Javascript 从php变量设置选项值

更改选择值时将提交表单

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提交表单后，它将重定向到同一页面。如何将所选选项值设置为我刚才选择的选项值？

用以下方式更改您的表单，尤其是

下拉列表

<form method="post" id="employee" action="change.php">
<select id="employeename" name="employeename" onchange="this.form.submit()">
 <?php 
while ($row = mysql_fetch_array($result))
{
    echo "<option value='".$row['name']."'>".$row['name']."</option>";
}
?>    
</select>
</form>

如果您在可用列表中找到所选键，则只需将其标记为选中。
您的意思是要在加载页面后显示所选的选项值。是的，这正是我想要的。如果（$_POST['employeename']===$row['name']）{echo”“，$row['name'].；}否则{echo”“，$row['name'].；}将此代码添加到可能重复的循环中
<form method="post" id="employee" action="change.php">
    <select id="employeename" name="employeename" onchange="this.form.submit()">
        <?php 
            while ($row = mysql_fetch_array($result)){
                $output = "<option value='".$row['name']."'";
                if($_POST['employeename'] == $row['name']){
                    $output .= " selected='selected'";
                }
                $output .= ">".$row['name']."</option>";
                echo $output;
            }
        ?>    
    </select>
</form>

    <form method="post" id="employee" action="change.php">
    <select id="employeename" name="employeename" onchange="this.form.submit()">
     <?php 
    while ($row = mysql_fetch_array($result))
    {
   if($_POST['employeename']==$row['name'])
        echo "<option value='".$row['name']."' selected>".$row['name']."</option>";
else 
   echo "<option value='".$row['name']."'>".$row['name']."</option>";
    }
    ?>    
    </select>
    </form>