我需要一个MySQL查询，将一个节点及其所有子节点移动到嵌套集中。我找到了this网站，但是这个功能看似不合逻辑 - 嵌套集模型中没有universeid或treeid，代码本身只比感觉要求的长。我在表中唯一的额外列是parent。
我不能再删除并添加节点，因为它会丢失其ID。
答案 0 :(得分：46)
我知道，这个话题已经很老了，但无论如何它仍然没有答案。我从谷歌来到这里，并没有找到这个问题的直接答案。
因此，经过一番研究后，我找到了相当简单的解决方案。
一切，我们需要移动我们的节点是：节点左右位置，新父节点右位置。然后，可以通过四个简单的步骤移动到新位置的节点：
这就是理论，现在 - MySQL中的这种算法实现（使用PHP的例子）：
-- step 0: Initialize parameters.
SELECT
@node_id := 1, --put there id of moving node
@node_pos_left := 0, --put there left position of moving node
@node_pos_right := 1, --put there right position of moving node
@parent_id := 2, --put there id of new parent node (there moving node should be moved)
@parent_pos_right := 4; --put there right position of new parent node (there moving node should be moved)
SELECT
@node_size := @node_pos_right - @node_pos_left + 1; -- 'size' of moving node (including all it's sub nodes)
-- step 1: temporary "remove" moving node
UPDATE `list_items`
SET `pos_left` = 0-(`pos_left`), `pos_right` = 0-(`pos_right`)
WHERE `pos_left` >= @node_pos_left AND `pos_right` <= @node_pos_right;
-- step 2: decrease left and/or right position values of currently 'lower' items (and parents)
UPDATE `list_items`
SET `pos_left` = `pos_left` - @node_size
WHERE `pos_left` > @node_pos_right;
UPDATE `list_items`
SET `pos_right` = `pos_right` - @node_size
WHERE `pos_right` > @node_pos_right;
-- step 3: increase left and/or right position values of future 'lower' items (and parents)
UPDATE `list_items`
SET `pos_left` = `pos_left` + @node_size
WHERE `pos_left` >= IF(@parent_pos_right > @node_pos_right, @parent_pos_right - @node_size, @parent_pos_right);
UPDATE `list_items`
SET `pos_right` = `pos_right` + @node_size
WHERE `pos_right` >= IF(@parent_pos_right > @node_pos_right, @parent_pos_right - @node_size, @parent_pos_right);
-- step 4: move node (ant it's subnodes) and update it's parent item id
UPDATE `list_items`
SET
`pos_left` = 0-(`pos_left`)+IF(@parent_pos_right > @node_pos_right, @parent_pos_right - @node_pos_right - 1, @parent_pos_right - @node_pos_right - 1 + @node_size),
`pos_right` = 0-(`pos_right`)+IF(@parent_pos_right > @node_pos_right, @parent_pos_right - @node_pos_right - 1, @parent_pos_right - @node_pos_right - 1 + @node_size)
WHERE `pos_left` <= 0-@node_pos_left AND `pos_right` >= 0-@node_pos_right;
UPDATE `list_items`
SET `parent_item_id` = @parent_id
WHERE `item_id` = @node_id;
请注意 - SQL代码中可能仍存在一些语法错误，因为我实际上在PHP中使用此算法，如下所示：
$iItemId = 1;
$iItemPosLeft = 0;
$iItemPosRight = 1;
$iParentId = 2;
$iParentPosRight = 4;
$iSize = $iPosRight - $iPosLeft + 1;
$sql = array(
// step 1: temporary "remove" moving node
'UPDATE `list_items`
SET `pos_left` = 0-(`pos_left`), `pos_right` = 0-(`pos_right`)
WHERE `pos_left` >= "'.$iItemPosLeft.'" AND `pos_right` <= "'.$iItemPosRight.'"',
// step 2: decrease left and/or right position values of currently 'lower' items (and parents)
'UPDATE `list_items`
SET `pos_left` = `pos_left` - '.$iSize.'
WHERE `pos_left` > "'.$iItemPosRight.'"',
'UPDATE `list_items`
SET `pos_right` = `pos_right` - '.$iSize.'
WHERE `pos_right` > "'.$iItemPosRight.'"',
// step 3: increase left and/or right position values of future 'lower' items (and parents)
'UPDATE `list_items`
SET `pos_left` = `pos_left` + '.$iSize.'
