Question

大家好，我试图将员工时间表从亚洲/马尼拉时区转换为美国/ los_angeles，但由于使用此代码进行转换，所以由于放假时间太长，我出现了错误

$datecon = new DateTime($date[$a].' '.$time[$a], new 
DateTimeZone('Asia/Manila'));
$datecon->setTimezone(new DateTimeZone('America/Los_Angeles'));

错误在这里

2019年4月7日，星期日|休息日致命错误：未捕获的异常： DateTime :: __ construct（）：无法解析时间字符串（2019年4月7日 OFF（关闭））在位置11（D）：在时区中找不到时区 C：\ xampp \ htdocs \ test \ index.php：38中的数据库堆栈跟踪：＃0 C：\ xampp \ htdocs \ test \ index.php（38）：DateTime-> __ construct（'04 / 07/2019 DAY ...'，Object（DateTimeZone））＃1 {main}被抛出 C：\ xampp \ htdocs \ test \ index.php在第38行

<table class="table table-bordered col-lg-6">
<thead class="thead-dark">
<tr>
    <td>Asia/Manila 
    </td>
    <td>America/Los_Angeles
    </td>
</tr>
</thead>
<tbody>
    <?php 
    $date = array('04/07/2019','04/08/2019','04/09/2019','04/10/2019','04/11/2019','04/12/2019','04/13/2019','04/14/2019','04/15/2019','04/16/2019','04/17/2019','04/18/2019','04/19/2019','04/20/2019');
    $time = array('DAY OFF','DAY OFF','12:00 AM','12:00 AM','12:00 AM','12:00 AM','12:00 AM','DAY OFF','11:00 PM','11:00 PM','11:00 PM','1:00 PM','1:00 PM','DAY OFF');
    $day = array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday','Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday','Sunday');
    $timezone = 'America/Los_Angeles';
    for($a=0; $a < 14; $a++){
    echo '<tr><td>';
    echo $day[$a].', '.$date[$a].'  | <b>'.$time[$a].'</b>';


    $datecon = new DateTime($date[$a].' '.$time[$a], new 
    DateTimeZone('Asia/Manila'));

    $datecon->setTimezone(new DateTimeZone('America/Los_Angeles'));



    echo '<td>';
    echo $day[$a].', '.$date[$a].'</td>';

    echo '</tr>';

    }
    ?>
</tbody>

Result

员工时间表时区转换

0 个答案: