php - Sql和Php，插入sql选择选项

Question

我在 php 中有一个表单，在此表单中有一个选择选项

<select id="birdname" style="width: 150px; height: 25px; border-radius: 5px;">
    <option>--Cambiar vendedor --</option>
    <?php
    $selectagent = "SELECT * FROM `" . $prefix . "user`";
    $resultforagents = mysql_query($selectagent);
    while ($agent = mysql_fetch_object($resultforagents)) {
        ?>
        <option value="<?= $agent->user_id ?>" ><?= $agent->username ?></option> 
<?php } ?>                          
</select>
<div id="stage" style="display:none;"></div>

显示用户名的名称，我需要将选定的用户名(引用而不是名称)插入到sql中

<?php
$userId = $_POST['birdname'];
$idtill = $_SESSION['caj'];
$cashid = $_SESSION['cash'];
$mvalue = number_format($givenamount, 2, '.', '');
$mdate = date("Y-m-d H:i:s");
$moption = '1';
$sql = "INSERT INTO `" . DB_PREFIX . "order_user`(`user_id`, `order_Id`, `till`, `value`, `operation`, `comment`, `date`,`cashid`) VALUES ('" . $userId . "','" . $lastorderidis . "','" . $cashid . "','" . $mvalue . "','" . $moption . "', '" . " " . "', '" . $mdate . "','" . $cashid . "')";
$common->showresultsql($sql, 'update', 1);

echo $lastorderidis;
?>

谢谢

Answer 1

将 name 标签添加到您的 select 值中，以获取 select 的值，如下所示:

<select id="birdname" name="birdname">

现在您将获得 $_POST['birdname'] 的值

更多信息请点击:get the selected index value of <select> tag in php