Question

我们有 scheduled_jobs 的表格，此表格包含以下两列：

重复类型：每周/每两周/每月

开始日期：重复的日期，即如果重复类型是每月，开始日期设置为2015-12-05，则会在每个月的第五天重复。

以下查询是当前正在使用的代码片段：

<?php
    foreach(range(1,100) as $num)
    {
        $numtimestwo = $num * 2;

        $query = "SELECT * FROM scheduled_jobs WHERE deleted != TRUE AND ((DATE_ADD(DATE(beginning_date), INTERVAL $num MONTH) = CURDATE() AND repeat_type = 'monthly') OR (DATE_ADD(DATE(beginning_date), INTERVAL $numtimestwo WEEK) = CURDATE() AND repeat_type = 'fortnightly') OR (DATE_ADD(DATE(beginning_date), INTERVAL $num WEEK) = CURDATE() AND repeat_type = 'weekly'))";

        // If result is not empty, process the scheduled job
    }
?>

但是我注意到了一些问题，这将最终停止工作，因为begin_date变得越来越老，看起来应该有更好的解决方案。

Answer 1

我使用PHP's diff method解决了DateTime对象的问题。

以下函数采用$ type（“每日/每周/每两周/每月/六个月/每年”）和$ beginning_date（日期字符串）。

例如。如果您的日期为“2010-01-01”，类型为“每月”，那么您希望在每个月的第一天返回true，这将有效。具有“每周”类型的日期“2015-06-05”将在该日期之后每7天返回一次。

function checkInterval($type, $beginning_date)
{
    // Get DateTime for today, and DateTime object for $beginngin_date input
    $today = date("Y-m-d");
    $today = new DateTime($today);
    $beginning_date = new DateTime($beginning_date);
    // Diff the two dates, and get the save variables for the difference in days, months, years, as well as total days passed ($interval->days)
    $interval = $beginning_date->diff($today);
    $difference_d = $interval->d;
    $difference_m = $interval->m;
    $difference_y = $interval->y;
    $difference_total = $interval->days;
    // If beginning date is today, do not proceed
    if($difference_total < 1)
    {
            return false;
    }
    // If daily, return true all days
    if($type == "daily")
    {
            return true;
    // If weekly, Return true on the day difference being a multiple of 7 days or 0
    } else if($type == "weekly") {
            if($difference_d == "7" || $difference_d == "14" || $difference_d == "21" || $difference_d == "28" || $difference_d == "0")
            {
                    return true;
            }
    // If fortnightly, Return true on the day difference between a multiple of 14 days or 0
    } else if ($type == "fortnightly") {
            if($difference_d == "14" || $difference_d == "28" || $difference_d == "0")
            {
                    return true;
            }
    // If monthly, only return true when it is same day every month (0)
    } else if ($type == "monthly") {
            if($difference_d === 0)
            {
                    return true;
            }
    // If biannually, return true when there's a 6 month difference, and it's the same day of the month
    } else if ($type == "six months") {
            if($difference_m == "6" && $difference_d == "0")
            {
                    return true;
            }
    // Return true when year difference is greater than 0, and day and month are equal (0)
    } else if ($type == "yearly") {
            if($difference_d == "0" && $difference_m == "0" && $difference_y > 0)
            {
                    return true;
            }
    // Else no matches, return false
    } else {
            return false;
    }
    return false;
}

此查询的更简单方法：如果{当前日期}是{每周/每两周/每月}重复/ {start_date}的倍数

1 个答案: