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Accepted
Grijesh Chauhan
Asked: 2012-12-07 07:36:41 +0800 CST

MySQL：树分层查询

  • 772

MySQL 中的一棵树中的子树

在我的 MYSQLDatabase COMPANY中，我有一个Table: Employee递归关联，一个员工可以是其他员工的老板。A self relationship of kind (SuperVisor (1)- SuperVisee (∞) ).

查询创建表： 

CREATE TABLE IF NOT EXISTS `Employee` (
  `SSN` varchar(64) NOT NULL,
  `Name` varchar(64) DEFAULT NULL,
  `Designation` varchar(128) NOT NULL,
  `MSSN` varchar(64) NOT NULL, 
  PRIMARY KEY (`SSN`),
  CONSTRAINT `FK_Manager_Employee`  
              FOREIGN KEY (`MSSN`) REFERENCES Employee(SSN)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

我插入了一组元组（查询）： 

INSERT INTO Employee VALUES 
 ("1", "A", "OWNER",  "1"),  

 ("2", "B", "BOSS",   "1"), # Employees under OWNER 
 ("3", "F", "BOSS",   "1"),

 ("4", "C", "BOSS",   "2"), # Employees under B
 ("5", "H", "BOSS",   "2"), 
 ("6", "L", "WORKER", "2"), 
 ("7", "I", "BOSS",   "2"), 
 # Remaining Leaf nodes   
 ("8", "K", "WORKER", "3"), # Employee under F     

 ("9", "J", "WORKER", "7"), # Employee under I     

 ("10","G", "WORKER", "5"), # Employee under H

 ("11","D", "WORKER", "4"), # Employee under C
 ("12","E", "WORKER", "4")

插入的行具有以下Tree-Hierarchical-Relationship： 

         A     <---ROOT-OWNER
        /|\             
       / A \        
      B     F 
    //| \    \          
   // |  \    K     
  / | |   \                     
 I  L H    C        
/     |   / \ 
J     G  D   E

我写了一个查询来查找关系： 

SELECT  SUPERVISOR.name AS SuperVisor, 
        GROUP_CONCAT(SUPERVISEE.name  ORDER BY SUPERVISEE.name ) AS SuperVisee, 
        COUNT(*)  
FROM Employee AS SUPERVISOR 
  INNER JOIN Employee SUPERVISEE ON  SUPERVISOR.SSN = SUPERVISEE.MSSN 
GROUP BY SuperVisor;

输出是： 

+------------+------------+----------+
| SuperVisor | SuperVisee | COUNT(*) |
+------------+------------+----------+
| A          | A,B,F      |        3 |
| B          | C,H,I,L    |        4 |
| C          | D,E        |        2 |
| F          | K          |        1 |
| H          | G          |        1 |
| I          | J          |        1 |
+------------+------------+----------+
6 rows in set (0.00 sec)

[问题]
而不是完整的层次树，我需要一个SUB-TREE点（选择性）例如：
如果输入参数是B那么输出应该如下......

+------------+------------+----------+
| SuperVisor | SuperVisee | COUNT(*) |
+------------+------------+----------+
| B          | C,H,I,L    |        4 |
| C          | D,E        |        2 |
| H          | G          |        1 |
| I          | J          |        1 |
+------------+------------+----------+

请帮助我。如果不查询，存储过程可能会有所帮助。
我试过了，但所有的努力都是徒劳的！

mysql stored-procedures

2 个回答

  1. Best Answer

    我已经使用存储过程解决了这种性质的问题：查找分层字段的最高级别：有与没有 CTE（2011 年 10 月 24 日）

    如果您查看我的帖子，您可以使用 GetAncestry 和 GetFamilyTree 函数作为从任何给定点遍历树的模型。

    更新 2012-12-11 12:11 EDT

    我从帖子中回顾了我的代码。我为你写了存储函数：

    DELIMITER $$

DROP FUNCTION IF EXISTS `cte_test`.`GetFamilyTree` $$
CREATE FUNCTION `cte_test`.`GetFamilyTree`(GivenName varchar(64))
RETURNS varchar(1024) CHARSET latin1
DETERMINISTIC
BEGIN

