目前我试图在MySQL Prepare Statement中使用Like，但问题是我没有办法在prepare语句中编写它。我用谷歌搜索了它，但没有写出来。这是我的代码。
$name = "Siqarah";
$statement = mysqli_prepare($con, "SELECT Institute_Name FROM institute WHERE Institute_Name LIKE ? " );
mysqli_stmt_bind_param($statement,"s", $name);
好的，我从这个链接得到答案
Correct way to use LIKE '%{$var}%' with prepared statements? [mysqli]