Currently I'm trying to Use Like in MySQL Prepare Statement but the issue is i'm not getting a way to write it in prepare statement. I've googled it but didn't get it write. Here is my code.
$name = "Siqarah";
$statement = mysqli_prepare($con, "SELECT Institute_Name FROM institute WHERE Institute_Name LIKE ? " );
mysqli_stmt_bind_param($statement,"s", $name);
ok I got the answer from this link
Correct way to use LIKE '%{$var}%' with prepared statements? [mysqli]