CODEWITHSUNDEEP
phpmysqlmariadb
Muhammad Ikram UL Mustafa
Muhammad Ikram UL Mustafa

Reputation: 370

Subject Update Failed Mysql

    Subject Update Failed!!You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '' at line 1
I am stuck here can anyone help me what I am missing in this code.The error is in Update Query.

Everything is ok, and I don't get any syntax error when I write the code (I am using a Dreamviwer code editor software. However, when I run it, I get this error: //Process the form

    $id= $current_subject["Id"];
$name=mysql_prep($_POST["Name"]);
$position=(int)$_POST["Position"];
$visible=(int)$_POST["Visible"];

$query="UPDATE subjects SET Name='{$name}',Position=$position,Visible=$visible WHERE Id={$id}";

$result= mysqli_query($conn, $query);
if($result && mysqli_affected_rows($conn)==1){
    //success
    $_SESSION["message"]="Subject updated.";
    redirect_to("manage_content.php");

}else{

    //Failure
   $message="Subject Update Failed" . $conn->error;

    }

Upvotes: 0

Views: 34

Answers (1)

svgrafov
svgrafov

Reputation: 2024

Most likely you mistyped the parameter name. Еcho your parameters first.

And use prepared statements to prevent SQL injections:

$dbh = new PDO('mysql:host=localhost;dbname=test', $user, $pass);
$query="UPDATE subjects SET Name = ? ,Position = ?,Visible = ? WHERE Id = ?";
$stmt = $dbh->prepare($query);
$stmt->bindParam(1, $name);
$stmt->bindParam(2, $position);
$stmt->bindParam(3, $visible);
$stmt->bindParam(4, $id);
$stmt->execute();
$stmt->fetchAll();

Further reading: PDO.

Upvotes: 1

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