    DECLARE rv,q,queue,queue_children,queue_names VARCHAR(1024);
    DECLARE queue_length,pos INT;
    DECLARE GivenSSN,front_ssn VARCHAR(64);

    SET rv = '';

    SELECT SSN INTO GivenSSN
    FROM Employee
    WHERE name = GivenName
    AND Designation <> 'OWNER';
    IF ISNULL(GivenSSN) THEN
        RETURN ev;
    END IF;

    SET queue = GivenSSN;
    SET queue_length = 1;

    WHILE queue_length > 0 DO
        IF queue_length = 1 THEN
            SET front_ssn = queue;
            SET queue = '';
        ELSE
            SET pos = LOCATE(',',queue);
            SET front_ssn = LEFT(queue,pos - 1);
            SET q = SUBSTR(queue,pos + 1);
            SET queue = q;
        END IF;
        SET queue_length = queue_length - 1;
        SELECT IFNULL(qc,'') INTO queue_children
        FROM
        (
            SELECT GROUP_CONCAT(SSN) qc FROM Employee
            WHERE MSSN = front_ssn AND Designation <> 'OWNER'
        ) A;
        SELECT IFNULL(qc,'') INTO queue_names
        FROM
        (
            SELECT GROUP_CONCAT(name) qc FROM Employee
            WHERE MSSN = front_ssn AND Designation <> 'OWNER'
        ) A;
        IF LENGTH(queue_children) = 0 THEN
            IF LENGTH(queue) = 0 THEN
                SET queue_length = 0;
            END IF;
        ELSE
            IF LENGTH(rv) = 0 THEN
                SET rv = queue_names;
            ELSE
                SET rv = CONCAT(rv,',',queue_names);
            END IF;
            IF LENGTH(queue) = 0 THEN
                SET queue = queue_children;
            ELSE
                SET queue = CONCAT(queue,',',queue_children);
            END IF;
            SET queue_length = LENGTH(queue) - LENGTH(REPLACE(queue,',','')) + 1;
        END IF;
    END WHILE;

    RETURN rv;

END $$

    它确实有效。这是一个示例：

    mysql> SELECT name,GetFamilyTree(name) FamilyTree
    -> FROM Employee WHERE Designation <> 'OWNER';
+------+-----------------------+
| name | FamilyTree            |
+------+-----------------------+
| A    | B,F,C,H,L,I,K,D,E,G,J |
| G    |                       |
| D    |                       |
| E    |                       |
| B    | C,H,L,I,D,E,G,J       |
| F    | K                     |
| C    | D,E                   |
| H    | G                     |
| L    |                       |
| I    | J                     |
| K    |                       |
| J    |                       |
+------+-----------------------+
12 rows in set (0.36 sec)

mysql>

    只有一个捕获。我为所有者添加了额外的一行

    • 所有者的 SSN 为 0
    • 所有者是他自己的老板，MSSN 为 0

    这是数据

    mysql> select * from Employee;
+-----+------+-------------+------+
| SSN | Name | Designation | MSSN |
+-----+------+-------------+------+
| 0   | A    | OWNER       | 0    |
| 1   | A    | BOSS        | 0    |
| 10  | G    | WORKER      | 5    |
| 11  | D    | WORKER      | 4    |
| 12  | E    | WORKER      | 4    |
| 2   | B    | BOSS        | 1    |
| 3   | F    | BOSS        | 1    |
| 4   | C    | BOSS        | 2    |
| 5   | H    | BOSS        | 2    |
| 6   | L    | WORKER      | 2    |
| 7   | I    | BOSS        | 2    |
| 8   | K    | WORKER      | 3    |
| 9   | J    | WORKER      | 7    |
+-----+------+-------------+------+
13 rows in set (0.00 sec)

mysql>
    • 5

  2. 您使用的称为Adjacency List Model。它有很多限制。当您想在特定位置删除/插入节点时，您会遇到问题。最好使用嵌套集模型。

    有详细的解释。不幸的是，mysql.com 上的文章不再存在。

    • 3

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