计算按行分组的左右总子节点数
我正在开展一个基于推荐（MLM）奖励人们的项目
我已经能够计算左侧和右侧的子节点总数，但现在我需要能够更新用户的等级，当他们有一定数量的用户时，两侧的某些等级低于他们。 （我将在下面更好地解释：
用户表
id | name | parentID| side | rank |
4 | Dan | | | starter|
5 | Chris | 4 | left | starter|
6 | James | 4 | right| starter|
7 | Tybe | 5 | left | starter|
8 | Rose | 5 | right| starter|
9 | Paul | 6 | left | starter|
10 | Zach | 6 | right| starter|
树状表
userID | left | right| leftCount | rightCount|
4 | 5 | 6 | 3 | 3 |
5 | 7 | 8 | 1 | 1 |
6 | 9 | 10 | 1 | 1 |
7 | | | 0 | 0 |
8 | | | 0 | 0 |
9 | | | 0 | 0 |
10 | | | 0 | 0 |
以下是从此表中生成的树
当用户注册
时，如何更新leftCount和rightCount $referralid; //who is referring the current user
$side; //the side the user is to fall under, either left or right
$temp_referralid = $referralid;
$temp_sideCount = $side.'Count'; //leftCount or rightCount
$temp_side = $side;
$total_count=1;
$i=1;
while($total_count>0){
$stmt = $db->prepare("SELECT * from tree WHERE id=:id");
$stmt->bindValue(':id', $temp_referralid);
$stmt->execute();
$r = $stmt->fetch(PDO::FETCH_ASSOC);
$current_temp_sideCount = ($r[$temp_sideCount]+1);
//This will add (+1) to the leftCount or rightCount
//of the referral depending on the side the user
//choose to fall under.
$sql ="UPDATE `tree` SET `$temp_sideCount`=:count WHERE `id` = :id";
$stmt = $db->prepare($sql);
$stmt->bindValue(':count', $current_temp_sideCount);
$stmt->bindValue(':id', $temp_referralid);
$stmt->execute();
//if the user has someone that referred them as
//only the last person at the top has no referral
if($temp_referralid!=""){
//change the $referralid to the person that referred the person
//referring this user. this is where the loop will go through
//to the last person maintaining the side upward
$next_referralid = $this->getReferringId($db, $temp_referralid);
$temp_side = $this->getReferringIdSide($db, $temp_referralid);
$temp_sideCount = $temp_side.'Count';
$temp_referralid = $next_referralid;
$i++;
}else{
$total_count=0;
}
}
}
使用的功能
getReferringId($db, $id) //gets the referringId of the user passed as
//param from the user table
getReferringIdSide($db, $id) //gets the side which the user
//is on (left or right) from the user table
这一切都运行得很好，但后来我需要实现一些东西，但我找不到最佳方法。
如果每个用户已达到某个阶段，我需要不断更改其排名。见下文和例子：
stage 1: starter //just registered
stage 2: grow // the person has leftCount=3 and rightCount=3
stage 3: growbig //the person has 7 - grow user on the left
//and 7- grow user on the right
state 4: growbigger //the person has 7 - growbig on left and 7 growbig
//on the right
到第2阶段，我没有问题，但向上是我无法掌握正确逻辑的地方
更新 例如：growbig的数量可以来自腿上的任何地方，它不应该只是直接节点，只是在每个方面都低于该用户的排名计数
更新：以更清晰的术语和规格重新询问
它是一棵二叉树（2：2），这意味着一个人只能在他们下面有两个人（一个在左边，另一个在右边。
上面的图片我的表看起来像这样
树状表
userID | left (userid) | right (userid)| rank
8 | 4 | 12 |
4 | 2 | 6 |
12 | 10 | 14 |
2 | 1 | 3 |
6 | 5 | 7 |
10 | 9 | 11 |
14 | 14 | 15 |
注意：对于用户而言，任何一方或两者都不得使用任何人。这意味着如果用户在他下面没有人，那么树（分支）就在那里结束，如果他有一个人直接在左边而没有在右边那意味着左分支可以继续增长。
