动态构建水平菜单 [英] Dynamically building horizontal menu

问题描述

我有这个JavaScript代码，它允许我创建一个带有子菜单的水平菜单，如下所示：

 < ul id = 菜单 > 
< li>菜单1 
< ul> 
< li>子菜单1< / li> 
< / ul> 
< / li> 
< / ul> 
  

我可以根据需要创建多个子菜单，问题是我使用PHP来抓取来自MySQL数据库的链接，不知道如何动态构建这些子菜单，而无需一次又一次手动检查子菜单。例如在MySQL表中：

字段：
Menu_ID
Menu_Name
Menu_Link
Menu_ParentID

因此，菜单ID只是一个自动增量，menu_parentid允许我为子菜单分配一个子菜单名称/链接。但为了做到这一点，我目前为此做了两个子菜单检查：

  $ query =SELECT * FROM site_menu WHERE Menu_ParentID = 0; 
 
 foreach（$ query AS $ q）
 {
 //运行结果
 $ query2 =SELECT * FROM site_menu WHERE Menu_ParentID = $ q ['' ID']; 
 
 foreach（$ query2 AS $ q2）
 {
 //运行结果
} 
} 
  

正如你所看到的，我必须查询两次才能得到第一个子菜单，如果有第三个子菜单会怎样？我必须运行3个查询吗？任何建议？

解决方案

也许一个函数或do..while循环可能是为了？概念验证：

$ p $ function menuQuery（$ id）
{
$ query =SELECT * FROM site_menu WHERE Menu_ParentID = $ id;
$ b $ if（$ query）{
foreach（$ query AS $ q）{
//遍历结果
menuQuery（$ q-> id） ;




//顶级菜单项的初始调用
menuQuery（0）;


I have this JavaScript code that allows me to create a horizontal menu with sub menus like so:

<ul id="menu">
<li>Menu 1
  <ul>
   <li>Sub Menu 1</li>
  </ul>
</li>
</ul>

I can create as many sub menus as I want, the problem is I'm using PHP to grab the links from a MySQL database and don't know how I can dynamically build these sub menus without manually checking a sub menu over and over again. For example in the MySQL table:

Fields: Menu_ID Menu_Name Menu_Link Menu_ParentID

So menu ID is just an auto increment and the menu_parentid allows me to assign a sub menu name/link to a parent menu. But in order to do it I currently do this for 2 sub menu checks:

$query = "SELECT * FROM site_menu WHERE Menu_ParentID = 0";

foreach($query AS $q)
{
//run through the results
$query2 = "SELECT * FROM site_menu WHERE Menu_ParentID = $q['id']";

foreach($query2 AS $q2)
{
//run through the results
}
}

As you can see I have to query twice to get just the first sub menu, what if there is a third sub menu? Do I have to run 3 queries? Any suggestions?

解决方案

Perhaps a function or a do..while loop may be in order? Proof of concept:

function menuQuery($id)
{
    $query = "SELECT * FROM site_menu WHERE Menu_ParentID = $id";

    if ($query) {
        foreach($query AS $q) {
            //run through the results
            menuQuery($q->id);
        }
    }
}

//initial call of top level menu items
menuQuery(0);

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