php - 使用 PHP+MySQL 显示带有子类别和 n-sub 类别的类别

Question

我的代码有问题。显示具有 n 个子类别的所有类别的最佳方式是什么？

我的 MySQL 数据库如下所示:

CREATE TABLE categories ( 
category_id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
parent_id INT UNSIGNED NOT NULL DEFAULT 0, 
name TEXT NOT NULL, 
user INT NOT NULL, 
PRIMARY KEY (category_id), 
INDEX parent (parent_id)
);

其中 user 是来自另一个表的 id(如果类别有 user = '2'，id = '2' 的用户可以看到那个类别)

category_id   parent_id   name                   user
1             0           Main category 1        2
2             1           Subcategory 1          2
3             1           Subcategory 2          2
4             3           SubSubcategory 1       2

我的 php 代码:

<?php 
    include('config.php'); //Mysql connect file
//get id from user

 if($_GET['id'] != $user_id) {

    $id = $_GET['id'];

//check to make sure the user is an admin, who can change mysql

$result = mysql_query("SELECT * FROM users WHERE username = '$username' AND admin = '1'");

$rowCheck = mysql_result($result, 0);



//if the query returns a number, we know the user is an admin. And here we can VIEW our categories

if($rowCheck > 0) { 

$user_result = mysql_query("SELECT * FROM users, categories WHERE id = '$id'");

     while($row = mysql_fetch_array($user_result)) { ?>     

        <ol>
        <li><a href=""><?php echo $row['name']; ?></a></li>
        </ol>

    <?php }

    } else {

        echo "You are not admin";
     }    
//bla bla bla etc.
...
 }

如何查看类

<ol>
 <li>
 <a href="">Main category 1</a>
      <ol>
      <li><a href="">Subcategory 1</a></li>
      <li><a href="">Subcategory 2</a>
          <ol>
              <li><a href="">Subsubcategory 1</a></li>
          </ol>
      </li>
      </ol>
 </li>
</ol>

Answer 1

试试这个

$result = mysql_query("SELECT * FROM users WHERE username = '$username' AND admin = '1'");

$rowCheck = mysql_result($result, 0);   

//if the query returns a number, we know the user is an admin. And here we can VIEW our categories

if($rowCheck > 0) { 

//First fetch only parent categories
$first_level_cats = mysql_query("SELECT * FROM categories WHERE user = '$id' AND parent_id = 0");
     echo '<ol>';
     while($row = mysql_fetch_array($first_level-cats)) { ?>     

        <li><a href=""><?php echo $row['name']; ?></a></li>

        <?php
         $second_level_cats = mysql_query("SELECT * FROM categories WHERE user = '$id' AND parent_id = $row['id']");
     echo '<ol>';
     while($row = mysql_fetch_array($second_level-cats)) { ?>     

        <li><a href=""><?php echo $row['name']; ?></a></li>              
        <?php 
          $third_level_cats = mysql_query("SELECT * FROM categories WHERE user = '$id' AND parent_id = $row['id']");
     echo '<ol>';
     while($row = mysql_fetch_array($sthird_level-cats)) { ?>     

        <li><a href=""><?php echo $row['name']; ?></a></li>              
        <?php
            }
          echo '</ol>'; //third level close
        ?>
          } 
         echo '</ol>'; // second level close
        ?>    
    <?php }
   echo '</ol>';  // first level close
    } else {

        echo "You are not admin";
     }    
//bla bla bla etc.
...
 }

我已尽力解决您的查询，根据您的架构交叉检查表名和列。

尝试使用Mysqli扩展名而不是 Mysql