内存在嵌套循环中使用数组耗尽

Question

这是我得到的错误.

致命错误:允许的内存大小耗尽.

我正在获取包含日期from和日期的数组till.我正在尝试获取介于两者之间的所有日期并将它们添加到新数组中.显然嵌套循环和多阵列令人筋疲力尽.

我需要一种不那么费力的方式来获取所有日期.

这是我的代码:

$query = "SELECT *
          FROM reservate
          ORDER BY from";
$result = $connect->query($query);

$reservations = array();
if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
        $reservations[] = $row;
    }
}

$connect->close();

//AVAILABILITY CHECK
$nonAvailable = array();
foreach($reservations as $reservation){
    $from = $reservation['from'];
    $till = $reservation['till'];

    while($from <= $till){
        $nonAvailable[] = $from;
        $from = date('Y-m-d', strtotime("+1 days"));
    }
}

Answer 1

看起来你做了一个无限循环¹.

// if $till is in the future, this is an infinite loop
while($from <= $till){
    // appending the same value of $from on each iteration
    $nonAvailable[] = $from;
    // as $from never changes value
    $from = date('Y-m-d', strtotime("+1 days"));
}

将1日期添加到当前值$from

while ($from <= $till){
     $nonAvailable[] = $from;
     // add 1 day to $from
     $from = date('Y-m-d', strtotime($from, "+1 days"));
}

仍然可以做出一些改进:

• 在下面的示例中,整个表不会复制到 reservation数组中,我们只从表中检索所需的列.
• 比较integers而不是strings.
• 手动 添加1日,$from而不是依靠strtotime它来做.

$query = "SELECT from, till
          FROM reservate
          ORDER BY from";
$result = $connect->query($query);

$nonAvailable = array();

if ($result->num_rows > 0) {
    while ($row = $result->fetch_assoc()) {
        $from = strtotime($row['from']);
        $till = strtotime($row['till']);

        while ($from <= $till) {
            $nonAvailable[] = date('Y-m-d', $from);
            $from += 60*60*24;
        }
    }
}

$connect->close();

这种算法可能会更有效率.例如,您是否需要每天保留一些详尽的清单？如果没有,那么只需要开始日期和结束日期即可.

_{1 给定足够的时间和内存,循环将退出,因为strtotime("+1 days")最终将返回大于的值$till.}