PHP：如何从关联数组中删除行答案

【问题标题】：PHP: How to delete rows from an associative arrayPHP：如何从关联数组中删除行
【发布时间】：2020-03-13 00:31:02
【问题描述】：

我有一个关联数组，其中键是日期时间类型的数据（间隔为 15 分钟）

    array:37 [▼
      "09:00" => Appointment {
                    #attributes: array:10 [▼
                        "id" => 1135
                        "startDateTime" => "2019-11-19 09:00:00"
                        "endDateTime" => "2019-11-19 09:45:00"
                        "duration" => 45
                    ]
                  }
      "09:15" => ""     // I want to delete this row -> 15 minutes
      "09:30" => ""     // I want to delete this row -> 30 minutes  end of the appointment
      "09:45" => ""
      "10:00" => Appointment {...duration => 60 ...}
      "10:15" => ""     // I want to delete this row -> 15 minutes
      "10:30" => ""     // I want to delete this row -> 30 minutes
      "10:45" => ""     // I want to delete this row -> 45 minutes
      "11:00" => ""     // I want to delete this row -> 60 minutes end of the appointment
      "11:15" => ""
      "11:30" => ""
      "11:45" => "" Appointment {...duration => 15 ...}
       ...
    ]

这个数组将提供一个表格，所以我想根据每个约会的持续时间删除后续行。我需要它，因为我想将约会跨越几行：

<td class="the-appointment" rowspan="{{ $appointment->duration / 15 }}">...

因此我需要从数组中删除后续行。

我这样做了：

    $index = -1;
    foreach ($row as  $key => $appointment) {
        if ($appointment) {
            $loops = $appointment->duration / 15;
        }
        for ($i = 1; $i < $loops; $i++) {
            unset($row[$index + 1]);
            $index++;
        }
    }

    array_push($calendar, $row);

但由于它是一个关联数组，我无法获取循环的索引。有没有更聪明的方法来做到这一点？

【问题讨论】：

array_keys 并将这些键用作值。 array_splice 是另一种解决方案，不需要密钥。

标签： php loops associative-array

【解决方案1】：

您可以使用for() 并使用持续时间跳过不需要的条目，而不是使用foreach()。结果放入$slots。

基本上，循环遍历时隙中的时间，如果设置了持续时间，则将循环计数器增加约会的长度（持续时间/15 并使用ceil() 将其向上取整）...

$row = ["09:00" => (object)["duration" => 40],
    "09:15" => "",
    "09:30" => "",
    "09:45" => "",
    "10:00" => (object)["duration" => 60],
    "10:15" => "",
    "10:30" => "",
    "10:45" => "",
    "11:00" => "",
    "11:15" => "",
    "11:30" => "",
    "11:45" => (object)["duration" => 45]
];

$slots = [];
$times = array_keys($row);
for ( $i = 0; $i < count($times); $i++  )    {
    $appointment = $row[$times[$i]];
    $slots[$times[$i]] = $appointment;
    if ( $appointment->duration ?? 0 > 0 )  {
        $i += ceil($appointment->duration / 15);
    }
}
print_r($slots);

有了测试数据，这给出了...

(
    [09:00] => stdClass Object
        (
            [duration] => 40
        )

    [10:00] => stdClass Object
        (
            [duration] => 60
        )

    [11:15] => 
    [11:30] => 
    [11:45] => stdClass Object
        (
            [duration] => 45
        )

)

【讨论】：

您好，感谢您的解决方案。我只是在这一行中更改了一个细节：$i += ceil($appointment->duration / 15)。解决方案是跳槽太多。如果约会从 10:00 开始，持续 45 分钟；以下时段应从 10:45 开始。所以我只是在 $i += ceil($appointment->duration / 15 - 1); 行中减去 1 个单位结果是完美的。感谢您的帮助！

【解决方案2】：

一些代码开始：

$duration = 0;
// I suggest to use array_filter and track `$duration` on each loop
$filtered = array_filter(
    $apps,
    function ($apm) use (&$duration) {
        if ($duration === 0) {
            if (!empty($apm->duration)) {
                $duration = $apm->duration - 15;
                return true;
            } else {
                return true;
            }
        } else {
            $duration -= 15;
            return false;    
        }
    }
);

工作小提琴here.

【讨论】：