Question

我在数据库中有3个表，我想加入这些表以构建3级嵌套列表（将用作菜单的一部分）。到目前为止，我已经想出如何从下表中的表中选择数据。我希望在php部分生成菜单的第3级有一些帮助。：

SQL表：

page_list：pageID（primary_key）page_name pageURL
level2_menu   ：GeneralID（primary_key）gener_name pageID（foreign_key）;
  level3_menu：deepID（primary_key）generalID（foreign_key）deep_title

使用：

<?php 
get_menu("SELECT page_list.pageID as pageID, page_list.page_name as page_name, page_list.pageURL as pageURL, level2_menu.GeneralID as GeneralID, level2_menu.gener_name as general_name, level3_menu.deepID as deepID, level3_menu.deep_title as deep_title
          FROM page_list
          LEFT OUTER JOIN level2_menu ON page_list.pageID = level2_menu.pageID
          LEFT OUTER JOIN level3_menu ON level2_menu.GeneralID = level3_menu.generalID
          UNION
          SELECT page_list.pageID as pageID, page_list.page_name as page_name, page_list.pageURL as pageURL, level2_menu.GeneralID as GeneralID, level2_menu.gener_name as general_name, level3_menu.deepID as deepID, level3_menu.deep_title as deep_title
          FROM page_list
          RIGHT OUTER JOIN level2_menu ON page_list.pageID = level2_menu.pageID
          RIGHT OUTER JOIN level3_menu ON level2_menu.GeneralID = level3_menu.generalID
          ORDER BY pageID desc
          ");

 ?>

这给出了以下sql结果：http://sqlfiddle.com/#!9/927a0

+--------------------+------------+----------+-----------+---------------+--------+------------+
| pageID Ascending 1 | page_name  | pageURL  | GeneralID | general_name  | deepID | deep_title |
+--------------------+------------+----------+-----------+---------------+--------+------------+
|                  7 | get quote  | 0        | NULL      | NULL          | NULL   | NULL       |
|                  4 | tjanster   | info.php | 9         | Title         | NULL   | NULL       |
|                  3 | products   | info.php | 5         | Title 2       | 8      | Subtitle   |
+--------------------+------------+----------+-----------+---------------+--------+------------+

除了获取1级列表和2级列表的数据外，还可以使用：

function get_menu($sql)
{
    include 'connect.php';
    $result = $conn->query($sql);
    if ($result->num_rows > 0)
        ;
    $current_album = "";
    echo "<div id='menu'><div class='menu_pos'><ul class='main-navigation'>\n";
    while ($row = $result->fetch_assoc()) {
        if ($current_album <> "" && $current_album <> $row['pageID'])
            echo "</ul></li>";

        if ($current_album <> $row['pageID']) {
            echo "<li id='qoute'><a href='#'>" . $row['page_name'] . "</a><ul>\n";
            $current_album = $row['pageID'];
        }

        if ($row['GeneralID'] == 'NULL') {
            echo "bob";
        } else {
            echo "<li><a href='#'>" . $row['general_name'] . "</a></li>\n";
        }

    }
    echo "</ul></div></div>";
}

这导致The first menu example shown here这是一个2级列表，但是我想要3级列表，就像第二个菜单示例一样，显示在jsfiddle中。

Answer 1

首先，这是一种艰难的做事方式。原始SQL / RAW HTML / RAW PHP。模板化（例如twig），DB助手（例如Doctrine）可以制造魔法。

但要回答你的问题 - 这可能有用：

首先：我建议修改SQL

Ben@Ben-AMMS MINGW64 /c/xampp/htdocs/testAlaKloon
$ git clone https://github.com/Diseasedfire/AlaBonteKoe.git
Cloning into 'AlaBonteKoe'...
remote: Counting objects: 3027, done.
remote: Compressing objects: 100% (66/66), done.
remote: Total 3027 (delta 27), reused 0 (delta 0), pack-reused 2955
Receiving objects: 100% (3027/3027), 37.23 MiB | 588.00 KiB/s, done.
Resolving deltas: 100% (494/494), done.
Checking connectivity... done.

Ben@Ben-AMMS MINGW64 /c/xampp/htdocs/testAlaKloon
$ vim README.MD

(I make edits to the readme)

Ben@Ben-AMMS MINGW64 /c/xampp/htdocs/testAlaKloon/AlaBonteKoe (master)
$ git commit -a -m "test remote commit"
[master 318b667] test remote commit
 1 file changed, 1 insertion(+)

Ben@Ben-AMMS MINGW64 /c/xampp/htdocs/testAlaKloon/AlaBonteKoe (master)
$ git remote add origin https://github.com/nebbii/AlaBonteKoe.git
fatal: remote origin already exists.

Ben@Ben-AMMS MINGW64 /c/xampp/htdocs/testAlaKloon/AlaBonteKoe (master)
$ git remote -v
origin  https://github.com/Diseasedfire/AlaBonteKoe.git (fetch)
origin  https://github.com/Diseasedfire/AlaBonteKoe.git (push)

Ben@Ben-AMMS MINGW64 /c/xampp/htdocs/testAlaKloon/AlaBonteKoe (master)
$ git push origin master
remote: Permission to Diseasedfire/AlaBonteKoe.git denied to nebbii.
fatal: unable to access 'https://github.com/Diseasedfire/AlaBonteKoe.git/':
The requested URL returned error: 403

第二：分割准备菜单和输出菜单。

SELECT 
page_list.pageID as pageID, page_list.page_name as page_name, page_list.pageURL as pageURL, 
level2_menu.GeneralID as GeneralID, level2_menu.gener_name as general_name, 
level3_menu.deepID as deepID, level3_menu.deep_title as deep_title
FROM page_list
    LEFT OUTER JOIN level2_menu ON page_list.pageID = level2_menu.pageID
    LEFT OUTER JOIN level3_menu ON level2_menu.GeneralID = level3_menu.generalID
ORDER BY pageID DESC, GeneralID DESC, deepID DESC

从数据库表构建php嵌套列表

1 个答案: