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我正在尝试创建一个动态列表,该列表将显示我的“城市”表的项目。但是我的代码显示的是下拉菜单而不是列表。任何人都可以通过告诉我如何从中构建动态列表来帮助我。我已经尝试过使用它,但没有找到有用的资源。

我的代码的一部分:

          <?php echo form_open('frontpage/index');?>
          <p><select name ='city_city_id'>
          <?php echo form_error('city_city_id'); ?>

<?php
$getType = mysql_query("SELECT city_id, cityname FROM city ORDER BY city_id");
while($type = mysql_fetch_object($getType)){
    echo "<option value=\"{$type->city_id}\">{$type->cityname} </option>",set_value('city_city_id');
}


?> 
    <p><input type="submit" value="submit" /></p>

       <?php echo form_close();?>   
4

2 回答 2

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您需要放入multiple="multiple"您的选择标签。

例如

<select multiple="multiple">
  <option>Volvo</option>
  <option>Saab</option>
  <option>Mercedes</option>
  <option>Audi</option>
</select>
于 2011-10-22T10:15:33.383 回答
0
<?php echo form_open('frontpage/index');?>
<p>
<?php echo form_error('city_city_id'); ?>
<select name="city_city_id" multiple="multiple">

<?php
$getType = mysql_query("SELECT city_id, cityname FROM city ORDER BY city_id");
while($type = mysql_fetch_object($getType)){
    echo "<option value=\"{$type->city_id}\">{$type->cityname} </option>",set_value('city_city_id');
}
?>
</select>
</p> 
<p><input type="submit" value="submit" /></p>

<?php echo form_close();?>   
于 2011-10-22T10:21:37.253 回答