问题描述
我有一个包含两个选择字段的表单，一个表示区域，另一个表示城市/村庄/等的名称。我有一个包含所有这些条目的数据库，格式如下：
I have a form that has two select fields, one representing the region and one the name of the city/village/etc. I have a database with all these entries in the following form:
id (int 11, ai)
region (varchar 50)
city(varchar 50)
我在这里找到了一个脚本这可以让你选择不同的选项，但它是用JS制作的，我不知道如何根据我的需要调整它。我需要一个数据库，因为我有13.000多个条目，并且在代码中手动输入它们并不是一个好主意。这是我读过的主题的链接，它的解决方案也在其中一条评论中。
I've found a script here on so that lets you select different options but it's made with JS and I have no idea how to adapt it to my needs. I need a database because I have 13.000+ entries and it's not really a good idea to enter them manually in the code. Here is the link to the topic that I've read, it's solution is in one of the comments as well.
。
如果我需要在降级之前编辑我的帖子，请告诉我。在此先感谢！
Please let me know if I need to edit my post before downrating. Thanks in advance!
推荐答案
当一个选择更改从服务器获取数据以提供其他选择时，只需使用ajax。 / p>
Just use ajax for this, when one select change fetch data from the server to feed other select.
<select class="select_one">
<?php /* render first select ?>
</select>
<select class="select_two"></select>
<script>
$(function() {
$('.select_one').change(function() {
var select = $('.select_two').empty();
$.get('script.php', {region: $(this).val()}, function(result) {
$.each(result, function(i, item) {
$('<option value="' + item.value + '">' + item.name + '</option>').
appendTo(select);
});
});
});
});
</script>
而你 script.php 应返回JSON来自db：
and you script.php should return JSON from db:
if (isset($_GET['region'])) {
$sql = new mysqli('localhost','username','password','database');
$region = mysqli_real_escape_string($sql,$_GET['region']);
$query = "SELECT * FROM cities WHERE region = $region";
$ret = $sql->query($query);
$result = array();
while ($row = $ret->fetch_assoc()) {
$result[] = array(
'value' => $row['id'],
'name' => $row['city']
);
}
echo json_encode($result);
}
这篇关于根据其他选择选项从DB中选择选项的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！