表没有填写xammp使用PHP代码

Question

对于以下代码，我没有得到预期的输出。我希望获取表中的数据并将其放在新表中。表格l3包含rollno，department，cgpa。表格l4必须填充rollno，department，{{ 1}}检查是否指定了所需的cgpa。

cgpa

new.php具有以下代码

<?php

include "new.php";

$cgpa = 4;

$sql = "SELECT ROLLNO, NAME, DEPARTMENT, CGPA FROM l3 WHERE CGPA >= '$cgpa'";

$result = mysqli_query($link, $sql);

if (mysqli_num_rows($result) > 0) {
    while($row = mysqli_fetch_assoc($result)) {
        $sql2 = "INSERT INTO l4 (ROLLNO,NAME,DEPARTMENT,CGPA)VALUES({$row['ROLLNO']},{$row['NAME']},{$row['DEPARTMENT']},{$row['CGPA']})";
        mysqli_query($link, $sql2);
    }
}

else
    echo "No records found.";

mysqli_close($link);
?>