我刚刚制作了“Closure表”组合查询分层数据的更新/添加/删除部分,该数据显示在此幻灯片中的第70页:http://www.slideshare.net/billkarwin/sql-antipatterns-strike-back
我的数据库如下所示:
表类别:
ID Name
1 Top value
2 Sub value1
表类别树:
child parent level
1 1 0
2 2 0
2 1 1
但是,我有一个问题是将完整的树从单个查询中恢复为多维数组。
以下是我想要回复的内容:
array (
'topvalue' = array (
'Subvalue',
'Subvalue2',
'Subvalue3)
);
);
更新 找到此链接,但我仍然很难将其转换为数组: http://karwin.blogspot.com/2010/03/rendering-trees-with-closure-tables.html
Update2: 我现在能够为每个类别添加深度,如果这可以有任何帮助。
答案 0 :(得分:9)
以下示例提供的内容比您要求的要多一些,但这是一种非常好的方式,并且仍然可以演示每个阶段信息的来源。
它使用以下表结构:
+--------+------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+------------------+------+-----+---------+----------------+
| id | int(10) unsigned | NO | PRI | NULL | auto_increment |
| parent | int(10) unsigned | NO | | NULL | |
| name | varchar(45) | NO | | NULL | |
+--------+------------------+------+-----+---------+----------------+
这是:
<?php
// Connect to the database
mysql_connect('localhost', 'root', '');
mysql_select_db('test');
echo '<pre>';
$categories = Category::getTopCategories();
print_r($categories);
echo '</pre>';
class Category
{
/**
* The information stored in the database for each category
*/
public $id;
public $parent;
public $name;
// The child categories
public $children;
public function __construct()
{
// Get the child categories when we get this category
$this->getChildCategories();
}
/**
* Get the child categories
* @return array
*/
public function getChildCategories()
{
if ($this->children) {
return $this->children;
}
return $this->children = self::getCategories("parent = {$this->id}");
}
////////////////////////////////////////////////////////////////////////////
/**
* The top-level categories (i.e. no parent)
* @return array
*/
public static function getTopCategories()
{
return self::getCategories('parent = 0');
}
/**
* Get categories from the database.
* @param string $where Conditions for the returned rows to meet
* @return array
*/
public static function getCategories($where = '')
{
if ($where) $where = " WHERE $where";
$result = mysql_query("SELECT * FROM categories$where");
$categories = array();
while ($category = mysql_fetch_object($result, 'Category'))
$categories[] = $category;
mysql_free_result($result);
return $categories;
}
}
在我的数据库中,我有以下行:
+----+--------+-----------------+
| id | parent | name |
+----+--------+-----------------+
| 1 | 0 | First Top |
| 2 | 0 | Second Top |
| 3 | 0 | Third Top |
| 4 | 1 | First Child |
| 5 | 1 | Second Child |
| 6 | 2 | Third Child |
| 7 | 2 | Fourth Child |
| 8 | 4 | First Subchild |
| 9 | 4 | Second Subchild |
+----+--------+-----------------+
因此脚本输出以下(冗长)信息:
Array
(
[0] => Category Object
(
[id] => 1
[parent] => 0
[name] => First Top
[children] => Array
(
[0] => Category Object
(
[id] => 4
[parent] => 1
[name] => First Child
[children] => Array
(
[0] => Category Object
(
[id] => 8
[parent] => 4
[name] => First Subchild
[children] => Array
(
)
)
[1] => Category Object
(
[id] => 9
[parent] => 4
[name] => Second Subchild
[children] => Array
(
)
)
)
)
[1] => Category Object
(
[id] => 5
[parent] => 1
[name] => Second Child
[children] => Array
(
)
)
)
)
[1] => Category Object
(
[id] => 2
[parent] => 0
[name] => Second Top
[children] => Array
(
[0] => Category Object
(
[id] => 6
[parent] => 2
[name] => Third Child
[children] => Array
(
)
)
[1] => Category Object
(
[id] => 7
[parent] => 2
[name] => Fourth Child
[children] => Array
(
)
)
)
)
[2] => Category Object
(
[id] => 3
[parent] => 0
[name] => Third Top
[children] => Array
(
)
)
)
如果您要从数据创建菜单,我建议创建一些递归函数:
