分層數據 Hierarchical Data 探索 (3.嵌套集合模型) 無限極分類

時間  2020-02-12
標籤 分層 數據 hierarchical data 探索 嵌套 集合 模型 無限 分類 欄目 快樂工作 简体版
原文   原文鏈接

分層數據Hierarchical Data探索（例如：無限級分類、多級菜單、省份城市）

引言

第一篇 分層數據Hierarchical Data探索(1.遞歸) 已經介紹了分層數據以及使用遞歸算法實現了無限極分類，可是遞歸即浪費時間，又浪費空間（內存）,尤爲是在數據量大的狀況下效率顯著降低。
第二篇 分層數據Hierarchical Data探索(2.鄰接表模型) 介紹了一種數據模型鄰接表模型來實現，但在檢索路徑的過程當中，除了本層外，每一層都會對應一個LEFT JOIN，那麼若是層數不定怎麼辦？或者層數過多？node

鄰接表模型的侷限性

用純SQL編碼實現鄰接表模型有必定的難度。在咱們檢索某分類的路徑以前，咱們須要知道該分類所在的層次。在刪除中間層的節點時，須要同時刪除該節點下的全部節點，不然會出現孤立節點。mysql

那麼，在MySQL中如何更好的處理分層數據呢？下面咱們來講一說嵌套集合模型算法

嵌套集合模型(Nested Set Model)

更多 嵌套集合模型（Nested Set Model）的介紹請見: wiki

在嵌套集合模型中，咱們將以一種新的方式來理解咱們的分層數據，再也不是線與點了，而是嵌套容器。下圖以嵌套容器的方式畫出了electronics分類圖：sql

經過集合的包含關係，嵌套結合模型能夠表示分層結構，每個分層能夠用一個Set來表示（一個圈），父節點所在的圈包含全部子節點所在的圈。segmentfault

爲了用MySQL來表示集合關係，須要定義連個字段 lft 和 rgt （表示一個集合的範圍）。electron

# 爲了模擬，咱們建立一個表category包含三個字段：id，title，lft，rgt以下：
CREATE TABLE category (
  id int(10) unsigned NOT NULL AUTO_INCREMENT PRIMARY KEY,
  title varchar(255) NOT NULL,
  lft int(10) NOT NULL,
  rgt int(10) NOT NULL
);

# 插入模擬數據
INSERT INTO category(title,lft,rgt) VALUES('Electronics',1,28);

INSERT INTO category(title,lft,rgt) VALUES('Laptops & PC',2,7);
 
INSERT INTO category(title,lft,rgt) VALUES('Laptops',3,4);
INSERT INTO category(title,lft,rgt) VALUES('PC',5,6);
 
INSERT INTO category(title,lft,rgt) VALUES('Cameras & photo',8,11);
INSERT INTO category(title,lft,rgt) VALUES('Camera',9,10);
 
INSERT INTO category(title,lft,rgt) VALUES('Phones & Accessories',12,27);
INSERT INTO category(title,lft,rgt) VALUES('Smartphones',13,20);
 
INSERT INTO category(title,lft,rgt) VALUES('Android',14,15);
INSERT INTO category(title,lft,rgt) VALUES('iOS',16,17);
INSERT INTO category(title,lft,rgt) VALUES('Other Smartphones',18,19);
 
INSERT INTO category(title,lft,rgt) VALUES('Batteries',21,22);
INSERT INTO category(title,lft,rgt) VALUES('Headsets',23,24);
INSERT INTO category(title,lft,rgt) VALUES('Screen Protectors',25,26);

select * from category;
+----+----------------------+-----+-----+
| id | title                | lft | rgt |
+----+----------------------+-----+-----+
|  1 | Electronics          |   1 |  28 |
|  2 | Laptops & PC         |   2 |   7 |
|  3 | Laptops              |   3 |   4 |
|  4 | PC                   |   5 |   6 |
|  5 | Cameras & photo      |   8 |  11 |
|  6 | Camera               |   9 |  10 |
|  7 | Phones & Accessories |  12 |  27 |
|  8 | Smartphones          |  13 |  20 |
|  9 | Android              |  14 |  15 |
| 10 | iOS                  |  16 |  17 |
| 11 | Other Smartphones    |  18 |  19 |
| 12 | Batteries            |  21 |  22 |
| 13 | Headsets             |  23 |  24 |
| 14 | Screen Protectors    |  25 |  26 |
+----+----------------------+-----+-----+
14 rows in set (0.00 sec)
  • 檢索分層路徑