WHERE `pos_left` >= "'.($iParentPosRight > $iItemPosRight ? $iParentPosRight - $iSize : $iParentPosRight).'"',
'UPDATE `list_items`
SET `pos_right` = `pos_right` + '.$iSize.'
WHERE `pos_right` >= "'.($iParentPosRight > $iItemPosRight ? $iParentPosRight - $iSize : $iParentPosRight).'"',
// step 4: move node (ant it's subnodes) and update it's parent item id
'UPDATE `list_items`
SET
`pos_left` = 0-(`pos_left`)+'.($iParentPosRight > $iItemPosRight ? $iParentPosRight - $iItemPosRight - 1 : $iParentPosRight - $iItemPosRight - 1 + $iSize).',
`pos_right` = 0-(`pos_right`)+'.($iParentPosRight > $iItemPosRight ? $iParentPosRight - $iItemPosRight - 1 : $iParentPosRight - $iItemPosRight - 1 + $iSize).'
WHERE `pos_left` <= "'.(0-$iItemPosLeft).'" AND i.`pos_right` >= "'.(0-$iItemPosRight).'"',
'UPDATE `list_items`
SET `parent_item_id` = "'.$iParentItemId.'"
WHERE `item_id`="'.$iItemId.'"'
);
foreach($sql as $sqlQuery){
mysql_query($sqlQuery);
}
请注意，该代码可能已经过优化，但为了更好的可读性，我将保留这些代码。如果在多用户系统中使用嵌套集，也请考虑表锁定。
希望我的信息能帮助任何人，他会在我之后寻找解决方案。任何评论和更正也是受欢迎的。
答案 1 :(得分：13)
这是一个解决方案，允许您将节点移动到树中的任何位置，作为兄弟或只有一个输入参数的子节点 - 节点的新左侧位置（newlpos）。
基本上有三个步骤：
在psuedo-sql中，它看起来像这样：
//
* -- create new space for subtree
* UPDATE tags SET lpos = lpos + :width WHERE lpos >= :newlpos
* UPDATE tags SET rpos = rpos + :width WHERE rpos >= :newlpos
*
* -- move subtree into new space
* UPDATE tags SET lpos = lpos + :distance, rpos = rpos + :distance
* WHERE lpos >= :tmppos AND rpos < :tmppos + :width
*
* -- remove old space vacated by subtree
* UPDATE tags SET lpos = lpos - :width WHERE lpos > :oldrpos
* UPDATE tags SET rpos = rpos - :width WHERE rpos > :oldrpos
*/
：distance变量是新旧位置之间的距离，：width是子树的大小，并且：tmppos用于跟踪在更新期间移动的子树。这些变量定义为：
// calculate position adjustment variables
int width = node.getRpos() - node.getLpos() + 1;
int distance = newlpos - node.getLpos();
int tmppos = node.getLpos();
// backwards movement must account for new space
if (distance < 0) {
distance -= width;
tmppos += width;
}
有关完整的代码示例，请参阅我的博客
http://www.ninthavenue.com.au/how-to-move-a-node-in-nested-sets-with-sql
如果你喜欢这个解决方案，请进行投票。
答案 2 :(得分：2)
我知道这是一个老问题，但我只是自己使用了答案，但对于SQL Server。如果有人想要它，这里是基于接受的答案的SQL Server存储过程的代码。
CREATE PROCEDURE [dbo].[Item_Move]
@id uniqueidentifier,
@destinationId uniqueidentifier
AS
BEGIN
SET NOCOUNT ON;
declare @moverLeft int,
@moverRight int,
@destinationRight int,
@node_size int
-- step 0: Initialize parameters.
