如何使用单个查询执行多个算术运算并将答案存储在同一表的两个不同列中

Question

这将从数据库中选择数据，并对其进行处理

    $name = mysqli_real_escape_string($connect, $_GET['name']);
                if (isset($_GET['name'])) {
                # code...

                        echo "</br> <br>";

                $count="SELECT id,employee_name, employee_salary, employee_age, class_id, (( employee_salary + employee_age + employee_allowance)/(100) * 100) AS staff_percentage, (employee_salary + employee_allowance + employee_age ) as staff_total FROM `employee`
                ";

                echo "<table>";
                echo "<tr><th>id</th><th>employee name</th><th>employee salary</th><th>employee age</th><th>employee allowance</th><th>staff total</th><th>staff total</th></tr>";

                    foreach ($connect->query($count) as $row) {
                    echo "<tr ><td>$row[id]</td><td>$row[employee_name]</td><td>$row[employee_salary]</td><td>$row[employee_age]</td><td>$row[employee_allowance]</td><td>$row[staff_percentage]%</td><td>$row[staff_total]</td></tr>";

在这个地方，我想捕获已经处理过的数据并将其发送到数据库，不仅是为了获得一个ID，还包括了所有的麻烦。

                    $sql = "UPDATE `employee` set employee_name = '" . $row["employee_name"] . "', employee_salary='" . $row["employee_salary"]."', employee_allowance='" . $row["employee_age"]."', class_id='" . $row["class_id"]. "', class_total;='" . $row["staff_total"]."', staff_percentage='" . $row["staff_percentage"]. "'";
                                $result = $connect->query($sql);

                                print "<script>alert('you have been registerded succesfully!');</script>";
                                        }

            echo "</table>";

            }

Answer 1

只需像这样在同一个查询中执行两个计算

Select  test, exam, 
        (test + exam) as total,
        ((test + exam)/100)*100 as percentage       
FROM school;

Answer 2

只需将查询的两个算术部分结合在一起即可：

SELECT test, exam, 
(test + exam) as total,
((test + exam)/100)*100 as percentage       
FROM school;