如何使用 PHP 从考勤记录中创建工资单报告

Question

我有一个名为：timesheets 的数据库表，其中包含以下列和数据：

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

如上所示，它记录了员工的每日出勤情况。我想要的是使用 PHP 从这些出勤记录创建一个工资单报告，该报告在单独的 html 表中显示每个部门的每个员工的总工作时间，如下所示：

财务部门工资单html表格

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

采购部门工资单html表格

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

请注意，我不知道所有员工的 ID，所以代码应该只列出使用 PHP / MYSQL 按部门分组的每个人

Answer 1

有几个视图可以完成这项工作：一个用于工作时间，一个用于员工，然后您可以加入它们并对您的过滤器进行编程，如下所示：

-- 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，在单个查询中（无视图）：

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，您应该使用循环和累积变量。首先，您应该使用这样的查询进行预过滤：

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

那么如果结果在一个名为 $rows 的多维数组中，在 php 中是这样的：

$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 );