SELECT
@moverLeft = leftExtent,
@moverRight = rightExtent
FROM
Item
WHERE
id = @id
SELECT
@destinationRight = rightExtent
FROM
Item
WHERE
id = @destinationId
SELECT
@node_size = @moverRight - @moverLeft + 1; -- 'size' of moving node (including all it's sub nodes)
-- step 1: temporary "remove" moving node
UPDATE Item
SET leftExtent = 0-(leftExtent), rightExtent = 0-(rightExtent), updatedDate = GETDATE()
WHERE leftExtent >= @moverLeft AND rightExtent <= @moverRight;
-- step 2: decrease left and/or right position values of currently 'lower' items (and parents)
UPDATE Item
SET leftExtent = leftExtent - @node_size, updatedDate = GETDATE()
WHERE leftExtent > @moverRight;
UPDATE Item
SET rightExtent = rightExtent - @node_size, updatedDate = GETDATE()
WHERE rightExtent > @moverRight;
-- step 3: increase left and/or right position values of future 'lower' items (and parents)
UPDATE Item
SET leftExtent = leftExtent + @node_size, updatedDate = GETDATE()
WHERE leftExtent >= CASE WHEN @destinationRight > @moverRight THEN @destinationRight - @node_size ELSE @destinationRight END;
UPDATE Item
SET rightExtent = rightExtent + @node_size, updatedDate = GETDATE()
WHERE rightExtent >= CASE WHEN @destinationRight > @moverRight THEN @destinationRight - @node_size ELSE @destinationRight END;
-- step 4: move node (and it's subnodes) and update it's parent item id
UPDATE Item
SET
leftExtent = 0-(leftExtent) + CASE WHEN @destinationRight > @moverRight THEN @destinationRight - @moverRight - 1 ELSE @destinationRight - @moverRight - 1 + @node_size END,
rightExtent = 0-(rightExtent) + CASE WHEN @destinationRight > @moverRight THEN @destinationRight - @moverRight - 1 ELSE @destinationRight - @moverRight - 1 + @node_size END,
updatedDate = GETDATE()
WHERE leftExtent <= 0-@moverLeft AND rightExtent >= 0-@moverRight;
UPDATE Item
SET parentId = @destinationId, updatedDate = GETDATE()
WHERE id = @id;
END
答案 3 :(得分：1)
在嵌套集设计中，移动子树非常昂贵且复杂。
您应该考虑使用不同的设计来表示树木。
例如，如果使用Path Enumeration设计，则将每个节点的直接祖先列表存储为连接字符串。
id path
1 1/
2 1/2/
3 1/3/
4 1/3/4/
5 1/3/5/
然后移动子树（比如节点3移动到节点2的子节点）：
UPDATE Tree t
JOIN Tree node2 ON (node2.id = 2)
JOIN Tree node3 ON (node3.id = 3)
SET t.path = CONCAT(node2.path, REPLACE(t.path, node3.path, node2.path))
WHERE t.path LIKE CONCAT(node3.path, '%');
答案 4 :(得分：0)
请参阅我博客中有关在MySQL中存储和使用分层数据的文章：
要在这样的表中移动整个分支，您只需要更新根parent（单行）
您需要创建一个功能：
CREATE FUNCTION hierarchy_connect_by_parent_eq_prior_id(value INT) RETURNS INT
NOT DETERMINISTIC
READS SQL DATA
BEGIN
DECLARE _id INT;
DECLARE _parent INT;
DECLARE _next INT;
DECLARE CONTINUE HANDLER FOR NOT FOUND SET @id = NULL;
SET _parent = @id;
SET _id = -1;
IF @id IS NULL THEN
RETURN NULL;
END IF;
LOOP
SELECT MIN(id)
