php - PHP逻辑每日更新

Question

有一种情况，如果员工从 join_date 起已经工作了一年，他将获得 12 个 Leave_balance，但下一年如何确定？因为它不能乘以年份差。例如

• join_date 2015年1月15日，今天日期是2016年1月15日，则更新leave_balance = Leave_balance + 12；
• 但是，在 2016 年 1 月 15 日至 2017 年 1 月 15 日之间，他使用了 5 天假期，因此他的 Leave_balance 将为 8，
• 2017 年 1 月 15 日，他将收到另一个休假余额配额 12，因此 Leave_balance 必须更新为 8 + 12 = 20。

我想的只是这样 if(join_date = 今年同一日期) 然后更新leave_balance

请帮忙

需要的表很简单，status_employee、join_date和leave_balance

$q = "select * from employee";
$r = mysql-query($q);
while($result = mysql_fetch_array($r))
{
    $datetime1 = new DateTime($result['join_date']);
    $interval = $datetime1->diff(new DateTime());
    $year= $interval->y;

       if($year > 0 && status_employee = 'Permanent Staff')
       {
             $leave_balance = $result['leave_balance'] + (12 * $year);
       }
}

仅供引用

更新

$q = "SELECT * FROM employee WHERE employee_type = 'Permanent' && DATEDIFF(.join_date , CURDATE()) >= 365";

$r = mysql_query($q);
while($res = mysql_fetch_array($r))
{
    $q = "SELECT * FROM holiday_transaction WHERE id_employee='".$res['id_employee']."'";
    $r = mysql_query($r);
    $res = mysql_fetch_array($r);
    $check = mysql_num_rows($r);

$id_employee = $res['id_employee'];
if($check > 0) //employee is on the transaction_table
{
    //check latest renewal date
    $q = "SELECT * FROM holiday_transaction WHERE id_employee = '$id_employee' && transaction_type='Renewal' && DATEDIFF(CURDATE() ,trn_date) >= 365";
    $r = mysql_query($q);
    $check = mysql_num_rows($r);

    if($check == 0) //0 means he qualify with conditions above that his latest renewal is already more than a year
    {
        $q = "INSERT INTO holiday_transaction(id_employee,holiday,transaction_type) VALUES('$id_employee','12','Renewal')";
        mysql_query($q)
    }
    else
    {
        //don't insert because it is not yet 365days since his latest renewal
    }

}
else //never been on the table and give 12 quota
{
    $q = "INSERT INTO holiday_transaction(id_employee,holiday,transaction_type) VALUES('$id_employee','12','Renewal')"; 
    mysql_query($q)
}

}

Answer 1

您不需要 cron 作业。当有人下次登录时，您可以只在 mysql 上运行存储过程或运行一些 php 代码。只需在向他们展示页面之前执行此操作即可。