Question

我正在尝试将1.3转换为1:30，但是当我使用带有日期功能的strtotime时，1.3变为1.03！什么是解决方案？

这就是我正在使用的并给我意想不到的结果：

<?php $time_from = 1.3; ?>
<?= date('h:i A', strtotime($time_from)); ?>

print_r或$ opening_hours_array：

阵列（ [sunday_from] =＆gt; 24.45 [sunday_to] =＆gt; 1.3 [monday_from] =＆gt; 2 [monday_to] =＆gt; 2.15 [tuesday_from] =＆gt; 3 [tuesday_to] =＆gt; 3.15 [wednesday_from] =＆gt; 4 [wednesday_to] =＆gt; 4.15 [thursday_from] =＆gt;五 [thursday_to] =＆gt; 5.15 [friday_from] =＆gt; 6 [friday_to] =＆gt; 6.15 [saturday_from] =＆gt; 7 [saturday_to] =＆gt; 1）

我想要做的就是向我们展示我们的开放：

 <?php
$days_of_week = array("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday");
            for ($index = 0; $index < count($days_of_week); $index++) {
                $pos = strpos($opening_hours_array[strtolower($days_of_week[$index]) . '_from'], '.');
                if ($pos === false) {
                    $time_from = $opening_hours_array[strtolower($days_of_week[$index]) . '_from'].'.00';
                }else{
                    $time_from = $opening_hours_array[strtolower($days_of_week[$index]) . '_from'];
                }
                $pos = strpos($opening_hours_array[strtolower($days_of_week[$index]) . '_to'], '.');
                if ($pos === false) {
                    $time_to = $opening_hours_array[strtolower($days_of_week[$index]) . '_to'].'.00';
                }else{
                    $time_to = $opening_hours_array[strtolower($days_of_week[$index]) . '_to'];
                }
}
?>

Answer 1

正如@Goikiu所建议的那样，下面的代码可以解决这个问题

$time_from = 1.3;
$time_from = (strlen($time_from)==3)?$time_from.'0':$time_from;
echo date('h:i A', strtotime($time_from));

But there are other situations like 

$time_from = 1 ; 

The code needs to check for those as well.

Answer 2

解决方案是：

$pos = strpos($opening_hours_array[strtolower($days_of_week[$index]) . '_from'], '.');
                if ($pos === false) {
                    $time_from = $opening_hours_array[strtolower($days_of_week[$index]) . '_from'].'.00';
                }else{
                    $time_from = $opening_hours_array[strtolower($days_of_week[$index]) . '_from'];
                    $time_from = (strlen($time_from)==3)?$time_from.'0':$time_from;
                }
                $pos = strpos($opening_hours_array[strtolower($days_of_week[$index]) . '_to'], '.');
                if ($pos === false) {
                    $time_to = $opening_hours_array[strtolower($days_of_week[$index]) . '_to'].'.00';
                }else{
                    $time_to = $opening_hours_array[strtolower($days_of_week[$index]) . '_to'];
                    $time_to = (strlen($time_to)==3)?$time_to.'0':$time_to;
                }

如何使用strtotime和date函数正确地将小数点时间转换为时间？

2 个答案: