Question

我有一个如下所示的数据库。我需要从子类别创建一个类别的路径。

例如：网站显示，父母＆gt;小孩＆gt;小孩＆gt;子

SELECT * FROM category ORDER BY category_id;
+-------------+----------------------+--------+
| category_id | name                 | parent |
+-------------+----------------------+--------+
|           1 | ELECTRONICS          |   NULL |
|           2 | TELEVISIONS          |      1 |
|           3 | TUBE                 |      2 |
|           4 | LCD                  |      2 |
|           5 | PLASMA               |      2 |
|           6 | PORTABLE ELECTRONICS |      1 |
|           7 | MP3 PLAYERS          |      6 |
|           8 | FLASH                |      7 |
|           9 | CD PLAYERS           |      6 |
|          10 | 2 WAY RADIOS         |      6 |
+-------------+----------------------+--------+

当我选择＆＃34; 7 | MP3播放器＆＃34;我想要显示如下。

电子＆gt;便携式电子设备＆gt; MP3播放器

我在这里使用哪种技术？

Answer 1

尝试以下代码。通过递归函数，它是可能的。

function fetchCategoryTree($parent = 0, $spacing = '', $user_tree_array = '') {

  if (!is_array($user_tree_array))
    $user_tree_array = array();

  $sql = "SELECT `category_id`, `name`, `parent` FROM `category` WHERE 1 AND `parent` = $parent ORDER BY cid ASC";
  $query = mysql_query($sql);
  if (mysql_num_rows($query) > 0) {
    while ($row = mysql_fetch_object($query)) {
      $user_tree_array[] = array("id" => $row->category_id, "name" => $spacing . $row->name);
      $user_tree_array = fetchCategoryTree($row->category_id, $spacing . '---', $user_tree_array);
    }
  }
  return $user_tree_array;
}

$categoryList = fetchCategoryTree();
?>
<select>
<?php foreach($categoryList as $cl) { ?>
  <option value="<?php echo $cl["id"] ?>"><?php echo $cl["name"]; ?></option>
<?php } ?>
</select>

如何在php上从数据库创建类别路径

1 个答案: