动态链接选择框 [英] Dynamic chained select box

问题描述

好的，因此我的页面上有2个选择框，第一个选择框的名称为"select"，并带有以下硬编码的选项:

Okay, so I have 2 select boxes on my page, the first with the name "select" with the following options hard-coded:

<select name="select" id="Testid">
        <option value="Misc">Misc...</option>
        <option value="Stunt">Stunt</option>
        <option value="deathmatches">Deathmatches</option>
        <option value="Car"> Car </option>
        <option value="Races">Races </option>
    </select>

但是第二个选择框(名为"deleteTp")我想根据第一个选择框的选择填充一个mysql表.每个选项值在我的数据库中都有一个与该值同名的特定表.

The second select box however (named "deleteTp") I want to populate with a mysql table depending on the selection of the first select box. Each option value has a specific table in my database with the same name as the value.

我知道如何用数据库表填充第二个选择框，但是我不确定如何根据用户的选择动态地执行此操作.

I know how to populate the second select box with the database table, however I'm not sure how to do this dynamically depending on the users choice.

请注意，我只知道基本的javascript.

Please note I only know basic javascript.

如果您想知道为什么要尝试这样做，那是因为第二个选择框将发布到外部页面上，以便从填充表中删除行.

And if you're wondering why I am trying to do this, it's because the second selectbox will be posted to an external page to delete the rows from the populated table.

谢谢您的时间.

推荐答案

$.ajax({
  type: 'post',
  url: "ajax.php."
  data: { id:'your id here' } //your id is your select box selected id.
}).done(function() { 

});

在php端:

$id = $_REQUEST['id']; 
$json = array();
$sql = mysql_query("select * from yourdb where id = $id");
while ($row = mysql_fetch_assoc($sql)) {
    $json[] = $row;
}
echo json_encode($json);

ajax ref:http://api.jquery.com/jQuery.ajax/

ajax ref : http://api.jquery.com/jQuery.ajax/

json参考:http://en.wikipedia.org/wiki/JSON

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