INTO @id
FROM t_hierarchy
WHERE parent = _parent
AND id > _id;
IF @id IS NOT NULL OR _parent = @start_with THEN
SET @level = @level + 1;
RETURN @id;
END IF;
SET @level := @level - 1;
SELECT id, parent
INTO _id, _parent
FROM t_hierarchy
WHERE id = _parent;
END LOOP;
END
并在查询中使用它：
SELECT CONCAT(REPEAT(' ', level - 1), CAST(hi.id AS CHAR)) AS treeitem, parent, level
FROM (
SELECT hierarchy_connect_by_parent_eq_prior_id(id) AS id, @level AS level
FROM (
SELECT @start_with := 0,
@id := @start_with,
@level := 0
) vars, t_hierarchy
WHERE @id IS NOT NULL
) ho
JOIN t_hierarchy hi
ON hi.id = ho.id
答案 5 :(得分：0)
我有一个存储过程将嵌套集中的节点移动到新的父节点。我在名为“somedb”的MySQL / InnoDB数据库中使用名为“category”的表。当然，如果目标是要移动的类别的子类别，则此过程会使事情搞砸，因此请确保您不要尝试在自身内部嵌入节点。我将把它作为练习留给读者，使这个程序安全。
CREATE PROCEDURE `somedb`.`moveCatParent` (IN cat_a VARCHAR(45), IN cat_b VARCHAR(45))
BEGIN
START TRANSACTION;
/* cat_b.lft + 1 is the destination. */
SELECT @destination := (lft + 1)
FROM category
WHERE name = cat_b;
SELECT @cat_a_width := ((rgt - lft) + 1)
FROM category
WHERE name = cat_a;
/* Rip this table a new cat_a sized hole inside cat_b. */
UPDATE category SET rgt = rgt + @cat_a_width WHERE rgt >= @destination;
UPDATE category SET lft = lft + @cat_a_width WHERE lft >= @destination;
SELECT @cat_a_lft := lft, @cat_a_rgt := rgt
FROM category
WHERE name = cat_a;
SELECT @diff := @destination - @cat_a_lft;
/* Move cat_a and all inhabitants to new hole */
UPDATE category SET rgt = rgt + @diff WHERE rgt BETWEEN @cat_a_lft AND @cat_a_rgt;
UPDATE category SET lft = lft + @diff WHERE lft BETWEEN @cat_a_lft AND @cat_a_rgt;
/* Close the gap created when we moved cat_a. */
UPDATE category SET rgt = rgt - @cat_a_width WHERE rgt >= @cat_a_lft;
UPDATE category SET lft = lft - @cat_a_width WHERE lft >= @cat_a_lft;
COMMIT;
END
答案 6 :(得分：0)
我相信通过两个额外的列来存储原始节点的左右值（以及所有后续子节点），可以简化算法。我用铅笔和纸做过例子，如果你发现算法上有任何漏洞，请告诉我。
目标节点（您正在移动的Node的新父节点）是tNode。目标节点的左值是tNode.L，右值是tNode.R。类似地，您正在移动的节点是mNode，mNode的左右值是mNode.L和mNode.R。两个额外的列是mNode.SL和mNode.SR
总而言之，我们有4列用于操纵R，L，SL和SR
计算
delta1 = (mNode.R - mNode.L) + 1
将mNode原始L和R保存到SL和SR列
- For All L between mNode.L and mNode.R
mNode.SL = mNode.L ; mNode.L = 0 ;
- For All R between mNode.L and mNode.R
mNode.SR = mNode.R ; mNode.R = 0 ;
Do For all Nodes
IF L > mNode.SR
L = L + delta1
IF R > mNode.SR
R = R + delta1
现在mNode与树分离，并且在没有mNode的情况下调整树。
计算
delta2 = (tNode.R - mNode.SL)
Do for all Nodes
IF L >= tNode.R
L = L + delta1
IF R >= tNode.R
R = R + delta1
现在我们已经调整了树（和目标节点）以接受已删除的节点数。
在tNode处附加mNode并重置SL / SR列值
Do for all Nodes
IF SL between mNode.SL and mNode.SR
L = mNode.SL + delta2 ; mNode.SL = 0 ;
IF SR between mNode.SL and mNode.SR
R = mNode.SR + delta2 ; mNode.SR = 0 ;
