动态填充下拉列表PHP MySQL [英] Dynamic Populated Drop Down List PHP MySQL

问题描述

我有两个下拉列表，第二个基于用户在第一个下拉列表中选择的列表。我将使用Javascript动态更改第二个下拉列表。但是现在问题来了。此信息将存储在用户的个人资料中。因此，理想情况下，我希望能够从MySQL数据库中提取此信息并将其显示在PHP页面上。当我从MySQl填充列表时，如何加载正确的动态菜单。我该如何以自己想要的方式完成此操作。我不认为javascript在这种情况下会行得通吗？

I have two Drop Down Lists, the second ones are based off which one the users selects in the first drop down. I was going to use Javascript to dynamically change the second drop down. But now here comes the problem. This information is going to be stored on a user's profile. So ideally I would like to be able to pull this information out of MySQL database and display it on a PHP page. How would I go about loading the proper dynamic menu when I populate the list from MySQl. How can I accomplish this the way I want. I dont think javascript would work in this case would it?

推荐答案

如果您不想刷新页面，那么答案是 JavaScript。现在，您有几个选择。就个人而言，我更喜欢预先加载更多内容，因此我保留了一个JSON对象，概述了各种可能性，然后将静态内容交换为静态内容。

If you don't want a page refresh, then the answer is "JavaScript". Now, you have a couple of options there. Personally, I prefer to load more up front, so I keep a JSON object which outlines the possibilities and then I swap static content for static content.

这意味着onchange我会有类似的东西：

This would mean that onchange I would have something like:

var first = document.getElementById( "id of first select" )
var items = possibilities[ 
    first.value // you can also do things with selectedIndex and options if needs.
];

var second = document.getElementById( "id of second select" )
for( var i in items )
{
    var opt = document.createElement('option');
    opt.setAttribute('value', items[ it ] );
    second.appendChild( opt );
} 

您的另一种选择是通过网络发送数据并使其正常工作通过AJAX。我没有足够的教学空间，我可以安全地将您链接到此。

Your other option is to send data "over the wire" and get it working through AJAX. Not having room for a tutorial, I think I can safely link you to this one.

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