用php构建mysql数据库报告

Question

How can I achieve in MySQL query a result like this:

Answer 1

First of all you wil have to get all employees from database table and then iterate through their IDs to get their attendance report in your desired format

$qry_employee = mysqli_query($connection, "SELECT employee_id FROM employee");
if($qry_employee)
{
    if(mysqli_num_rows($qry_employee) > 0)
    {
        while ($employees = mysqli_fetch_array($qry_employee))
        {
            $qry_attendance = mysqli_query($connection, "SELECT e.employee_id, e.employee_name, 
                                a.am_id, a.am_out, a.pm_in, a.pm_out, a.ot_in, a.ot_out, a.date 
                                FROM employee e 
                                JOIN attendance a ON e.ID = a.ID 
                                WHERE a.date BETWEEN '2015-08-01' AND 2015-08-30' 
                                AND e.employee_id = ".$employees['employee_id']."
                                ORDER BY e.ID ASC");

            // your report generation here
        }
    }
    else
    {
        echo "No result found to display.";
    }
}

Answer 2

SELECT `DATE`,`am IN`,`am OUT`,`pm IN`,`pm OUT`,`OT IN`,`OT OUT` 
FROM `EMPLOYEE ATTENDANCE LOGS` 
WHERE `ID No.` = '3' 
AND `Dept.` = `ADMIN/OFFICE` 
AND `Name` = 'GOJO CRUZ, CHARLOTTE'

Answer 3

See Andrews answer. The reason you can only see one Employee is because in his Query, ID No. = 3.

To produce this for all employees, you need a list of Employee IDs, and for each of them run this query.

As an alternative, you could run it without ID restrictions, but you will have to group the data by ID in your Application Code, then