如何使用php&与子菜单创建动态菜单MySQL的 [英] how to create dynamic menu with sub-menu with php & mysql

问题描述

我想要一个动态菜单，该菜单使用php和mysql从表中获取数据.

I want to have a dynamic menu that feeds from a table using php and mysql.

我的桌子看起来像这样:

My table looks like this:

sec_id  sec_name    sec_group
1   section 1   group 1
2   section 2   group 1
3   section 3   group 2
4   section 4   group 1
5   section 5   group 3

我可以执行查询以获取并显示唯一的sec_group值，但无法找出将sec_name包含在每个sec_group中的方法

I can do a query to get and display unique sec_group values but can't figure out a way to include sec_name into each sec_group

//Query by unique sec_group
$qry_secs="SELECT DISTINCT sec_group FROM tbl_user_sec ORDER BY sec_id ASC";
$result_secs = mysql_query($qry_secs);

//echo values
while($row_secs = mysql_fetch_assoc($result_secs)){
 echo '<ul><li><a href="#">'.$row_secs['sec_group'].'</a></li></ul>';
}

最终，HTML应该喜欢下面的代码.

Eventually, the HTML should like the code below.

<ul>
<li><a href="#">Group 1</a>
  <ul>
  <li><a href="#">Section 1</a></li>
  <li><a href="#">Section 2</a></li>
  <li><a href="#">Section 4</a></li>
  </ul>
</li>

<li><a href="#">Group 2</a>
  <ul>
  <li><a href="#">Section 3</a></li>
  </ul>
</li>

<li><a href="#">Group 3</a>
  <ul>
  <li><a href="#">Section 5</a></li>
  </ul>
</li>
</ul>

有什么想法吗?

推荐答案

$q = mysql_query("SELECT sec_id, sec_name, sec_group FROM tbl_user_sec ORDER BY sec_id");

// prepare data 
$groups = Array();
while($w = mysql_fetch_assoc($q)) {
  if(!isset($groups[$w['sec_group']])) $groups[$w['sec_group']] = Array();
  $groups[$w['sec_group']][] = $w;
}

// display data
echo "<ul>";
foreach($groups as $group_name => $sections) {
  echo '<li><a href="#">'.$group_name.'</a><ul>';
  foreach($sections as $section) {
    echo '<li><a href="#">'.$section['sec_name'].'</a>';
  }
  echo '</ul></li>';
}
echo "</ul>";

如果您不在乎按sec_id

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