计算 2 个日期范围之间的交集答案

【问题标题】：Calculate intersection between 2 date ranges计算 2 个日期范围之间的交集
【发布时间】：2022-01-24 12:19:24
【问题描述】：

我想计算特定周内的空闲天数（缺勤）。我使用返回以下数据的 API：

    {
    "count": 1,
    "data": [
        {
            "id": "11ec62ff1df2654d8bd6f1d234a6c496",
            "type": "HOLIDAY",
            "from": "2021-12-22",
            "to": "2021-12-23",
            "resourceId": "11ec46d6547a00728be3e1ed8ff29535",
            "createdAt": "2021-12-22T08:14:00"
        }
    ],
    "success": true
    }

这些是假期和疾病数据。我有一份每周报告，我需要计算那一周的缺勤天数。我需要找到一种简单的方法来计算一周中的缺勤天数。

我尝试过使用https://www.php.net/manual/de/datetime.format.php 并将其转换为“z”格式，但它看起来并不优雅，从性能角度来看，我认为它不是最好的。


//The week range 
$weekStart = new DateTime("2021-12-20");
$weekEnd = new DateTime("2021-12-24");

//The Planned absence
$absenceStart = new DateTime("2021-12-22");
$absenceEnd = new DateTime("2021-12-23");

//Specify the DateInterval for calculating the period
$interval = DateInterval::createFromDateString('1 day');

//Need to add the interval to the end date in order to consider the end as well
$weekEnd->add($interval);
$absenceEnd->add($interval);

//Getting the 2 periods week and absence
$weekPeriod = new DatePeriod($weekStart, $interval, $weekEnd);
$absencePeriod = new DatePeriod($absenceStart, $interval, $absenceEnd);


$weekArray = array();
$absenceArray = array();

//put the day number format('z') into an array of the week
foreach ($weekPeriod as $i => $dt) {
    $weekArray[$i] = $dt->format('z');
}

//put the day number format('z') into an array of the absence
foreach ($absencePeriod as $i => $dt) {
    $absenceArray[$i] = $dt->format('z');
}

//get the intersection between both arrays
$ergebnis = array_intersect($weekArray, $absenceArray);

//calculate the number of entries
echo "The employee has <b>".count($ergebnis)."</b> free days in the week from 2021-12-20 until 2021-12-24";

这是返回正确的信息。

The employee has 2 free days in the week from 2021-12-20 until 2021-12-24

任何人都可以建议是否有更好的方法，或者我是否可以至少对其进行调整以使其更优雅和性能更好？

非常感谢

【问题讨论】：

标签： php arrays datetime

【解决方案1】：

如果不需要该信息，您可以消除对缺席数组和第二周数组循环的循环，而不是循环遍历周数组、缺席数组，然后再次遍历周数组 (array_intersect)。所以你只有 1 个循环而不是 3 个应该更高效。

//The week range 
$weekStart = new DateTime("2021-12-20 00:00:00");
$weekEnd = new DateTime("2021-12-24 23:59:59");

//The Planned absence
$absenceStart = new DateTime("2021-12-22 00:00:00");
$absenceEnd = new DateTime("2021-12-23 23:59:59");

//Specify the DateInterval for calculating the period
$interval = DateInterval::createFromDateString('1 day');

//Getting the available period
$weekPeriod = new DatePeriod($weekStart, $interval, $weekEnd);

$availableDayCount = 0;

//put the day number format('z') into an array of the week
foreach ($weekPeriod as $dt) {
    // Filter out days the employee is absent.
    if($dt < $absenceStart || $dt > $absenceEnd) {
        $availableDayCount += 1;
    }
}

//calculate the number of entries
echo "The employee has <b>" . $availableDayCount . "</b> free days in the week from " . $weekStart->format('Y-m-d') . " until " . $weekEnd->format('Y-m-d');

此外，它似乎没有提供正确的信息（也许我错了？）。你的问题是 2 天，但我数 3：

2021-12-20
2021-12-21
2021-12-24

（22号和23号缺席）

如果这不正确，您可以将 $weekEnd 更改为 2021-12-23（我添加了一天的开始/结束时间，以便计算一整天）。

【讨论】：