显示月/日/年-不工作 - show month / day / year - not works

show month / day / year - not works

本文关键字：工作显示 | 更新日期: 2023-09-27

我必须设置日期、月份和年份，但这就是我在这里遇到问题的原因:

问题现在的样子。(问题是，在日期和年之前会有"array ...."。)"ArrayTirsdag 11 june 2013"

想要这样:《2013年6月11日第一天》

echo "<td>" . 
                $days = array("Mandag","Tirsdag", "Onsdag","Torsdag","Fredag","Lørdag","Søndag");
                $months = array("Jan", "Feb", "Mar", "Apr", "Maj", "Jun", "Jul", "Aug", "Sep", "Okt", "Nov", "Dec");
                $timestamp = strtotime($c->pubDate);
                $dayInWeek = $days[date('N', $timestamp)-1];
                $month = $months[date('n', $timestamp)-1];
                $dayInMonth = date('d', $timestamp);
                $year = date('Y', $timestamp);
                echo $dayInWeek . ' ' .$dayInMonth. ' '.$month.' '.$year
                . "</td>";

你的问题是，当你需要一个;时，你有一个.，所以你不是在回应<td>，而是在试图回应你所有的逻辑。因此，将第一行更改为echo "<td>";，然后就设置好了。您也可以将开头的<td>保存到末尾，像这样:

// date logic
$days       = array("Mandag", "Tirsdag", "Onsdag", "Torsdag", "Fredag", "Lørdag", "Søndag");
$months     = array("Jan", "Feb", "Mar", "Apr", "Maj", "Jun", "Jul", "Aug", "Sep", "Okt", "Nov", "Dec");
$timestamp  = strtotime($c->pubDate);
$dayInWeek  = $days[date('N', $timestamp)-1];
$month      = $months[date('n', $timestamp)-1];
$dayInMonth = date('d', $timestamp);
$year       = date('Y', $timestamp);
echo "<td>$dayInWeek $dayInMonth $month $year</td>";

你有。代替第一次回显时的逗号。

echo "<td>"; 
                $days = array(1=>"Mandag","Tirsdag", "Onsdag","Torsdag","Fredag","Lørdag","Søndag");
                $months = array(1=>"Jan", "Feb", "Mar", "Apr", "Maj", "Jun", "Jul", "Aug", "Sep", "Okt", "Nov", "Dec");
                $timestamp = strtotime($c->pubDate);
                $dayInWeek = $days[date('N', $timestamp)];
                $month = $months[date('n', $timestamp)];
                $dayInMonth = date('d', $timestamp);
                $year = date('Y', $timestamp);
                echo $dayInWeek . ' ' .$dayInMonth. ' '.$month.' '.$year
              . "</td>";