dongwuxie5112 2011-05-03 15:15
浏览 16
已采纳

MySQL 1到很多。 仅显示第一个表中的1个结果

EXAMPLE CODE

<?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>
";

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

OUTPUT

Services
- service 1

Services
- service 2 ...etc

Products
- product 1

Products
- product 2 ...etc


I would like

Services
- service 1
- service 2

Products
- product 1
- product 2

  • 写回答

2条回答 默认 最新

  • dongtuo4132 2011-05-03 15:27
    关注
    <?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>
    ";
        }
    
    //SUB MENU (show all related results)
        echo "       <ul>
    ";
        echo "        <li><a href=\"$sub_href\">$sub_link_name</a></li>
    ";
        echo "       </ul>
    ";
        if ($p_link_name !== $link_name) {
             echo "      </li>
    ";
        }
        $p_link_name = $link_name;
    }
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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