在小时数数组中查找最近的下一小时[关闭]

Question

我想在一个小时数组中找到下一个最近的小时，我有下面的代码可以在前后两个方向上搜索最近的小时。

我需要做的是只搜索下一个小时，在下面的示例中应该是 14:00 而不是 03:50。

条件详情：

• 持续时间：24H - 1D
• 开始时间：00:00
• $timeslots 和 $expected_time 中的值都是动态的

$timeslots = ["01:00","01:30","02:00","02:30","00:30","01:00","01:30","02:00","02:30","03:50", "14:00"];
$expected_time = "04:00";
$timestamp = strtotime($expected_time);
$diff = null;
$index = null;
        foreach ($timeslots as $key => $time) {
            $currDiff = abs($timestamp - strtotime($time));
           if (is_null($diff) || $currDiff < $diff) {
                $index = $key;
                $diff = $currDiff;
            }
        }
echo $timeslots[$index];

Answer 1

在开始迭代之前对数组进行排序总是一个好主意。

然后，一旦我们在数组中找到匹配项，我们就会中断循环并输出我们找到的内容

$timeslots = ["01:00","01:30","02:00","02:30","00:30","01:00","01:30","02:00","02:30","03:50", "14:00", "15:00", "15:30"];
sort($timeslots); // first we sort the array, very important

$expected_time = "04:00";
$timestamp = strtotime($expected_time);
$diff = PHP_INT_MAX;
$index = null;

foreach ($timeslots as $key => $time) {
    
    $currDiff = $timestamp - strtotime($time);

    // Add a condition to check if the difference was negative or not - because it'll only be the "next" hour when the difference is negative
    if ($currDiff < $diff && $currDiff < 0) {
        $index = $key;
        $diff = $currDiff;
        break;
    }
}

if(!is_null($index))
    echo $timeslots[$index];
else
    echo 'Not found';

https://3v4l.org/kGdP8

Answer 2

您的代码似乎正确。它不会返回 下一个 小时，因为您使用时差的绝对值进行比较，而是这样做：

$timeslots = ["01:00","01:30","02:00","02:30","00:30","01:00","01:30","02:00","02:30","03:50", "14:00"];

$expected_time = "04:00";
$timestamp = strtotime($expected_time);
$diff = null;
$index = null;

foreach ($timeslots as $key => $time) {
    
    // Don't use abs function here - to get negative values as well
    $currDiff = $timestamp - strtotime($time);

    // Add a condition to check if the difference was negative or not - because it'll only be the "next" hour when the difference is negative
    if ((is_null($diff) || abs($currDiff) < $diff) && $currDiff < 0) {
        $index = $key;
        $diff = abs($currDiff);
    }
}

if(!is_null($index))
    echo $timeslots[$index];
else
    echo 'Not found';