如何在PHP中检查MySQL结果是否返回空？

如何检查mysql结果是否为空.如果mysql查询结果为空,则不应该删除condtion.
如果MysqL结果数据那里& in else条件我的错误我的消息在那里,但它没有显示任何错误消息.
???我尝试了以下代码,但没有在屏幕上显示任何警报或回显消息.

<?PHP
$sql = "select * from hall_search_data_1 where rent BETWEEN '".$_SESSION['amount1']."' AND '".$_SESSION['amount2']."'";
$res = MysqL_query($sql);
if(!empty($res)){
while($row=MysqL_fetch_row($res))
{
// here my data 
}
}else{
echo "no results found";
echo "<br>";
echo "<script>alert('No any result found..!!');</script>";
echo "no results found";
}
?>

解决方法:

    <?PHP
    $servername = "localhost";
    $username = "username";
    $password = "password";
    $dbname = "myDB";
    $conn = new MysqLi($servername, $username, $password, $dbname);
    if ($conn->connect_error) {
        die("Connection Failed: " . $conn->connect_error);
    }
    $sql = "SELECT id, firstname, lastname FROM MyGuests";
    $result = $conn->query($sql);
    if ($result->num_rows > 0) {
        while ($row = $result->fetch_assoc()) {
            echo "id: " . $row["id"] . " - Name: " . $row["firstname"] . " " . $row["lastname"] . "<br>";
        }
    } else {
        echo "0 results";
    }
    $conn->close();
    ?>

（编辑：北几岛）

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