这是我的代码到目前为止,现在我想结合这个我能做什么?如何结合两个foreach循环合并为一个
<table>
<tr>
<th>Imported Files</th>
<th>Report Files</th>
</tr>
<?php
$dir = str_replace("/var/www/13_Feb/subscriber-files/","","/var/www/13_Feb/subscriber-files/u*.[cC][sS][vV]");
$dir2 = str_replace("/var/www/13_Feb/subscriber-files/","","/var/www/13_Feb/subscriber-files/r*.[cC][sS][vV]");
foreach(glob($dir) as $file) {
echo "<tr>";
echo "<td>". $file."</td>";
}
foreach(glob($dir2) as $file) {
echo "<td>". $file."</td>";
}
echo "</tr>";
?>
</table>
我想打印这样
echo "<tr>";
echo "<td>". $file1."</td>";
echo "<td>". $file2."</td>";
}
echo "</tr>";
即在同一个TD我能做些什么帮助我
update:-
我想在打印此以TD
$dir = str_replace("/var/www/13_Feb/subscriber-files/","","/var/www/13_Feb/subscriber-files/u*.[cC][sS][vV]");
$dir2 = str_replace("/var/www/13_Feb/subscriber-files/","","/var/www/13_Feb/subscriber-files/r*.[cC][sS][vV]");
每个目录中的文件数是否相同?你是说你想在dir2中的第一个文件旁边的dir中显示第一个文件吗? – 2012-02-15 14:19:47