因爲子節點的 lft 值總在父節點的 lft 和 rgt 值之間，因此能夠經過父節點鏈接到子節點上來檢索整棵樹函數

SELECT node.id,node.title,node.lft,node.rgt
FROM category AS node,
     category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND parent.title = 'Electronics'
ORDER BY node.lft;
+----+----------------------+-----+-----+
| id | title                | lft | rgt |
+----+----------------------+-----+-----+
|  1 | Electronics          |   1 |  28 |
|  2 | Laptops & PC         |   2 |   7 |
|  3 | Laptops              |   3 |   4 |
|  4 | PC                   |   5 |   6 |
|  5 | Cameras & photo      |   8 |  11 |
|  6 | Camera               |   9 |  10 |
|  7 | Phones & Accessories |  12 |  27 |
|  8 | Smartphones          |  13 |  20 |
|  9 | Android              |  14 |  15 |
| 10 | iOS                  |  16 |  17 |
| 11 | Other Smartphones    |  18 |  19 |
| 12 | Batteries            |  21 |  22 |
| 13 | Headsets             |  23 |  24 |
| 14 | Screen Protectors    |  25 |  26 |
+----+----------------------+-----+-----+
14 rows in set (0.05 sec)

不像以前鄰接表模型的例子，這個查詢語句無論樹的層次有多深都能很好的工做。在BETWEEN的子句中咱們沒有去關心node的rgt值，是由於使用node的rgt值得出的父節點老是和使用lft值得出的是相同的。網站

  • 檢索全部葉子節點

檢索出全部的葉子節點，使用嵌套集合模型的方法比鄰接表模型的LEFT JOIN方法簡單多了。若是你仔細得看了category表，你可能已經注意到葉子節點的左右值是連續的。要檢索出葉子節點，咱們只要查找知足 rgt=lft+1 的節點：this

SELECT id,title,lft,rgt
FROM category
WHERE rgt = lft + 1;
+----+-------------------+-----+-----+
| id | title             | lft | rgt |
+----+-------------------+-----+-----+
|  3 | Laptops           |   3 |   4 |
|  4 | PC                |   5 |   6 |
|  6 | Camera            |   9 |  10 |
|  9 | Android           |  14 |  15 |
| 10 | iOS               |  16 |  17 |
| 11 | Other Smartphones |  18 |  19 |
| 12 | Batteries         |  21 |  22 |
| 13 | Headsets          |  23 |  24 |
| 14 | Screen Protectors |  25 |  26 |
+----+-------------------+-----+-----+
9 rows in set (0.00 sec)

查詢

  • 檢索單一路徑

在嵌套集合模型中，咱們能夠不用多個自鏈接就能夠檢索出單一路徑：編碼

SELECT parent.id,parent.title,parent.lft,parent.rgt
FROM category AS node,
     category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.title = 'PC'
ORDER BY parent.lft;

+----+--------------+-----+-----+
| id | title        | lft | rgt |
+----+--------------+-----+-----+
|  1 | Electronics  |   1 |  28 |
|  2 | Laptops & PC |   2 |   7 |
|  4 | PC           |   5 |   6 |
+----+--------------+-----+-----+
3 rows in set (0.00 sec)
  • 檢索節點的深度

咱們已經知道怎樣去呈現一棵整樹，可是爲了更好的標識出節點在樹中所處層次，咱們怎樣才能檢索出節點在樹中的層級呢？咱們能夠在以前的查詢語句上增長COUNT函數和GROUP BY子句來實現：

SELECT node.title,(COUNT(parent.title) - 1) AS lev
FROM category AS node,
     category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.title
ORDER BY node.lft;

