extract($_REQUEST);
    if($role =="staff"){
        $table="multiusers_staff";
        $rurl="staff/trainer.php";
    }
    else($role=="student"){
        $table="multiusers_student";
        $rurl="student/suexecute.php";
    }
    $mail  = $role."_email";
    $uname = $role."_username";
    $pass  = $role."_password";
    echo $check1="SELECT * FROM $table WHERE $mail ='$uemail' AND $pass ='$upwd' ";
    $login1 = $con->query($check1);
    $login_result1 = $login1->num_rows;
    $row1 = mysqli_fetch_array($login1);
    if(is_array($row1)) {
        $_SESSION['s_usersid']      = $row1[0];
        $_SESSION['s_userrole']     = $role;
        $_SESSION['s_usersemail']   = $row1[$mail];
        $_SESSION['s_usersname']    = $row1[$uname];
    }
    else{ 
        echo $con->error;
    }
    if(isset($_SESSION["s_usersemail"])) {
        echo "<script>window.location.assign('".$rurl."')</script>";
    }
?>

check this post https://stackoverflow.com/questions/18050071/php-parse-syntax-errors-and-how-to-solve-them — Junius L, Dec 09 '17 at 04:54
You need to debug your code from top to bottom, you can put die before conditions and check up till which part your code is working or enclose all braces so that extra brace can come out. — Amit Gupta, Dec 09 '17 at 05:51
dont use `extract($_REQUEST);` your asking for trouble, e.g: `?con=foobar` — Lawrence Cherone, Dec 09 '17 at 06:00
or worse `?role=foobar&table=multiusers_staff WHERE id=1;--` — Lawrence Cherone, Dec 09 '17 at 06:02

score 0 · Answer 1 · answered Dec 09 '17 at 04:56

0

else($role=="student") change this to else if ($role=="student")

else can not have a condition Php Doc

answered Dec 09 '17 at 04:56

Vijay Arun

414
8
15

score 0 · Answer 2 · answered Dec 09 '17 at 05:58

Your error is in else

else never have condition and you put condition here.

You have two ways

use else only.
use elseif($role=="student").

Code is as follows

    <?php
    session_start();
    include("dbconfig.php");
    $table ="";
    $rurl="";
    extract($_REQUEST);
    if($role =="staff"){
        $table="multiusers_staff";
        $rurl="staff/trainer.php";
    }
    else{ //Or use elseif with condition `elseif($role=="student"){`
        $table="multiusers_student";
        $rurl="student/suexecute.php";
    }
    $mail  = $role."_email";
    $uname = $role."_username";
    $pass  = $role."_password";
    echo $check1="SELECT * FROM $table WHERE $mail ='$uemail' AND $pass ='$upwd' ";
    $login1 = $con->query($check1);
    $login_result1 = $login1->num_rows;
    $row1 = mysqli_fetch_array($login1);
    if(is_array($row1)) {
        $_SESSION['s_usersid']      = $row1[0];
        $_SESSION['s_userrole']     = $role;
        $_SESSION['s_usersemail']   = $row1[$mail];
        $_SESSION['s_usersname']    = $row1[$uname];
    }
    else{ 
        echo $con->error;
    }
    if(isset($_SESSION["s_usersemail"])) {
        echo "<script>window.location.assign('".$rurl."')</script>";
    }
?>

Parse error: syntax error, unexpected '{' in C:\xampp\htdocs\multiusers\users\user_Validate.php on line 13

2 Answers2