我正在创建一个页面,我想在其中显示从数据库中捕获的信息。 我想显示的项目之一是下拉列表菜单中的选定值,但默认为选定的封面
我使用以下函数来显示下拉列表:
//function to display the cover in the dropdown
//function to display the cover in the dropdown
public function coverdropdown($companyid, $coverid)
{
$mydata = mysql_query("SELECT a.`coverid`, a.`cover_name`, a.`sum_insured`, a.`info`
FROM covertype a WHERE a.`companyid`=$companyid");
while($record = mysql_fetch_array($mydata))
{
echo '<option value="'.$record['coverid'].'" if ($record['coverid']==$coverid) {selected=selected}">'.$record['cover_name'].' - R'.$record['sum_insured'].'</option>';
}
}
我希望下拉列表显示选定的封面。我知道它是这样的:
<option value="" selected="selected">XYZ cover</option>
但是现在我不确定如何将其添加到此行中,因为我添加的 if 语句不起作用,其语法不正确:
while($record = mysql_fetch_array($mydata))
{
echo '<option value="'.$record['coverid'].'" if ($record['coverid']==$coverid) {selected=selected}">'.$record['cover_name'].' - R'.$record['sum_insured'].'</option>';
}
您应该能够在<option>
之前分配一个变量。类似于:
while($record = mysql_fetch_array($mydata)) {
// One thing to note: In XHTML the selected attribute should probably be defined as <option selected="selected">.
// In regular HTML <option selected> is ok
$selected = ($record['coverid'] == $coverid)? ' selected="selected"':'';
echo '<option value="'.$record['coverid'].'"'.$selected.'>'.$record['cover_name'].' - R'.$record['sum_insured'].'</option>';
}
这必须完成工作:
while($record = mysql_fetch_array($mydata)){
if($record['coverid']==$coverid)
echo '<option value="'.$record['coverid'].' selected>'.$record['cover_name'].'</option>';
else
echo '<option value="'.$record['coverid'].'>'.$record['cover_name'].'</option>';
}