Question

我有以下数组

 Array
(
    [0] => Array
        (
        [from_time] => 15:00
        [to_time] => 17:15
    )

[1] => Array
    (
        [from_time] => 10:00
        [to_time] => 12:15
    )

[2] => Array
    (
        [from_time] => 09:00
        [to_time] => 11:15
    )

[3] => Array
    (
        [from_time] => 09:00
        [to_time] => 11:15
    )

[4] => Array
    (
        [from_time] => 14:00
        [to_time] => 16:15
    )

[5] => Array
    (
        [from_time] => 15:00
        [to_time] => 17:15
    )

)

我想在这个阵列中获得共同的时间。

如果我在这个数组中找到共同的时间，我的预期结果应该是

Array
    (
        [0] => Array
            (
            [from_time] => 10:00
            [to_time] => 11:15
        )

    [1] => Array
        (
            [from_time] => 15:00
            [to_time] => 16:15
        )
)

我尝试使用以下代码但未获得准确的结果。

foreach ($booking_time as $time1) {
    foreach ($booking_time as $time2) {
        {
          if(($time1['from_time'] > $time2['from_time'] && $time1['from_time'] < $time2['to_time'])|| ( $time1['to_time'] > $time2['from_time']  && $time1['to_time'] < $time2['to_time'])){
                echo $time1['from_time'];
                 echo $time1['to_time']; 
        }
        }
    }
}

Answer 1

快速示例（我认为它可以更短，但为了快速理解它可以）：

$intervals = [
    ['from_time' => '15:00', 'to_time' => '17:15'],
    ['from_time' => '10:00', 'to_time' => '12:15'],
    ['from_time' => '09:00', 'to_time' => '11:15'],
    ['from_time' => '09:00', 'to_time' => '11:15'],
    ['from_time' => '14:00', 'to_time' => '16:15'],
    ['from_time' => '15:00', 'to_time' => '17:15'],
];

$overlaps = [];

foreach ($intervals as $interval) {
    $key = null;
    foreach ($overlaps as $_key => $_intervals) {
        foreach ($_intervals as $_interval) {
            if (
                ($_interval['from_time'] <= $interval['from_time'] && $interval['from_time'] <= $_interval['to_time']) 
                ||
                ($_interval['from_time'] <= $interval['to_time'] && $interval['to_time'] <= $_interval['to_time'])
            ) {
                $key = $_key;
                break 2;
            }
        }
    }
    if (is_null($key)) {
        $key = count($overlaps);
    }
    $overlaps[$key][] = $interval;
}
foreach ($overlaps as &$overlap) {
    $from = '00:00';
    $to = '23:59';
    foreach ($overlap as $_interval) {
        $from = max($from, $_interval['from_time']);
        $to = min($to, $_interval['to_time']);
    }
    $overlap = ['from_time' => $from, 'to_time' => $to];
}
unset($overlap);

print_r($overlaps);

输出：

Array
(
    [0] => Array
        (
            [from_time] => 15:00
            [to_time] => 16:15
        )
    [1] => Array
        (
            [from_time] => 10:00
            [to_time] => 11:15
        )
)

Answer 2

你想要增加或+ = [任何数字]

 $latest = aray_pop($booking_time); // this will grab the last element from the array
    for($i = 0; $i < $latest; $i += 15 // this is the number you want to increment by) {
        $booking_time[$key] = $i;

    }

Answer 3

用于比较A和B的常规叠加

X) A(start) < B(start) < A(end) < B(end)
Y) B(start) < A(start) < B(end) < A(end)

（并且通常也需要完整的包含物）

B(start) <= A(start) < A(end) <= B(end)
A(start) <= B(start) < B(end) <= A(end)

你实际上写下了这一点，但你忽略了A和B在你的来源中切换位置（A将在另一次运行时为B和BA - ＆gt;你只需要检查X 或 Y ）根据您检查的情况，您必须调整变量/键选择。

此外，您没有很好地处理包含物（包括它们或适当地包括它们）。

简单示例：

a = 10-20 b = 15-25

你希望结果是15-20

让我们假设运行：time1 a，time2 b

$time1['to_time'] > $time2['from_time']  && 
$time1['to_time'] < $time2['to_time']

匹配并打印

10-20 (time1 (a) from and to -- you wanted to print time2 from and time1 to)

另一次运行将指定：time2 a和time1 b

$time1['from_time'] > $time2['from_time'] && 
$time1['from_time'] < $time2['to_time'])

匹配并打印

15-25 (time1 (b) from and to -- you wanted to print time1 from and time2 to)

这是一个工作示例

foreach($array as $key => $time1) {
    foreach($array as $key2 => $time2) {
        if (($time1["from_time"] < $time2["from_time"]) && ($time2["from_time"] < $time1["to_time"]) && ($time2["to_time"] > $time1["to_time"])) {
            $overlap[] = array("from_key" => $time2["from_time"], "to_time" => $time1["to_time"]);
        }

        //inclusion
        if (($time1["from_time"] >= $time2["from_time"]) && ($time2["to_time"] >= $time1["to_time"]) && ($key !== $key2)) {
            $overlap[] = array("from_time" => $time1["from_time"], "to_time" => $time1["to_time"]);
        }

    }
}

免责声明：没有双重条目的检查。如果你不想要包含，你需要一种方法来处理A（开始）＆lt; B（开始）＆lt; A（结束）== B（结束）等。

如何获得时间与给定多次之间的间隔重叠？

3 个答案: