小编iko*_*kon的帖子

if(isset($_GET["id"])){
        $sql=mysql_query("SELECT * FROM aMovie WHERE aName= '{$_GET['id']}'"); 
        $row=mysql_fetch_object($sql);  
}

<input type = "text" name = "name" value = "<?php echo $row->aC; ?>"/> 
<select name = "name" >
        <option value = "" <?php echo ($row->aC== "Deadpool") ? 'selected = "selected"': '';?>">Deadpool</option>
        <option value = "" <?php echo ($row->aC == "BATMAN VS SUPERMAN") ? 'selected = "selected"': '';?>">BATMAN VS SUPERMAN</option>
</select>

假设aMovie是我的表名,在我的表中有aName和aC.但是,我想显示一个匹配aC ["Deadpool"或"Batman Vs Superman"]的名称,并在下拉按钮中显示它.它仅适用于输入类型,但不适用于下拉按钮.