根据每个用户的增长情况对其进行评级的逻辑以及他们如何设法平衡每一方的增长是一个问题。
逻辑
等级1：主管：用户必须在其右侧分支上有3个用户，在左侧分支上有3个用户。
等级2：控制器：用户必须有7位左侧是“主管”的用户和7位右侧成为主管的用户。
等级3：高级管理员：用户必须有7个用户在左侧分支上是“控制器”，右侧还有7个“控制器”。
等级4：大使：用户必须有7个用户，左边是高级控制员，右边有7个高级控制器
等级5：高级大使：用户必须有7位用户，他们是左侧的大使，右侧是大使。
等级6：大使：用户必须拥有7位双方高级大使的用户。
说明： 让我挑选等级并解释它：
排名：大使 - 如果ID 3的用户右侧有45个人，右侧分支的7个人是高级管制员，左侧有67个人，7个已经是高级管制员，那么用户ID 3应升级为“大使”
@blag
1 个答案:
答案 0 :(得分：3)
我更有可能采取这个问题（使用http://mikehillyer.com/articles/managing-hierarchical-data-in-mysql/）：
新表架构是：
'id'| 'name' | 'left'| 'right'| 'rank'
4 | 'Dan' | 1 | 14 | 'starter'
5 | 'Chris'| 2 | 7 | 'starter'
6 | 'James'| 8 | 13 | 'starter'
7 | 'Tybe' | 3 | 4 | 'starter'
8 | 'Rose' | 5 | 6 | 'starter'
9 | 'Paul' | 9 | 10 | 'starter'
10 | 'Zach' | 11 | 12 | 'starter'
完整版：
注意，我使用以下值来避免使用更大的数据集
-- on each side
set @grow = 1 // -- 1 child R & L
set @growbig = 2 // -- 2 grow child R & L
<强> PROCEDURE
CREATE PROCEDURE p(IN p_parent varchar(15), IN p_children varchar(15))
BEGIN
IF NOT EXISTS (
select parent.*, count(distinct dc.id) direct_child
from u parent
left join u as dc on(parent.left=dc.left-1 or parent.right=dc.right+1)
WHERE parent.name = p_parent
group by parent.id
having count(distinct dc.id) =2
)
THEN
SET @myLeft =(SELECT `left` FROM u WHERE name = p_parent);
UPDATE u SET `right` = `right` + 2 WHERE `right` > @myLeft;
UPDATE u SET `left` = `left` + 2 WHERE `left` > @myLeft;
INSERT INTO u(`name`, `left`, `right`)
VALUES(p_children, @myLeft + 1, @myLeft + 2);
END IF;
END
//
call p('Tybe','Marjorie') //
call p('Tybe','Vernon') //
call p('Rose','Marc') //
call p('Rose','Peter') //
call p('Marc','Lily') //
call p('Marc','Ignotus') //
call p('Ignotus','Aragorn') //
call p('Ignotus','Arwen') //
call p('Peter','Chase') //
call p('Peter','Foreman') //
call p('Chase','Pétunia') //
call p('Chase','Dudley') //
call p('Foreman','Bob') //
call p('Foreman','Taub') //
call p('Paul','Lisa') //
call p('Paul','Bilbo') //
call p('Lisa','House') //
call p('Lisa','Gandalf') //
call p('Gandalf','Frodo') //
call p('Gandalf','Legolas') //
call p('House','Thirteen') //
call p('House','Wilson') //
call p('Thirteen','Ginny') //
call p('Thirteen','Harry') //
call p('Wilson','Kutner') //
call p('Wilson','Master') //
call p('Master','Adams') //
call p('Master','Park') //
call p('Zach','Ary') //
<强>生长
set @grow = 1 //
update u
set rank = 'grow'
where u.id in (
select id from (
select id
from (
select p.id id, p.name name, lc.id lcid, null rcid
from u p
inner join u l on (p.left = l.left-1 and p.right <> l.right+1)
inner join u lc on (lc.left >= l.left and lc.right <= l.right)
inner join u r on (p.right = r.right+1 and p.left <> r.left-1)
union all
select p.id id, p.name name, null lcid, rc.id rcid
from u p
inner join u l on (p.left = l.left-1 and p.right <> l.right+1)
inner join u r on (p.right = r.right+1 and p.left <> r.left-1)
inner join u rc on (rc.left >= r.left and rc.right <= r.right)