function outputCategories($categories, $startingLevel = 0)
{
$indent = str_repeat(" ", $startingLevel);
foreach ($categories as $category)
{
echo "$indent{$category->name}\n";
if (count($category->children) > 0)
outputCategories($category->children, $startingLevel+1);
}
}
$categories = Category::getTopCategories();
outputCategories($categories);
将输出以下内容:
First Top
First Child
First Subchild
Second Subchild
Second Child
Second Top
Third Child
Fourth Child
Third Top
享受
答案 1 :(得分:9)
好的,我编写了扩展Zend Framework DB表,行和行集类的PHP类。无论如何,我一直在开发这个,因为我在几个星期内就PHP Tek-X发表关于分层数据模型的演讲。
我不想将我的所有代码发布到Stack Overflow,因为如果我这样做,它们会隐式获得Creative Commons的许可。 更新:我已将代码提交至Zend Framework extras incubator,我的演示文稿位于slidehare Models for Hierarchical Data with SQL and PHP。
我将用伪代码描述解决方案。我正在使用动物分类学作为测试数据,从ITIS.gov下载。该表格为longnames
:
CREATE TABLE `longnames` (
`tsn` int(11) NOT NULL,
`completename` varchar(164) NOT NULL,
PRIMARY KEY (`tsn`),
KEY `tsn` (`tsn`,`completename`)
)
我为分类法层次结构中的路径创建了一个闭包表:
CREATE TABLE `closure` (
`a` int(11) NOT NULL DEFAULT '0', -- ancestor
`d` int(11) NOT NULL DEFAULT '0', -- descendant
`l` tinyint(3) unsigned NOT NULL, -- levels between a and d
PRIMARY KEY (`a`,`d`),
CONSTRAINT `closure_ibfk_1` FOREIGN KEY (`a`) REFERENCES `longnames` (`tsn`),
CONSTRAINT `closure_ibfk_2` FOREIGN KEY (`d`) REFERENCES `longnames` (`tsn`)
)
给定一个节点的主键,你可以这样得到它的所有后代:
SELECT d.*, p.a AS `_parent`
FROM longnames AS a
JOIN closure AS c ON (c.a = a.tsn)
JOIN longnames AS d ON (c.d = d.tsn)
LEFT OUTER JOIN closure AS p ON (p.d = d.tsn AND p.l = 1)
WHERE a.tsn = ? AND c.l <= ?
ORDER BY c.l;
closure AS p
的加入是包含每个节点的父ID。
查询很好地利用了索引:
+----+-------------+-------+--------+---------------+---------+---------+----------+------+-----------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+-------------+-------+--------+---------------+---------+---------+----------+------+-----------------------------+
| 1 | SIMPLE | a | const | PRIMARY,tsn | PRIMARY | 4 | const | 1 | Using index; Using filesort |
| 1 | SIMPLE | c | ref | PRIMARY,d | PRIMARY | 4 | const | 5346 | Using where |
| 1 | SIMPLE | d | eq_ref | PRIMARY,tsn | PRIMARY | 4 | itis.c.d | 1 | |
| 1 | SIMPLE | p | ref | d | d | 4 | itis.c.d | 3 | |
+----+-------------+-------+--------+---------------+---------+---------+----------+------+-----------------------------+
鉴于我在longnames
中有490,032行,在closure
中有4,299,883行,它运行得非常好:
+--------------------+----------+
| Status | Duration |
+--------------------+----------+
| starting | 0.000257 |
| Opening tables | 0.000028 |
| System lock | 0.000009 |
| Table lock | 0.000013 |
| init | 0.000048 |
| optimizing | 0.000032 |
| statistics | 0.000142 |
| preparing | 0.000048 |
| executing | 0.000008 |
| Sorting result | 0.034102 |
| Sending data | 0.001300 |
| end | 0.000018 |
| query end | 0.000005 |
| freeing items | 0.012191 |
| logging slow query | 0.000008 |
| cleaning up | 0.000007 |
+--------------------+----------+
现在我对上面的SQL查询的结果进行后处理,根据层次结构(伪代码)将行排序为子集:
while ($rowData = fetch()) {
$row = new RowObject($rowData);
$nodes[$row["tsn"]] = $row;
if (array_key_exists($row["_parent"], $nodes)) {
$nodes[$row["_parent"]]->addChildRow($row);
} else {
$top = $row;
}
}
return $top;
我还为行和行设定了类。 Rowset基本上是一个行数组。 A Row包含行数据的关联数组,并且还包含其子项的Rowset。叶子节点的子行为空。
行和行集还定义了名为toArrayDeep()
的方法,这些方法以递归方式将其数据内容转储为普通数组。
然后我可以像这样使用整个系统:
// Get an instance of the taxonomy table data gateway
$tax = new Taxonomy();
// query tree starting at Rodentia (id 180130), to a depth of 2
$tree = $tax->fetchTree(180130, 2);
// dump out the array
var_export($tree->toArrayDeep());
