unexpected '=' in the code, Tried to execute two SQL queries in one php file

Question

I m trying to get the existing "usercouponinhand" value from "userpaytoget" table and add it to received 'id' value as from "GET" tag. Then update it to the same "userpaytoget" table "usercouponinhand" column.

But unfortunately i see this "Unexpected =" error on the line result = $conn->query($sql);

The code follows:

<?php include('usercoupondelete.php'); ?>
<?php
$servername  = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "";

$mobile = $_SESSION['mobile'];
$date = date('M-d,Y H:i:s');
$date2 = date('M-d,Y');

$conn = new mysqli ($servername, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT * FROM userpaytoget WHERE mobile = '$mobile' ";
result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {

$usercouponinhand = $row["usercouponinhand"];
$couponvalue = $_GET["id"];
$totalvalue = $couponvalue + $usercouponinhand ;

$sql2 = "UPDATE userpaytoget SET usercouponinhand = '$totalvalue', date = '$date', date2 = '$date2'
WHERE mobile = '$mobile'";

if ($conn->query($sql2) === TRUE) {
echo  '<a href="usercoupondelete"></a>';
}
else {
echo "ERROR" . $sql2 . "<br>" . $conn->error;
}

}
 } else {
echo "None";
} 
$conn->close();
?>

Any Help is greatly appreciated..

Possible duplicate of [PHP Parse/Syntax Errors; and How to solve them?](http://stackoverflow.com/questions/18050071/php-parse-syntax-errors-and-how-to-solve-them) — chris85, Jan 13 '17 at 06:26
@HarishKolliparat I did vaguely here's a link, http://php.net/manual/en/mysqli.quickstart.prepared-statements.php. Parameterize it and use the prepare/execute and you'll be good. — chris85, Jan 13 '17 at 06:31
@NN PDO and mysqli both support prepared statements, either can be used incorrectly (as demonstrated here). — chris85, Jan 13 '17 at 06:32

score 0 · Accepted Answer · answered Jan 13 '17 at 06:21

0

You are missing a $

It should be $result = $conn->query($sql);

answered Jan 13 '17 at 06:21

dobaniashish

66
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5

unexpected '=' in the code, Tried to execute two SQL queries in one php file

1 Answers1