【发布时间】:2011-04-18 14:46:01
【问题描述】:
我有以下代码:
<form action="" method="POST">
<?php
$count = isset($_POST['count']) ? $_POST['count'] : 1;
if($count > 11) $count = 11;
?>
<table>
<!-- Keeps track of the current number of rows -->
<input type="hidden" name="count" value="<?php echo $count+1; ?>"/>
<?php for($i = 0; $i < $count; $i++):
// Loop through all rows gathering the data here, and then creating the fields below
$val0 = isset($_POST['field'][$i]['0']) ? $_POST['field'][$i]['0'] : '';
$val1 = isset($_POST['field'][$i]['1']) ? $_POST['field'][$i]['1'] : '';
$val2 = isset($_POST['field'][$i]['2']) ? $_POST['field'][$i]['2'] : '';
?>
<tr>
<td><input name="field[<?php echo $i; ?>][0]" value="<?php echo $val0; ?>"/></td>
<td><input name="field[<?php echo $i; ?>][1]" value="<?php echo $val1; ?>"/></td>
<td><input name="field[<?php echo $i; ?>][2]" value="<?php echo $val2; ?>"/></td>
</tr>
<?php endfor; ?>
</table>
<input type="submit" value="click me" />
如何将字段设置为下拉菜单,当您按下提交时,将下拉菜单作为文本而不是作为下拉菜单回显?
【问题讨论】:
标签: php drop-down-menu html-table