完成所有这些步骤后，我们应该将mNode移到tNode下。
答案 7 :(得分：0)
感谢将lft和rgt转换为负面对手的想法。 我在这里发布了一个更通用的方法：Move node in Nested Sets tree。
queryBatch（）函数将查询包含在事务中。
答案 8 :(得分：0)
这很简单，首先定义一个存储过程：
CREATE DEFINER=`root`@`localhost` PROCEDURE `move_item`(
IN itemId BIGINT, IN kind SMALLINT,
IN newSiblingId BIGINT UNSIGNED, IN newSiblingKind SMALLINT, IN newParentId BIGINT UNSIGNED,
IN jobId BIGINT UNSIGNED, IN companyId BIGINT UNSIGNED,
OUT outSucess SMALLINT UNSIGNED)
proc_label:BEGIN
接下来我们需要一些局部变量：
DECLARE oldLeft, oldRight, newLeft, newRight, itemWidth, moveBy INT UNSIGNED DEFAULT 0;
set outSucess =0;
现在让我们的老左派＆amp;正确并获得宽度
SELECT `LFT`, `RGT` into oldLeft, oldRight from `nodes` where `ID`=itemId LIMIT 1;
SET itemWidth = oldRight - oldLeft + 1;
现在乘以-1“
将它们”从树中取出“UPDATE `nodes` SET `RGT`=`RGT`* -1, `LFT`=`LFT`* -1 WHERE ``LFT` BETWEEN oldLeft and oldRight;
下一部分没有必要，因为没有它就可以使用树，但它很整洁;缩小差距：
-- Update right
UPDATE `nodes` SET `RGT` = `RGT` - itemWidth WHERE `RGT` > oldRight;
-- Update left
UPDATE `nodes` SET `LFT` = `LFT` - itemWidth WHERE `LFT` > oldRight;
现在找到新位置：
SELECT (`RGT`+1) into newLeft from `nodes` where `ID`=newSiblingId LIMIT 1;
-- No sibling, so make it last in parent
IF (newLeft = 0) AND (newParentId != 0) THEN
SELECT `RGT` into newLeft from `nodes` WHERE `ID`=newParentId LIMIT 1;
END IF;
-- If no previous sibling or parent, set to first item in tree
IF (newLeft=0) OR (newLeft=NULL) THEN SET newLeft=1;
END IF;
现在腾出一些空间：
-- Update right
UPDATE `nodes` SET `RGT` = `RGT` + itemWidth WHERE `RGT` >= newLeft;
-- Update left
UPDATE `nodes` SET `LFT` = `LFT` + itemWidth WHERE `LFT` >= newLeft;
最后将从树中移出的节点移回* -1，当你在它时，将它们移动到正确的位置：
SET moveBy = OldLeft - NewLeft;
UPDATE `nodes` SET `RGT`=(`RGT`* -1)-moveBy, `LFT`=(`LFT`* -1)-moveBy WHERE `LFT` < 0;
set outSucess =1;
未经过测试，粘贴，调整和简化工作程序。
答案 9 :(得分：0)
我知道这篇文章已经过时了但是我发布了这个解决方案，每个人都可以在这里找到解决方案。我发现这个@ sedna-soft.de。我测试了id并且工作得很好
-- moves a subtree before the specified position
-- if the position is the rgt of a node, the subtree will be its last child
-- if the position is the lft of a node, the subtree will be inserted before
-- @param l the lft of the subtree to move
-- @param r the rgt of the subtree to move
-- @param p the position to move the subtree before
SET @r: , @l: , @p:
update tree
set
lft = lft + if (@p > @r,
if (@r < lft and lft < @p,
@l - @r - 1,
if (@l <= lft and lft < @r,
@p - @r - 1,
0
)
),
if (@p <= lft and lft < @l,
@r - @l + 1,
if (@l <= lft and lft < @r,
@p - @l,
0
)
)
),
rgt = rgt + if (@p > @r,
if (@r < rgt and rgt < @p,
@l - @r - 1,
if (@l < rgt and rgt <= @r,
@p - @r - 1,
0
)
),
if (@p <= rgt and rgt < @l,
@r - @l + 1,
if (@l < rgt and rgt <= @r,
@p - @l,
0
)
)
)
where @r < @p or @p < @l;
答案 10 :(得分：0)
