PHP/MySQL构建树菜单 [英] PHP / MySQL build tree menu

问题描述

我正在尝试用PHP和MySQL从数据库中构建一个不被接受的列表菜单树.

I am trying to build an un-oredered list menu tree from my database in PHP and MySQL.

我有一个从数据库返回的页面对象数组.每个页面对象都有parent_id属性，如果没有父对象，则将其设置为null.页面对象如下所示:

I have an array of page objects I am returning from the db. Each page object has parent_id attribute, which is set to null if it doesn't have a parent. Here's what the page objects look like:

page object
  id
  title
  parent_id

如果可能的话，我不想递归地执行它，而只打一次数据库，因为我将在几乎每个请求上构建菜单.我想创建一个函数，我可以将我的对象数组传递给它，它将返回html列表.

If possible I would like to not do it recursively and only hit the database once, since I am going to be building the menu on almost every request. I want to create a function that I can just pass my array of objects to and it will return the html list.

推荐答案

我喜欢@mario的解决方案，并且通过防止多余的<ul>对其进行了改进.我只建议对您的SQL查询执行ORDER BY以获得所需顺序的菜单(甚至建议将权重/顺序列添加到架构中.

I like @mario's solution, and have improved on it with the prevention of the excess <ul>. I would just recommend doing an ORDER BY on your SQL query to get the menu in the order you want (might even recommend a weight/sequence column be added to the schema.

数据设置:

$menu = array( // Presumed to have been coming from a SQL SELECT, populated for demo.
  array('id'=>1,'title'=>'Menu 1',          'parent_id'=>null),
  array('id'=>2,'title'=>'Sub 1.1',         'parent_id'=>1),
  array('id'=>3,'title'=>'Sub 1.2',         'parent_id'=>1),
  array('id'=>4,'title'=>'Sub 1.3',         'parent_id'=>1),
  array('id'=>5,'title'=>'Menu 2',          'parent_id'=>null),
  array('id'=>6,'title'=>'Sub 2.1',         'parent_id'=>5),
  array('id'=>7,'title'=>'Sub Sub 2.1.1',   'parent_id'=>6),
  array('id'=>8,'title'=>'Sub 2.2',         'parent_id'=>5),
  array('id'=>9,'title'=>'Menu 3',          'parent_id'=>null),
);

处理:

function has_children($rows,$id) {
  foreach ($rows as $row) {
    if ($row['parent_id'] == $id)
      return true;
  }
  return false;
}
function build_menu($rows,$parent=0)
{  
  $result = "<ul>";
  foreach ($rows as $row)
  {
    if ($row['parent_id'] == $parent){
      $result.= "<li>{$row['title']}";
      if (has_children($rows,$row['id']))
        $result.= build_menu($rows,$row['id']);
      $result.= "</li>";
    }
  }
  $result.= "</ul>";

  return $result;
}
echo build_menu($menu);

输出:

<ul>
  <li>Menu 1<ul>
    <li>Sub 1.1</li>
    <li>Sub 1.2</li>
    <li>Sub 1.3</li>
  </ul></li>
  <li>Menu 2<ul>
    <li>Sub 2.1<ul>
      <li>Sub Sub 2.1.1</li>
    </ul></li>
    <li>Sub 2.2</li>
  </ul></li>
  <li>Menu 3</li>
</ul>

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