Question

我创建了2个下拉菜单，其中第二个菜单应该对第一个选择做出反应。

我已经测试了MySQL查询和工作没有任何问题。出于某种原因，似乎页面getter.php未被激活＆＃39;。有什么建议吗？

炫魅

<?php
require_once('includes/db_connect.php');

echo '<select id="first-choice">
  <option>Please choose here first</option>';

  $sql_lev = "SELECT
                    id,
                    klantnaam
                    FROM adressen
                    ORDER BY klantnaam ASC ";

    if(!$res_lev = mysqli_query($mysqli, $sql_lev)) { include('includes/error_database.php'); die; }

    while($row_lev = mysqli_fetch_array($res_lev)) {

        echo '<option value="'.$row_lev['id'].'">'.$row_lev['klantnaam'].'</option>';
    }
    echo '
</select>

<br>

<select id="second-choice">
  <option>Please choose from above</option>
</select>';
?>

<script type="text/javascript">
$("#first-choice").change(function() {
  $("#second-choice").load("getter.php?choice=" + $("#first-choice").val());
});
</script>

getter.php

<?php
require_once('includes/db_connect.php');

$choice = mysqli_real_escape_string($mysqli, $_GET['choice']);

echo '<option value="">Choose here now</option>';

  $sql_cnt = "SELECT
                    id,
                    naam
                    FROM contactpersoon
                    WHERE klant_id = ".$choice."
                    ORDER BY naam ASC ";

    if(!$res_cnt = mysqli_query($mysqli, $sql_cnt)) { include('includes/error_database.php'); die; }

    while($row_cnt = mysqli_fetch_array($res_cnt)) {

        echo '<option value="'.$row_cnt['id'].'">'.$row_cnt['naam'].'</option>';
    }
?>

Answer 1

用于$("#first-choice option:selected")

的选择对象

<script type="text/javascript">
$("#first-choice").change(function() {
  $("#second-choice").load("getter.php?choice=" + $("#first-choice option:selected").val());
});
</script>

动态下拉菜单结构无响应

1 个答案: