我想从数据库中回显1个以上的id,如何在索引页中显示1个以上id?
我的功能:
function query($sql) {
global $conn;
$results = mysql_query($sql, $conn);
$rows = array();
if(!is_bool($results))
while($row = mysql_fetch_assoc($results))
$rows['id'] = $row;
return $rows;
}
我的索引页面:
<?php foreach($funds as $fund) { ?>
<td width="87%" valign="top"><ul id="funds" class="bodycopy2" style="margin:0;">
<li><?php echo $fund['fund']; ?><span class="price"><div align="center" class="h1"><strong><br />R123.45</strong> <br />as at <br />1 January 2012</div></span></li>
<li><?php echo $fund['fund']; ?><span class="price"><div align="center" class="h1"> <strong><br />R200.45</strong> <br />as at <br />1 January 2012</div></span> </li>
<li><?php echo $fund['fund']; ?><span class="price"><div align="center" class="h1"> <strong><br />R150.00</strong> <br />as at <br />1 January 2012</div></span></li>
</td>
</tr>
</table>
<?php } ?>
您可以使用一个id数组,如:
$rows['ids'][] = $row;
然后你可以使用通过ID
foreach ($rows as $id)
如果只保留id,则直接使用$rows[] = $row
。
您总是覆盖同一个元素$rows['id'] = $row;
。你应该做
$rows['id'][] = $row;
要显示id,您应该在$rows 上迭代
foreach($rows['id'] as $id){
echo $id;
}