警告:date_format()期望参数1为DateTime

Question

我使用以下脚本从mysql数据库中提取日历信息并将其显示在页面上.我试图从标准的Mysql日期格式重新格式化日期,但是当从数据库中检索它时,我收到以下错误:

Warning: date_format() expects parameter 1 to be DateTime, string given in C:\easyphp\www\twinfocus\managecalendar.php on line 24

Warning: date_format() expects parameter 1 to be DateTime, string given in C:\easyphp\www\twinfocus\managecalendar.php on line 24

Warning: date_format() expects parameter 1 to be DateTime, string given in C:\easyphp\www\twinfocus\managecalendar.php on line 24

Warning: date_format() expects parameter 1 to be DateTime, string given in C:\easyphp\www\twinfocus\managecalendar.php on line 24

数据库(如您所见,日期存储正确):

剧本 :

<?php
     $sql2 = <<<SQL
        SELECT *
        FROM `calendar`
    SQL;
    if(!$result2 = $db->query($sql2)){ die('There was an error running the query [' . $db->error . ']');}
    echo '<table class="admintable"> <thead>';
    echo '<tr><th>Client Names</th><th>Email</th><th>Tel</th><th>Wedding Date</th><th>Date Created</th><th>Start Time</th><th>End Time</th><th>Price</th><th>Location</th><th>Other Info</th><th>Edit</th><th>Delete</th></tr></thead>';
    while($row2 = $result2->fetch_assoc()){ 
    $weddingdate = $row2['weddingdate'];
    $formattedweddingdate = date_format($weddingdate, 'd-m-Y');
    echo '<tr><td>'.$row2['name'].'</td><td>'.$row2['email'].'</td><td>'.$row2['tel'].'</td><td style="min-width:70px;">'.$formattedweddingdate.'</td><td style="min-width:70px;">'.$row2['datecreated'].'</td><td>'.$row2['starttime'].'</td><td>'.$row2['endtime'].'</td><td>&pound;'.$row2['price'].'</td><td>'.$row2['location'].'</td><td style="min-width:400px;">'.$row2['otherinfo'].'</td><td><a href="managecalendar.php?&key='.$key.'&editwedding='.$row2['id'].'">Edit</a></td><td><a href="calenderdelete.php?&key='.$key.'&delwedding='.$row2['id'].'">Delete</a></td></tr>';}
    echo '</table>';

?>

Answer 1

最好的方法是使用DateTime对象来转换日期.

$myDateTime = DateTime::createFromFormat('Y-m-d', $weddingdate);
$formattedweddingdate = $myDateTime->format('d-m-Y');

注意:它仅支持PHP 5> = 5.3.0.

Answer 2

您需要将DateTime对象传递给此func.见手册: php

string date_format ( DateTime $object , string $format )

您可以尝试使用:

date_format (new DateTime($time), 'd-m-Y');

或者您也可以使用:

$date = date_create('2000-01-01');
echo date_format($date, 'Y-m-d H:i:s');

Answer 3

你为什么不这样试试呢:

$Weddingdate = new DateTime($row2['weddingdate']);
$formattedweddingdate = date_format($Weddingdate, 'd-m-Y');

或者您也可以这样做:

$Weddingdate = new DateTime($row2['weddingdate']);
echo $Weddingdate->format('d-m-Y');