+----------------------+-----+
| title                | lev |
+----------------------+-----+
| Electronics          |   0 |
| Laptops & PC         |   1 |
| Laptops              |   2 |
| PC                   |   2 |
| Cameras & photo      |   1 |
| Camera               |   2 |
| Phones & Accessories |   1 |
| Smartphones          |   2 |
| Android              |   3 |
| iOS                  |   3 |
| Other Smartphones    |   3 |
| Batteries            |   2 |
| Headsets             |   2 |
| Screen Protectors    |   2 |
+----------------------+-----+
14 rows in set (0.01 sec)
若是當前MySQL版本是5.7或者以上可能會出現 1055 的報錯，下面是是解決辦法 
報錯：
ERROR 1055 (42000): Expression #1 of ORDER BY clause is not in GROUP BY clause and contains nonaggregated column 'test.node.lft' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by

緣由：In 5.7 the sqlmode is set by default to:
ONLY_FULL_GROUP_BY,NO_AUTO_CREATE_USER,STRICT_TRANS_TABLES,NO_ENGINE_SUBSTITUTION

解決：To remove the clause ONLY_FULL_GROUP_BY you can do this:
SET sql_mode=(SELECT REPLACE(@@sql_mode,'ONLY_FULL_GROUP_BY',''));

This supposed you need to make that GROUP BY with non aggregated columns.

咱們能夠根據 lev 值來縮進分類名字，使用 CONCAT 和 REPEAT 字符串函數:

SELECT CONCAT( REPEAT(' ', COUNT(parent.title) - 1), node.title) AS name,(COUNT(parent.title) - 1) AS lev
FROM category AS node,
     category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.title
ORDER BY node.lft;
+-----------------------+-----+
| name                  | lev |
+-----------------------+-----+
| Electronics           |   0 |
|  Laptops & PC         |   1 |
|   Laptops             |   2 |
|   PC                  |   2 |
|  Cameras & photo      |   1 |
|   Camera              |   2 |
|  Phones & Accessories |   1 |
|   Smartphones         |   2 |
|    Android            |   3 |
|    iOS                |   3 |
|    Other Smartphones  |   3 |
|   Batteries           |   2 |
|   Headsets            |   2 |
|   Screen Protectors   |   2 |
+-----------------------+-----+
14 rows in set (0.01 sec)
  • 檢索子樹的深度
SELECT node.title, (COUNT(parent.title) - (sub_tree.lev + 1)) AS lev
FROM category AS node,
    category AS parent,
    category AS sub_parent,
    (
        SELECT node.title, (COUNT(parent.title) - 1) AS lev
        FROM category AS node,
        category AS parent
        WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.title = 'Phones & Accessories'
        GROUP BY node.title
        ORDER BY node.lft
    ) AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
    AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
    AND sub_parent.title = sub_tree.title
GROUP BY node.title
ORDER BY node.lft;

這個查詢語句能夠檢索出任一節點子樹的深度值，包括根節點。這裏的深度值跟你指定的節點有關。

  • 檢索節點的直接子節點

能夠想象一下，你在零售網站上呈現電子產品的分類。當用戶點擊分類後，你將要呈現該分類下的產品，同時也需列出該分類下的直接子分類，而不是該分類下的所有分類。爲此，咱們只呈現該節點及其直接子節點，再也不呈現更深層次的節點。
要實現它很是的簡單，在先前的查詢語句上添加 HAVING 子句：

SELECT node.title, (COUNT(parent.title) - (sub_tree.lev + 1)) AS lev
FROM category AS node,
    category AS parent,
    category AS sub_parent,
    (
        SELECT node.title, (COUNT(parent.title) - 1) AS lev
        FROM category AS node,
        category AS parent
        WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.title = 'Phones & Accessories'
        GROUP BY node.title
        ORDER BY node.lft
    ) AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
    AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
    AND sub_parent.title = sub_tree.title
GROUP BY node.title
HAVING lev <= 1
ORDER BY node.lft;

若是你不但願呈現父節點，你能夠更改 HAVING lev <= 1 爲 HAVING lev = 1。

新增節點

  • 添加同一層次的節點

到如今，咱們已經知道了如何去查詢咱們的樹，是時候關注一下如何增長一個新節點來更新咱們的樹了。
當咱們想要在 Laptops & PC 和 Cameras & photo節點之間新增一個節點，新節點的 lft 和 rgt 的 值爲8和9，全部該節點的右邊節點的lft和rgt值都將加2，以後咱們再添加新節點並賦相應的lft和rgt值。我使用了鎖表（LOCK TABLES）語句來隔離查詢：

LOCK TABLE category WRITE;

