我的代码
<?php
include('ConnectToDb.php');
$query = "SELECT * FROM News WHERE NewsFlag = 1 ORDER BY PostDate DESC";
$arrCount = -1;
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
$ID=$row['ID'];
$PostDate = $row['PostDate'];
$NewsHeader = stripslashes($row['NewsHeader'])
;
$NewsStart = stripslashes($row['NewsStart'])
;
echo "<hr>";
echo "<div>". date('j F Y',strtotime($PostDate)). "</div>";
echo "<p>";
$news_id = strval(sprintf("%1$04d",$ID));
$array = scanImageFolder("newsImages/newsThumbs",$news_id);
if(count($array)>0) {
echo "<img src='". $array[0]. "' alt='' />";
}
echo "<h2 style='text-align:center'><u><a href='latestnews_full.php?ID=$ID'>". $NewsHeader. "</a></u></h2>";
echo "<div style='text-align:left'><h3>";
echo $NewsStart. " ......<a href='latestnews_full.php?ID=$ID'>(more)</a><br />";
echo "<div style='text-align:center'>";
echo "</div>";
echo "</h3></div>";
}
?>
很好地在四行上显示我的数据,日期在顶部,然后是图片,标题和描述。
但是,我想将数据显示为像这样的表
<table style="width: 100%">
<tr>
<td colspan="2">postDate here</td>
</tr>
<tr>
<td rowspan="2">picture here</td>
<td>newsHeader here</td>
</tr>
<tr>
<td>newsStart here</td>
</tr>
</table>
我不确定如何正确回显表格单元格,到目前为止我的所有尝试都产生了白页。有人可以赐教我吗?
答案 0 :(得分:1)
我建议你将你的逻辑与你的可呈现部分分开。准备好提取结果,分配var等后关闭PHP标记,然后:
<table style="width: 100%">
<?php
//yourcode
//...
//...
$NewsStart = stripslashes($row['NewsStart']);
$news_id = strval(sprintf("%1$04d",$ID));
$array = scanImageFolder("newsImages/newsThumbs",$news_id);
?>
<tr>
<td colspan="2"><?= date('j F Y',strtotime($PostDate)) ?></td>
</tr>
<tr>
<?php
if(count($array)>0) {
?>
<td rowspan="2"><img src='<?= $array[0] ?>' alt='' /></td>
<?php } ?>
<td><?= $NewsHeader ?></td>
</tr>
<tr>
<td><?= $NewsStart ?> </td>
</tr>
<?php } ?>
</table>
然而,它再次不那么清楚,我建议使用模板引擎。如果你愿意我可以发布一个代码,将代码分配给Smarty,并在Smarty模板中输出你的演示文稿