) p
group by p.id
having
count(distinct lcid) >= @grow
and count(distinct rcid) >= @grow
) z
)
//
<强> growbig
set @growbig = 2 //
update u
set rank = 'growbig'
where u.id in (
select id from (
select id
from (
select p.id id, p.name name, lc.id lcid, null rcid
from u p
inner join u l on (p.left = l.left-1 and p.right <> l.right+1)
inner join u lc on (lc.rank ='grow' and lc.left >= l.left and lc.right <= l.right)
inner join u r on (p.right = r.right+1 and p.left <> r.left-1)
union all
select p.id id, p.name name, null lcid, rc.id rcid
from u p
inner join u l on (p.left = l.left-1 and p.right <> l.right+1)
inner join u r on (p.right = r.right+1 and p.left <> r.left-1)
inner join u rc on (rc.rank ='grow' and rc.left >= r.left and rc.right <= r.right)
) p
group by p.id
having
count(distinct lcid) >= @growbig
and count(distinct rcid) >= @growbig
) z
)
//
查询1 ：
-- output parent that have both right and left child
select parent.*, count(distinct dc.id) direct_child
from u parent
left join u as dc on(parent.left=dc.left-1 or parent.right=dc.right+1)
group by parent.id
having count(distinct dc.id) =2
<强> Results ：
| id | name | left | right | rank | direct_child |
|----|----------|------|-------|---------|--------------|
| 4 | Dan | 1 | 72 | growbig | 2 |
| 5 | Chris | 2 | 35 | grow | 2 |
| 6 | James | 36 | 71 | grow | 2 |
| 7 | Tybe | 3 | 8 | grow | 2 |
| 8 | Rose | 9 | 34 | growbig | 2 |
| 9 | Paul | 37 | 66 | grow | 2 |
| 13 | Marc | 24 | 33 | grow | 2 |
| 14 | Peter | 10 | 23 | grow | 2 |
| 16 | Ignotus | 25 | 30 | grow | 2 |
| 19 | Chase | 17 | 22 | grow | 2 |
| 20 | Foreman | 11 | 16 | grow | 2 |
| 25 | Lisa | 40 | 65 | grow | 2 |
| 27 | House | 47 | 64 | grow | 2 |
| 28 | Gandalf | 41 | 46 | grow | 2 |
| 31 | Thirteen | 58 | 63 | grow | 2 |
| 32 | Wilson | 48 | 57 | grow | 2 |
| 36 | Master | 49 | 54 | grow | 2 |
查询2 ：
-- show the tree
SELECT CONCAT( REPEAT('|...', COUNT(parent.name) - 1), node.id, ' ', node.name,' /',node.rank) AS name
FROM u AS node,
u AS parent
WHERE node.left BETWEEN parent.left AND parent.right
GROUP BY node.name
ORDER BY node.left
<强> Results ：
| name |
|-----------------------------------------------|
|4 Dan /growbig |
||...5 Chris /grow |
||...|...7 Tybe /grow |
||...|...|...12 Vernon /starter |
||...|...|...11 Marjorie /starter |
||...|...8 Rose /growbig |
||...|...|...14 Peter /grow |
||...|...|...|...20 Foreman /grow |
||...|...|...|...|...24 Taub /starter |
||...|...|...|...|...23 Bob /starter |
||...|...|...|...19 Chase /grow |
||...|...|...|...|...22 Dudley /starter |
||...|...|...|...|...21 Pétunia /starter |
||...|...|...13 Marc /grow |
||...|...|...|...16 Ignotus /grow |
||...|...|...|...|...18 Arwen /starter |
||...|...|...|...|...17 Aragorn /starter |
||...|...|...|...15 Lily /starter |
||...6 James /grow |
||...|...9 Paul /grow |
||...|...|...26 Bilbo /starter |
||...|...|...25 Lisa /grow |