输出如下:
array (
'tsn' => '180130',
'completename' => 'Rodentia',
'_parent' => '179925',
'_children' =>
array (
0 =>
array (
'tsn' => '584569',
'completename' => 'Hystricognatha',
'_parent' => '180130',
'_children' =>
array (
0 =>
array (
'tsn' => '552299',
'completename' => 'Hystricognathi',
'_parent' => '584569',
),
),
),
1 =>
array (
'tsn' => '180134',
'completename' => 'Sciuromorpha',
'_parent' => '180130',
'_children' =>
array (
0 =>
array (
'tsn' => '180210',
'completename' => 'Castoridae',
'_parent' => '180134',
),
1 =>
array (
'tsn' => '180135',
'completename' => 'Sciuridae',
'_parent' => '180134',
),
2 =>
array (
'tsn' => '180131',
'completename' => 'Aplodontiidae',
'_parent' => '180134',
),
),
),
2 =>
array (
'tsn' => '573166',
'completename' => 'Anomaluromorpha',
'_parent' => '180130',
'_children' =>
array (
0 =>
array (
'tsn' => '573168',
'completename' => 'Anomaluridae',
'_parent' => '573166',
),
1 =>
array (
'tsn' => '573169',
'completename' => 'Pedetidae',
'_parent' => '573166',
),
),
),
3 =>
array (
'tsn' => '180273',
'completename' => 'Myomorpha',
'_parent' => '180130',
'_children' =>
array (
0 =>
array (
'tsn' => '180399',
'completename' => 'Dipodidae',
'_parent' => '180273',
),
1 =>
array (
'tsn' => '180360',
'completename' => 'Muridae',
'_parent' => '180273',
),
2 =>
array (
'tsn' => '180231',
'completename' => 'Heteromyidae',
'_parent' => '180273',
),
3 =>
array (
'tsn' => '180213',
'completename' => 'Geomyidae',
'_parent' => '180273',
),
4 =>
array (
'tsn' => '584940',
'completename' => 'Myoxidae',
'_parent' => '180273',
),
),
),
4 =>
array (
'tsn' => '573167',
'completename' => 'Sciuravida',
'_parent' => '180130',
'_children' =>
array (
0 =>
array (
'tsn' => '573170',
'completename' => 'Ctenodactylidae',
'_parent' => '573167',
),
),
),
),
)
重新评论有关计算深度 - 或每条路径的确实长度。
假设您刚刚在表中插入了一个包含实际节点的新节点(上例中的longnames
),新节点的id由MySQL中的LAST_INSERT_ID()
返回,否则你可以以某种方式得到它。
INSERT INTO Closure (a, d, l)
SELECT a, LAST_INSERT_ID(), l+1 FROM Closure
WHERE d = 5 -- the intended parent of your new node
UNION ALL SELECT LAST_INSERT_ID(), LAST_INSERT_ID(), 0;
答案 2 :(得分:2)
我喜欢icio的答案,但我更喜欢拥有数组数组,而不是对象数组。这是他的脚本修改为无需制作对象的工作:
<?php
require_once('mysql.php');
echo '<pre>';
$categories = Taxonomy::getTopCategories();
print_r($categories);
echo '</pre>';
class Taxonomy
{
public static function getTopCategories()
{
return self::getCategories('parent_taxonomycode_id = 0');
}
public static function getCategories($where = '')
{
if ($where) $where = " WHERE $where";
$result = mysql_query("SELECT * FROM taxonomycode $where");
$categories = array();
// while ($category = mysql_fetch_object($result, 'Category'))
while ($category = mysql_fetch_array($result)){
$my_id = $category['id'];
$category['children'] = Taxonomy::getCategories("parent_taxonomycode_id = $my_id");
$categories[] = $category;
}
mysql_free_result($result);
return $categories;
}
}
我认为值得注意的是,我的答案和icios都没有直接解决您的问题。它们都依赖于主表中的父id链接,并且不使用闭包表。但是,递归查询数据库肯定是要做的事情,但是不是递归传递父ID,而是必须传入父ID和深度级别(每次递归时应该增加1)以便查询在每个级别可以使用parent + depth从闭包表中获取直接父信息,而不是将其放在主表中。
HTH, -FT
答案 3 :(得分:1)
如果希望输出为无序列表,可以按如下方式更改outputCategories方法(基于数组中的ftrotters数组):
public function outputCategories($categories, $startingLevel = 0)
{
echo "<ul>\n";
foreach ($categories as $key => $category)
{
if (count($category['children']) > 0)
{
echo "<li>{$category['name']}\n";
$this->outputCategories($category['children'], $startingLevel+1);
echo "</li>\n";
}
else
{
echo "<li>{$category['name']}</li>\n";
}
}
echo "</ul>\n";
}
答案 4 :(得分:0)
很抱歉,但我认为您不能从您的(或任何)数据库查询中获取多维数组。