文本框中的Onchange值未在数据库中传递

时间:2016-09-24 07:12:18

标签: php drop-down-menu

我有这个脚本来获取下拉列表的文本

<script>
function streetvalue(data) {

var i = street.selectedIndex;
if (i != -1) {
document.getElementById("street2").value = street.options[i].text;
}

}
</script>

这是我的HTML

<tr>
        <td width="120" height="30" class="label"><label for="accno"><strong>Street/Alley</strong></label></td>
        <td height="30" class="content">
        <select name="street" id="street" style="width:200px;"  onchange="streetvalue(this)" required>
        <option value="" label="Select your street">Select your   street</option>
        <?php 
        $sql = "SELECT * FROM tblstreet ORDER BY lname asc";
        $result = dbQuery($sql);
        while($row = dbFetchAssoc($result)) {
        echo '<option value="'.$row['lid'].'">'.$row['lname'].'</option>';  
        }
        ?>
        </select>
        <input type="text" id="street2">

从下拉列表中选择项目时,它正常工作,文本框会复制它。但是当我点击提交并将其保存到我的数据库时,我的数据库中始终显示的值是&#34; street2&#34;而不是我在下拉列表中选择的文字。我错过了什么吗?谢谢你的帮助!

enter image description here

当按下提交按钮时,这是我的功能

function doRegister()
{
$fname  = $_POST['firstname'];
$mname  = $_POST['middlename'];
$lname  = $_POST['lastname'];
$email  = $_POST['email'];  
$phone  = $_POST['phone'];
$cphone = $_POST['cphone'];
$dob    = $_POST['dob'];
$job_type = $_POST['intext4'];
$street2    = $_POST['street2'];

$sql = "INSERT INTO tbl_address (user_id, blknum, apartment, street,    street2, subdivision, city, zipcode, res_type, address) 
        VALUES ($insert_id, '$blknum', '$apart', '$street', 'street2',    '$subdi', '$city', '$zip', '$typeres', '$address')";
dbQuery($sql);

街道字段是我的下拉菜单,street2字段是下拉菜单中的复制文本。谢谢!

0 个答案:

没有答案