SELECT @myRight := rgt FROM category WHERE title = 'Laptops & PC';

UPDATE category SET rgt = rgt + 2 WHERE rgt > @myRight;
UPDATE category SET lft = lft + 2 WHERE lft > @myRight;

INSERT INTO category(title, lft, rgt) VALUES('Game Consoles', @myRight + 1, @myRight + 2);

UNLOCK TABLES;

咱們能夠檢驗一下新節點插入的正確性：
SELECT CONCAT( REPEAT(' ', COUNT(parent.title) - 1), node.title) AS name,(COUNT(parent.title) - 1) AS lev
FROM category AS node,
     category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.title
ORDER BY node.lft;

+-----------------------+-----+
| name                  | lev |
+-----------------------+-----+
| Electronics           |   0 |
|  Laptops & PC         |   1 |
|   Laptops             |   2 |
|   PC                  |   2 |
|  Game Consoles        |   1 |
|  Cameras & photo      |   1 |
|   Camera              |   2 |
|  Phones & Accessories |   1 |
|   Smartphones         |   2 |
|    Android            |   3 |
|    iOS                |   3 |
|    Other Smartphones  |   3 |
|   Batteries           |   2 |
|   Headsets            |   2 |
|   Screen Protectors   |   2 |
+-----------------------+-----+
15 rows in set (0.00 sec)
  • 添加葉子節點

若是咱們想要在葉子節點下增長節點，咱們得稍微修改一下查詢語句。讓咱們在 Camera 葉子節點下添加 SLR 節點：

LOCK TABLE category WRITE;

SELECT @myLeft := lft FROM category WHERE title = 'Camera';

UPDATE category SET rgt = rgt + 2 WHERE rgt > @myLeft;
UPDATE category SET lft = lft + 2 WHERE lft > @myLeft;

INSERT INTO category(title, lft, rgt) VALUES('SLR', @myLeft + 1, @myLeft + 2);

UNLOCK TABLES;

刪除節點

最後刪除節點。刪除節點的處理過程跟節點在分層數據中所處的位置有關，刪除一個葉子節點比刪除一個子節點要簡單得多，由於刪除子節點的時候，咱們須要去處理孤立節點。

  • 刪除葉子節點

刪除一個葉子節點的過程正好是新增一個葉子節點的逆過程，咱們在刪除節點的同時該節點右邊全部節點的左右值和該父節點的右值都會減去該節點的寬度值：

LOCK TABLE category WRITE;


SELECT @myLeft := lft, @myRight := rgt, @myWidth := rgt - lft + 1 FROM category WHERE title = 'Game Consoles';


DELETE FROM category WHERE lft BETWEEN @myLeft AND @myRight;


UPDATE category SET rgt = rgt - @myWidth WHERE rgt > @myRight;
UPDATE category SET lft = lft - @myWidth WHERE lft > @myRight;

UNLOCK TABLES;
  • 刪除子節點以及整顆子樹
LOCK TABLE category WRITE;


SELECT @myLeft := lft, @myRight := rgt, @myWidth := rgt - lft + 1 FROM category WHERE title = 'Cameras & photo';


DELETE FROM category WHERE lft BETWEEN @myLeft AND @myRight;


UPDATE category SET rgt = rgt - @myWidth WHERE rgt > @myRight;
UPDATE category SET lft = lft - @myWidth WHERE lft > @myRight;

UNLOCK TABLES;
  • 刪除該節點，而不刪除該節點的子節點
LOCK TABLE category WRITE;


SELECT @myLeft := lft, @myRight := rgt, @myWidth := rgt - lft + 1 FROM category WHERE title = 'Cameras & photo';


DELETE FROM category WHERE lft = @myLeft;


UPDATE category SET rgt = rgt - 1, lft = lft - 1 WHERE lft BETWEEN @myLeft AND @myRight;
UPDATE category SET rgt = rgt - 2 WHERE rgt > @myRight;
UPDATE category SET lft = lft - 2 WHERE lft > @myRight;

UNLOCK TABLES;

在這個例子中，咱們對該節點全部右邊節點的左右值都減去了2（由於不考慮其子節點，該節點的寬度爲2），對該節點的子節點的左右值都減去了1（彌補因爲失去父節點的左值形成的裂縫）

參考資源

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