Question

我有一个PHP脚本，该脚本正在从数据库中返回值。特别是跟踪号和创建跟踪号的时间戳。

我试图弄清楚如何格式化此时间戳，下面的值是如何返回时间戳。

20180828115208

 <div class="resul-count" align="center">
Found <?php echo $query->num_rows; ?> Results.
</div><br>

    if($query->num_rows){
        while($r = $query->fetch_object()){
        ?>
        <div class="result">
                <tr>
<td><?php echo $r->ShipToCompanyorName;   ?></td>
<td><?php echo $r->ShipToAttention; ?></td>
<td><?php echo $r->ShipToAddress1; ?></td>
<td><?php echo $r->ShipToCityorTown;   ?></td>
<td><?php echo $r->ShipToStateProvinceCounty; ?></td>
<td><?php echo $r->ShipToPostalCode; ?></td>
<td><?php echo $r->ShipmentInformationShipmentProcessTimestamp; ?></td>
<td><a href="http://wwwapps.ups.com/WebTracking/track?track=yes&trackNums=<?= $r->PackageTrackingNumber; ?>"><?= $r->PackageTrackingNumber;?></a></td>

Answer 1

您可以使用DateTime::createFromFormat

$date = DateTime::createFromFormat('YmdHis', $r->ShipmentInformationShipmentProcessTimestamp);
echo $date->format('Y-m-d'); //output 2018-08-28 or choose whatever format you want

以字符串形式返回值时，用PHP格式化日期

1 个答案: