如何創建一個就業報告出來使用PHP

Question

的考勤記錄我有數據庫表命名爲：時間表它具有下列列和數據：

employee_id employee_name year month day timein timeout department 
1   dave   2016 09  15 8  4  finance 
1   dave   2016 09  16 8  4  finance 
1   dave   2016 09  17 8  4  finance 
2   frank   2016 09  15 8  4  purchase 
2   frank   2016 09  16 8  4  purchase 

它記錄員工日常考勤，你可以往上看。我要的是創造這些考勤記錄了使用PHP的就業報告顯示的工作總小時數在一個單獨的HTML表中的每個部門，像這樣的僱員：

財務部門的工資HTML表格

employe_id | employee_name | total_working_hours 
-----------+----------------+--------------------- 
1   | dave   | 24 (8 hrs * 3 days) 

採購部門工資單的HTML表格

employe_id | employee_name | total_working_hours 
-----------+----------------+--------------------- 
1   | frank   | 16 

請注意，我不知道所有emplo的ID的所以代碼應該只列出使用PHP/MYSQL按部門分組的所有人

Answer 1

有幾個視圖可以完成這項工作：一個用於工作時間，一個用於員工，然後您可以加入他們並設置過濾器，如這樣的：

-- Query the working hours 
CREATE VIEW 
vw_working_hours 
AS 
SELECT 
    employee_id, 
    year, 
    month, 
    SUM((timeout+12) - timein) AS total_working_hours 
FROM 
    timesheets 
GROUP BY 
    employee_id, 
    year, 
    month; 

-- Query the employees 
CREATE VIEW 
    vw_employees 
AS 

SELECT 
    DISTINCT 
    employee_id, 
    employee_name, 
    department 
FROM 
    timesheets; 

-- This query is the actual report 
-- just had to filter by department 
-- in your script 
SELECT 
    wh.employee_id, 
    emp.employee_name, 
    wh.total_working_hours 
FROM 
    vw_working_hours AS wh 
JOIN 
    vw_employees AS emp 
ON 
    wh.employee_id = emp.employee_id 
WHERE 
    wh.year = 2016 
    AND 
     wh.month = '09' 
    AND 
     emp.department = 'finance'; 

或者，在一個單一的查詢（無意見）：

SELECT 
    wh.employee_id, 
    emp.employee_name, 
    wh.total_working_hours 
FROM 
    (
    SELECT 
     employee_id, 
     year, 
     month, 
     SUM((timeout+12) - timein) AS total_working_hours 
    FROM 
     timesheets 
    GROUP BY 
     employee_id, 
     year, 
     month 

    ) AS wh 
JOIN 
    (
    SELECT 
     DISTINCT 
     employee_id, 
     employee_name, 
     department 
    FROM 
     timesheets 

    ) AS emp 
ON 
    wh.employee_id = emp.employee_id 
WHERE 
    wh.year = 2016 
    AND 
     wh.month = '09' 
    AND 
     emp.department = 'finance'; 

在PHP中，你應該使用一個循環和積累的變量。首先，你應該預過濾器這樣的查詢：

SELECT 
    * 
FROM 
    timesheets 
WHERE 
    department = 'finance' 
    AND 
    year = 2016 
    AND 
    month = '09' 
    ORDER BY 
    employee_id; 

然後，如果結果是在一個名爲$行，東西在PHP這樣的多維數組：

$employee_id = $rows[0]['employee_id']; 
    $employee_name = $rows[0]['employee_name']; 
    $accumulated = 0; 

    foreach($rows as $row) { 
     $total_working_hours = ($row['timeout'] + 12) - $row['timein']; 
     if ($row['employee_id'] == $employee_id) { 
     // Same employee, acumulate 
     $accumulated += $total_working_hours; 
     } else { 
     // Another employee, pass the acumulation to result 
     $rowTmp['employee_id'] = $employee_id; 
     $rowTmp['employee_name'] = $employee_name; 
     $rowTmp['total_working_hours'] = $accumulated; 
     $result[] = $rowTmp; 
     // Updated the accumulation variables 
     // new employee 
     $employee_id = $row['employee_id']; 
     $employee_name = $row['employee_name']; 
     // reset accumulated 
     $accumulated = $total_working_hours; 
     } 
    } 

    // at the end, updates the las result: 
    $rowTmp['employee_id'] = $employee_id; 
    $rowTmp['employee_name'] = $employee_name; 
    $rowTmp['total_working_hours'] = $accumulated; 
    $result[] = $rowTmp; 

    // Should pass result to HTML table 
    print_r($result);