见下面的代码
$num_rows = MysqL_num_rows($result41);
for ($i = 1; $i <= MysqL_num_rows($result41); $i++)
{
$row = MysqL_fetch_array($result41);
$upload_id = $row ['upload_id'];
$file = $row ['FILE_NAME'];
echo"
<td>
<a href='image.PHP?id=$upload_id&gallery=$id'><center>
<img src='uploads/$file' alt='$name gallery' title='$name Album' class='resize'>
</td>";
}
if ($i % 0 == 4) {
echo '</tr>'; // it's time to move to next row
}
我的问题是,在显示4列之后如何移动到另一行? (每行只有4张图片)
我的脚本中有if($i%0 == 4),但似乎没有工作?
谢谢
解决方法:
像这样 :
for ($i = 1; $i <= MysqL_num_rows($result41); $i++)
{
$row = MysqL_fetch_array($result41);
$upload_id = $row ['upload_id'];
$file = $row ['FILE_NAME'];
if($i % 4 == 0) echo "<tr>";
echo"
<td>
<a href='image.PHP?id=$upload_id&gallery=$id'><center>
<img src='uploads/$file' alt='$name gallery' title='$name Album' class='resize'>
</td>";
if($i % 4 == 3) echo "</tr>";
}
然后开始循环(for)为0?