从另一个表中获取相关选项，并与当前结果合并 - grab associated options from another table and merge with current results

grab associated options from another table and merge with current results

我有3张表

站点:ID，名称
assoc: type, type_id, site_id
选项:ID, name

我想查询选择*网站，然后从assoc选择相关的选项，并从选项表中抓取名称。

代码我想缩短为一个查询:

     $getsites = mysql_query("SELECT * FROM sites")or die(mysql_error());
     while($row = mysql_fetch_array($getsites)){
     echo $row['name'];
     $getassoc = mysql_query("SELECT * FROM assoc WHERE type='options' AND site_id = '$row[ID]'")or die(mysql_error());
      echo'<ul>';
      while($subrow = mysql_fetch_array($getassoc)){
      $getoption = mysql_query("SELECT * FROM options WHERE ID = '$subrow[assoc_id]'")or die(mysql_error());
      $option = mysql_fetch_assoc($getoption);
      echo '<li>'.$option['name'].'</li>';
   }
   echo'</ul><br/>';
   }

它被称为join:

select options.name
from sites
inner join assoc on sites.id = assoc.site_id
inner join options on options.id = assoc.assoc_id
where assoc.type = 'options'

问题:assoc表中没有assoc_id