php - MySQL 1 到多。仅显示第一个表中的 1 个结果

Question

示例代码

<?php
require_once('class_library/pdo.php');
$pdoConnection = new sdb('web_structure');
$select = $pdoConnection->query("SELECT main_menu.href AS main_href
    , main_menu.link_name AS main_link
    , sub_menu.href AS sub_href
    , sub_menu.link_name AS sub_link
  FROM main_menu
  LEFT JOIN sub_menu ON main_menu.id = sub_menu.main_menu_id 
  ORDER BY main_menu.position ASC");

while($row = $select->fetch())
{
    $href = $row["main_href"];
    $link_name = $row["main_link"];
    $sub_href = $row["sub_href"];
    $sub_link_name = $row["sub_link"];  

//MAIN MENU (display only once)
    echo "      <li><a href=\"$href\">$link_name</a>\n";

//SUB MENU (show all related results)
    echo "       <ul>\n";
    echo "        <li><a href=\"$sub_href\">$sub_link_name</a></li>\n";
    echo "       </ul>\n";
    echo "      </li>\n";
}
?>

输出

服务
- 服务 1

服务
- 服务 2 ...等等

产品
- 产品 1

产品
- 产品 2 ...等等

---

我愿意

服务
- 服务 1
- 服务 2

产品
- 产品 1
- 产品 2

Answer 1

<?php
require_once('class_library/pdo.php');
$pdoConnection = new sdb('web_structure');
$select = $pdoConnection->query("SELECT main_menu.href AS main_href
    , main_menu.link_name AS main_link
    , sub_menu.href AS sub_href
    , sub_menu.link_name AS sub_link
  FROM main_menu
  LEFT JOIN sub_menu ON main_menu.id = sub_menu.main_menu_id 
  ORDER BY main_menu.position ASC, main_menu.id ASC");

$p_link_name = '';
while($row = $select->fetch())
{
    $href = $row["main_href"];
    $link_name = $row["main_link"];
    $sub_href = $row["sub_href"];
    $sub_link_name = $row["sub_link"];  

//MAIN MENU (display only once)
    if ($p_link_name !== $link_name) {
        echo "      <li><a href=\"$href\">$link_name</a>\n";
    }

//SUB MENU (show all related results)
    echo "       <ul>\n";
    echo "        <li><a href=\"$sub_href\">$sub_link_name</a></li>\n";
    echo "       </ul>\n";
    if ($p_link_name !== $link_name) {
         echo "      </li>\n";
    }
    $p_link_name = $link_name;
}
?>