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像这样转换 MySQL 表的最干净的方法是什么:

     id | fullindi                              | parent | rank
---------------------------------------------------------------
      1 | LHUILLIER Pierre (ca 1700 - 1745)     |    0   |   0
      9 | LHUILLIER Claude (ca 1729 - 1806)     |    1   |   1
  10357 | LHUILLIER Joseph (ca 1730 - 1738)     |    1   |   2
      7 | LHUILLIER François (ca 1731 - 1794)   |    1   |   3
      3 | LHUILLIER Antoine (1736 - av. 1797)   |    1   |   4
      4 | LHUILLIER Anne Marie (1737 - ____)    |    1   |   5
   4903 | LHUILLIER Dominique (1740 - ____)     |    1   |   6
      5 | LHUILLIER Thérèse (1741 - ____)       |    1   |   7
      8 | LHUILLIER Augustin (ca 1743 - ____)   |    1   |   8
      6 | LHUILLIER Joseph (1745 - ap. 1804)    |    1   |   9
    322 | LHUILLIER N... (1749 - ____)          |    9   |   1
    323 | LHUILLIER Marianne (1751 - ____)      |    9   |   2
    324 | LHUILLIER François (1752 - ____)      |    9   |   3
    325 | LHUILLIER Augustin (1754 - av. 1810)  |    9   |   4
    326 | LHUILLIER Léopold (1757 - av. 1819)   |    9   |   5
    327 | LHUILLIER Nicolas (1758 - ____)       |    9   |   6
    328 | LHUILLIER N... (1760 - ____)          |    9   |   7
    329 | LHUILLIER Claude (1765 - ____)        |    9   |   8
   4643 | LHUILLIER Jean Baptiste (1766 - 1836) |    9   |   9
    331 | LHUILLIER Marie Jeanne (1767 - 1823)  |    9   |  10
   etc

到这样的嵌套表:

     id | fullindi                              | posleft | posright
--------------------------------------------------------------------
      1 | LHUILLIER Pierre (ca 1700 - 1745)     |    0    |   848
      9 | LHUILLIER Claude (ca 1729 - 1806)     |    1    |   1
    322 | LHUILLIER N... (1749 - ____)          |    2    |   3
    323 | LHUILLIER Marianne (1751 - ____)      |    4    |   5
    324 | LHUILLIER François (1752 - ____)      |    6    |   7
    325 | LHUILLIER Augustin (1754 - av. 1810)  |    8    |   9
   etc

我确切地说它需要独立于深度(最大 = 20 个级别)和项目数量(超过 1.000 个项目)。

任何帮助将不胜感激。

此致。

4

1 回答 1

1

有一个以前的问题Here

如果有人在 php 中执行此操作,您可能可以从中获取逻辑来获得您需要的解决方案。

我在一个糟糕的网站上发现了这个,并且 SQL 都在一行中,所以它需要一些格式。我几乎原样留下了这个示例,所有的功劳都应该归功于出色的 Joe Celko,他多年来一直在写关于 sql 的文章。

     CREATE TABLE Tree (
    child CHAR(10) NOT NULL, 
    parent CHAR(10), 
    CONSTRAINT PK_Tree PRIMARY KEY CLUSTERED(child))

     -- insert the sample data for testing 

     INSERT INTO Tree(child,parent) VALUES ('Albert', NULL)
     INSERT INTO Tree(child,parent) VALUES ('Bert', 'Albert') 
     INSERT INTO Tree(child,parent) VALUES ('Chuck', 'Albert') 
     INSERT INTO Tree(child,parent) VALUES ('Donna', 'Chuck') 
     INSERT INTO Tree(child,parent) VALUES ('Eddie', 'Chuck') 
     INSERT INTO Tree(child,parent) VALUES ('Fred', 'Chuck') 


CREATE TABLE Stack (
    StackID int IDENTITY(1,1),
    stack_top INTEGER NOT NULL, 
    child VARCHAR(10) NOT NULL, 
    lft INTEGER NOT NULL, 
    rgt INTEGER, 
    CONSTRAINT PK_Stack PRIMARY KEY CLUSTERED(StackID))


    DECLARE @lft_rgt INTEGER, @stack_pointer INTEGER, @max_lft_rgt INTEGER

    SET @max_lft_rgt = 2 * (SELECT COUNT(*) FROM Tree)

    INSERT INTO Stack 
    SELECT 1, child, 1, @max_lft_rgt 
    FROM Tree 
    WHERE parent IS NULL

    SET @lft_rgt = 2

    SET @Stack_pointer = 1

    DELETE FROM Tree WHERE parent IS NULL

    -- The Stack is now loaded and ready to use 

    WHILE (@lft_rgt < @max_lft_rgt) 
        BEGIN 
            IF EXISTS (SELECT * FROM Stack AS S1, Tree AS T1 WHERE S1.child = T1.parent AND S1.stack_top = @stack_pointer) 
                BEGIN 
                    -- push when stack_top has subordinates and set lft value 
                    INSERT INTO Stack 
                    SELECT (@stack_pointer + 1), 
                    MIN(T1.child), 
                    @lft_rgt, 
                    NULL 
                    FROM Stack AS S1, 
                    Tree AS T1 
                    WHERE S1.child = T1.parent AND S1.stack_top = @stack_pointer

                     -- remove this row from Tree 
                     DELETE FROM Tree 
                     WHERE child = (SELECT child FROM Stack WHERE stack_top = @stack_pointer + 1)

                     SET @stack_pointer = @stack_pointer + 1 
                END 
        -- push 
        ELSE 
            BEGIN 
                -- pop the Stack and set rgt value 
                UPDATE Stack SET rgt = @lft_rgt, stack_top = -stack_top 
                WHERE stack_top = @stack_pointer 

                SET @stack_pointer = @stack_pointer - 1
            END

            -- pop 
        SET @lft_rgt = @lft_rgt + 1
    END

您应该能够通过更改列名等来使用它来整理您的列表。

再一次,这不是我的工作,再次感谢 Joe Celko(我很长时间以来一直是嵌套集合模型的粉丝,并且有一些生产系统正在使用它)。如果有,我一直找不到 Joe 的博客(如果你在外面,请在此处发表评论并获得所有功劳。

于 2011-05-15T15:27:58.803 回答