问题描述
+-------+-------+-----------+
| id | fname | parent_id |
+=======+=======+===========+
| sk001 | aa | null |
+-------+-------+-----------+
| sk002 | ss | ssk001 |
+-------+-------+-----------+
| sk003 | dd | sk001 |
+-------+-------+-----------+
| sk004 | ff | sk002 |
+-------+-------+-----------+
| sk005 | gg | sk002 |
+-------+-------+-----------+
| sk006 | hh | sk005 |
+-------+-------+-----------+
| sk007 | jj | sk006 |
+-------+-------+-----------+
| sk008 | kk | sk006 |
+-------+-------+-----------+
| sk009 | ll | sk004 |
+-------+-------+-----------+
| sk010 | mm | sk005 |
+-------+-------+-----------+
我有这样的invoice_order表
+-------+-------+-----------+--------+
| id | fname | parent_id | amount |
+=======+=======+===========+========+
| sk001 | aa | null | 100 |
+-------+-------+-----------+--------+
| sk002 | ss | ssk001 | 400 |
+-------+-------+-----------+--------+
| sk002 | dd | sk001 | 225 |
+-------+-------+-----------+--------+
| sk004 | ff | sk002 | 50 |
+-------+-------+-----------+--------+
| sk005 | gg | sk002 | 59 |
+-------+-------+-----------+--------+
| sk006 | hh | sk005 | 77 |
+-------+-------+-----------+--------+
| sk007 | jj | sk006 | 89 |
+-------+-------+-----------+--------+
| sk004 | ff | sk002 | 87 |
+-------+-------+-----------+--------+
| sk009 | ll | sk004 | 45 |
+-------+-------+-----------+--------+
| sk010 | mm | sk005 | 56 |
+-------+-------+-----------+--------+
在这里,我必须计算ID的个人总数和团队总数, 示例(树):
sk001
|
|--sk002
| |--sk004
| | |--sk009
| | |--xy
| |
| |--sk005
|
|--sk003
例如:sk002
的团队总数应包含(sk004,sk009,xy)
团队中某人的总购买金额
我刚刚尝试过这样的事情:
SELECT (SUM(amount))/2 as indteamtotal
from (SELECT * from invoice_order
order by id,parent_id) products_sorted,(SELECT @pv := 'sk002') initialisation
where find_in_set(parent_id,@pv) > 0
and length(@pv := concat(@pv,',id)) order by parent_id asc
但是我没有得到正确的团队总数 有没有可能使用MysqL过程
解决方法
如果您使用的是MySQL 8+,则可以使用CTE
with recursive tree as (
select id from invoice_order where id = 'sk002'
union
select i.id from invoice_order i join tree on tree.id = i.parent_id
)
select sum(amount) from invoice_order where id in (select id from tree);
链接到fiddle