which part of my code is wrong in mysqli_real_escape_string();

Question

I have the mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in....& Undefined variable: link in..... error.i use the $link in my code.but i dont where is my code is wrong. this is my code:

    function Check_Get($value)
{
    include 'connect.php';
    $return1 = mysqli_real_escape_string($link,$value);
    $return2 = htmlspecialchars($return1);
    $return3 = intval($return2);
    return $return3;
}
-------------------------------
if(isset($_POST['login'])){
    if($_POST['username']=="" || $_POST['password']=="" || $_POST['email']=="")
    {
        $security->Redirect("index","empty=1020");
    }
    else{

    }
}
-------------------------------
    if(isset($_GET['empty']))
    {
        $security->Check_Get($_GET['empty']);
        $template->message("please fill all field","red");
    }
-------------------------------

this is my connection code:

$Server_name = "localhost";
$Server_username = "root";
$Server_password = "";
$db_name = "news";
$link = mysqli_connect($Server_name,$Server_username,$Server_password) or 
        exit ("Error in Connection to Server");
if($link)
{
    if(mysqli_select_db($link,$db_name))
    {
        mysqli_query($link,"set names utf8");
        mysqli_query($link,"set charset utf8");
        $result = mysqli_query($link,$sql);
        if(!$result)
        {
            echo "Error in Query";  
        }
        return $result;
    }
    else
    {
        echo "Error in Connection to DataBase";
    }
}
else
{
    echo "Error in Connection to Server";
}

;

i think you did not give the connection object in it.. `mysqli_real_escape_string($conn,$data);` — phpfresher, Nov 27 '15 at 12:34
that isn't enough code here. You need to post an example of your connection. Plus, this smells more like a variable scope issue. — Funk Forty Niner, Nov 27 '15 at 12:47
and these are not needed `$return2 = htmlspecialchars($return1); $return3 = intval($return2);` you're already escaping. — Funk Forty Niner, Nov 27 '15 at 12:48
don't put your connection code in comments, edit your question — Funk Forty Niner, Nov 27 '15 at 12:48

score 3 · Accepted Answer · answered Nov 27 '15 at 12:33

I believe you will be using the below to create a connection object using:

$conn = mysqli_connect();

And then you need to pass that object here:

$user = mysqli_real_escape_string($conn, $user);

I believe you are putting something like the below, without the first parameter:

$user = mysqli_real_escape_string($user);

which part of my code is wrong in mysqli_real_escape_string();

1 Answers1