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我在让我的 php 表单正常运行时遇到了一些困难。我正在尝试使用多个下拉框，它们都引用数据库并从各自的值中填充自己。

正如您在下面的代码中看到的那样，我正在尝试使用“电子邮件”和“类型”框中的结果创建查询。任何人都可以帮忙吗？

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title></title>
</head>

<body>

<?php
    $host = "****";
    $dbname = "****";
    $username = "****";
    $password = "****";

@   $connection = mysql_connect($host, $username, $password);
@   mysql_select_db($dbname,$connection);

    if (!$connection) {
    die('=( Could not connect: ' . mysql_error());
    }
    echo '<b>Connected successfully to $dbname</b>';
    mysql_close($connection);

    /*
    Check to see if connection was successful
    */



?>

    <form name="survey" method="post" action="scripts/report_script.php" onSubmit="">

        Email:

        <select name="email">
<!-- Begin email selection -->
            <?php 
                mysql_connect($host, $username, $password);
                mysql_select_db($dbname);
                $email_query = "SELECT sgl_userID,sgl_user_email FROM tbl_sgl_user";
                $email_result = mysql_query($email_query);
                while ($row = mysql_fetch_array($email_result)){

                    echo '<option value="'.$row['sgl_userID'].'">'.$row['sgl_user_email'].'</option>';
                }

            ?>
        </select>
<!-- End email selection -->
<!--Start Survey Type selection-->
            <select name="type">
                <option value="government">Government</option>
                <option value="sport">Sport</option>
                <option value="school">School</option>
            </select>
<!--End Survey Type selection-->
<!--Start Date selection-->
            <select name="date">
                <?php
                    mysql_connect($host, $username, $password);
                    mysql_select_db($dbname);
                    $date_query = "SELECT survey_date FROM tbl_survey WHERE survey_name='$type' AND sgl_userID='$email'";
                    $date_result = mysql_query($date_query);
                    while ($row = mysql_fetch_array($date_result)){
                        echo '<option value="'.$row['survey_date'].'">'.$row['survey_date'].'</option>';
                    }
                ?>
            </select>
<!--End Date selection-->
        <input type="Submit" name="Submit" value="Submit" />
    </form>


</body>
</html>
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