||...|...|...|...28 Gandalf /grow |
||...|...|...|...|...30 Legolas /starter |
||...|...|...|...|...29 Frodo /starter |
||...|...|...|...27 House /grow |
||...|...|...|...|...32 Wilson /grow |
||...|...|...|...|...|...36 Master /grow |
||...|...|...|...|...|...|...38 Park /starter |
||...|...|...|...|...|...|...37 Adams /starter |
||...|...|...|...|...|...35 Kutner /starter |
||...|...|...|...|...31 Thirteen /grow |
||...|...|...|...|...|...34 Harry /starter |
||...|...|...|...|...|...33 Ginny /starter |
||...|...10 Zach /starter |
||...|...|...39 Ary /starter |
查询3 ：
-- show parent + child data
select *,(l.right - l.left -1)/2 l , (r.right - r.left -1)/2 r from u p
inner join u l on (p.left = l.left-1 and p.right <> l.right+1)
inner join u r on (p.right = r.right+1 and p.left <> r.left-1)
<强> Results ：
| id | name | left | right | rank | id | name | left | right | rank | id | name | left | right | rank | l | r |
|----|----------|------|-------|---------|----|---------|------|-------|---------|----|----------|------|-------|---------|----|----|
| 4 | Dan | 1 | 72 | growbig | 5 | Chris | 2 | 35 | grow | 6 | James | 36 | 71 | grow | 16 | 17 |
| 5 | Chris | 2 | 35 | grow | 7 | Tybe | 3 | 8 | grow | 8 | Rose | 9 | 34 | growbig | 2 | 12 |
| 6 | James | 36 | 71 | grow | 9 | Paul | 37 | 66 | grow | 10 | Zach | 67 | 70 | starter | 14 | 1 |
| 7 | Tybe | 3 | 8 | grow | 12 | Vernon | 4 | 5 | starter | 11 | Marjorie | 6 | 7 | starter | 0 | 0 |
| 8 | Rose | 9 | 34 | growbig | 14 | Peter | 10 | 23 | grow | 13 | Marc | 24 | 33 | grow | 6 | 4 |
| 13 | Marc | 24 | 33 | grow | 16 | Ignotus | 25 | 30 | grow | 15 | Lily | 31 | 32 | starter | 2 | 0 |
| 16 | Ignotus | 25 | 30 | grow | 18 | Arwen | 26 | 27 | starter | 17 | Aragorn | 28 | 29 | starter | 0 | 0 |
| 14 | Peter | 10 | 23 | grow | 20 | Foreman | 11 | 16 | grow | 19 | Chase | 17 | 22 | grow | 2 | 2 |
| 19 | Chase | 17 | 22 | grow | 22 | Dudley | 18 | 19 | starter | 21 | Pétunia | 20 | 21 | starter | 0 | 0 |
| 20 | Foreman | 11 | 16 | grow | 24 | Taub | 12 | 13 | starter | 23 | Bob | 14 | 15 | starter | 0 | 0 |
| 9 | Paul | 37 | 66 | grow | 26 | Bilbo | 38 | 39 | starter | 25 | Lisa | 40 | 65 | grow | 0 | 12 |
| 25 | Lisa | 40 | 65 | grow | 28 | Gandalf | 41 | 46 | grow | 27 | House | 47 | 64 | grow | 2 | 8 |
| 28 | Gandalf | 41 | 46 | grow | 30 | Legolas | 42 | 43 | starter | 29 | Frodo | 44 | 45 | starter | 0 | 0 |
| 27 | House | 47 | 64 | grow | 32 | Wilson | 48 | 57 | grow | 31 | Thirteen | 58 | 63 | grow | 4 | 2 |
| 31 | Thirteen | 58 | 63 | grow | 34 | Harry | 59 | 60 | starter | 33 | Ginny | 61 | 62 | starter | 0 | 0 |
| 32 | Wilson | 48 | 57 | grow | 36 | Master | 49 | 54 | grow | 35 | Kutner | 55 | 56 | starter | 2 | 0 |
| 36 | Master | 49 | 54 | grow | 38 | Park | 50 | 51 | starter | 37 | Adams | 52 | 53 | starter | 0 | 0 |
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