php - 如何使用隐藏输入发送选择的值

Question

我在尝试通过隐藏输入发送所选选项的值以将值插入 mysql 表时遇到困难:

<form  action="interface.php" method="post">
VLAN: <select name="vlan" onchange="document.getElementById('vlan_id').value=this.options[this.options.selectedIndex].value;">
<option value="0">Выберите оборудование</option>
<?php
    $res = mysql_query('select `vlan_id`,`number` from `vlan`'); 
    while($row = mysql_fetch_assoc($res)){
        printf("<option value='%s'>%s</option>", $row['vlan_id'], $row['number']);
    }
?>
</select>
<input type="hidden" id="vlan_id name="vlan_id" value="vlan_id" />

接口(interface).php

 $val3 = $_POST['vlan'];
 $val4 = $_POST['vlan_id'];

 $result_insert =  mysql_query ("INSERT INTO `ADRESSA`.`interface` ( `vlan`,`vlan_id`) VALUES ('$val3','$val4');") 
            or die ("<b>Query failed:</b> " . mysql_error());

最诚挚的问候!

Answer 1

您缺少有关隐藏 ID 属性的引用，请尝试更改

<input type="hidden" id="vlan_id name="vlan_id" value="vlan_id" />

至

<input type="hidden" id="vlan_id" name="vlan_id" value="vlan_id" />

注意:- mysql_* 已弃用，使用 mysqli_* 或 PDO

还可以通过转义字符串来防止 sql 注入(inject):-

$val3 = mysql_real_escape_string($_POST['vlan']);
$val4 = mysql_real_escape_string($_POST['vlan_id']);

获取值和 ID 更改:-

onchange="document.getElementById('vlan_id').value=this.options[this.options.selectedIndex].text;"

您将得到:- Array ( [vlan] => 1 [vlan_id] => ss ) 还根据值更改查询