# Get Left and Right offsets of both source node (Drag Node) and target node (Drop off Node).
SELECT lft,rgt INTO @sourceNodeLft,@sourceNodeRgt FROM treetest WHERE id=_sourceNodeId;
SELECT lft,rgt INTO @targetNodeLft,@targetNodeRgt FROM treetest WHERE id=_targetNodeId;
# Determine node order direction
SET @direction := IF(@targetNodeLft<@sourceNodeLft,'UP','DOWN');
# Determine with of source node (Drag Node)
SET @width := @sourceNodeRgt - @sourceNodeLft + 1;
# Mark all displaced nodes with negative lft and rgt
UPDATE treetest SET lft = 0-lft, rgt = 0-rgt
WHERE lft >= @targetNodeLft AND rgt <= @targetNodeRgt;
UPDATE treetest SET lft = 0-lft, rgt = 0-rgt
WHERE lft >= @sourceNodeLft AND rgt <= @sourceNodeRgt;
# Update left and right offsets of inner nodes between source (Drag Node)and target (Drop off) node
UPDATE treetest SET lft = (lft + IF(@direction = 'UP',@width,0-@width)),
rgt = (rgt + IF(@direction = 'UP',@width,0-@width))
WHERE lft > IF(@direction = 'UP', @targetNodeLft, @sourceNodeLft)
AND rgt < IF(@direction = 'UP', @sourceNodeLft, @targetNodeLft);
# Update source (Drag) Node and its children offsets
SET @sourceOffset := IF(@direction = 'UP',@targetNodeLft - @sourceNodeLft, @targetNodeRgt - @width - @sourceNodeLft+1);
UPDATE treetest SET lft = 0 - lft + @sourceOffset,
rgt = 0 - rgt + @sourceOffset
WHERE (0-lft) >= @sourceNodeLft AND (0-rgt) <= @sourceNodeRgt;
# Update target (Drop off) node and its children offsets
SET @targetOffset := IF(@direction = 'UP', 0 - @width,@width);
UPDATE treetest SET lft = 0 - (lft + @targetOffset),
rgt = 0 - (rgt + @targetOffset)
WHERE (0-lft) >= @targetNodeLft AND (0-rgt) <= @targetNodeRgt;
答案 11 :(得分：0)
已经有很多答案，但是我觉得我的方法对某人可能有用。 基于Roger Keays的回答（非常感谢！），我为mySQL数据库编写了存储过程：
-- to move target before specified node
CREATE DEFINER=`root`@`%` PROCEDURE `move_before`(IN target_id int, before_id int)
BEGIN
SELECT @new_pos := lft FROM dirs WHERE id = before_id;
CALL move(target_id, @new_pos);
END
-- to move target after specified node
CREATE DEFINER=`root`@`%` PROCEDURE `move_after`(IN target_id int, after_id int)
BEGIN
SELECT @new_pos := rgt + 1 FROM dirs WHERE id = after_id;
CALL move(target_id, @new_pos);
END
-- to move target to the specified node
CREATE DEFINER=`root`@`%` PROCEDURE `move_in`(IN target_id int, parent_id int)
BEGIN
SELECT @new_pos := rgt FROM dirs WHERE id = parent_id;
CALL move(target_id, @new_pos);
END
--main procedure to move target before position
CREATE DEFINER=`root`@`%` PROCEDURE `move`(in target_id int, in new_pos int)
BEGIN
SELECT @oldlft := lft, @oldrgt := rgt
FROM dirs
WHERE target_id = id;
SET @width := @oldrgt - @oldlft +1;
SET @distance := new_pos - @oldlft;
SET @tmppos := @oldlft;
IF (@distance <0)
THEN
SELECT @distance := @distance - @width;
SELECT @tmppos := @tmppos + @width;
END IF;
-- create new space for subtree
UPDATE dirs SET lft = lft + @width WHERE lft >= new_pos;
UPDATE dirs SET rgt = rgt + @width WHERE rgt >= new_pos;
-- move subtree into new space
UPDATE dirs SET lft = lft + @distance, rgt = rgt + @distance
WHERE lft >= @tmppos AND rgt < @tmppos + @width;
-- remove old space vacated by subtree
UPDATE dirs SET lft = lft - @width WHERE lft > @oldrgt;
UPDATE dirs SET rgt = rgt - @width WHERE rgt > @oldrgt;
END
答案 12 :(得分：-1)
$ row是一个数组，表示我必须移动的行; 它一定是这样的：
Array ( [lft] => 5 [rgt] => 10 [width] => 6 )
$ row2是一个代表命运节点的数组;
Array ( [id] => 5 [lft] => 2 [rgt] => 17 [width] => 16 )
...
mysql_query("UPDATE entryCategory SET rgt = rgt + %d - %d, lft = lft + %d - %d WHERE rgt <= %d and lft >= %d;",$row2["rgt"],$row["lft"],$row2["rgt"],$row["lft"],$row["rgt"],$row["lft"]);
mysql_query("UPDATE entryCategory SET rgt = rgt + %d WHERE id=%d;",$row["width"],$row2["id"]);
mysql_query("UPDATE entryCategory SET rgt = rgt - %d, lft = lft - %d WHERE rgt > %d and lft > %d;",$row["width"],$row["width"],$row["rgt"],$row["rgt"]);