Question

如何计算此类产品？

TABLE CATEGORY

category_id 
title 
lft 
rght 
parent 
level

表格产品

product_id 
category_id 
title

DATA 表类别

|1|Electronics|1|16|0|0
|2|Televisions|2|7|1|1
|3|LCD|3|4|2|2
|4|PLASMA|5|6|2|2
|5|Players|8|15|1|1
|6|Mp3 players|9|10|5|2
|7|CD players|11|12|5|2
|8|DVD players|13|14|5|2
|9|Furniture|17|18|0|0

表产品

|1|4|Plasma 1
|2|4|Plasma 2
|3|4|Plasma 3
|4|4|Plasma 4
|5|4|Plasma 5
|6|3|LCD 1
|7|3|LCD 2
|8|3|LCD 3
|9|6|MP3 1
|10|6|MP3 2
|11|7|CD 1
|12|7|CD 2
|13|8|DVD 1
|14|8|DVD 2
|15|8|DVD 3

我想算一下这些例子。

Electronics (15)
Televisions (8)
LCD (3)
PLASMA (5)
Players (7)
Mp3 players (2)
CD players (2)
DVD players (3)
Furniture (0)

并使用

SQL

select parent.title, parent.level, count(product.title) as sum_products  
from category as node, category as parent, product  
where node.lft between parent.lft and parent.rght   
and node.category_id = product.category_id  
group by parent.title   
order by node.lft

和这些结果

Electronics (15)
Televisions (8)
LCD (3)
PLASMA (5)
Players (7)
Mp3 players (2)
CD players (2)
DVD players (3)

我如何像样品一样制作出来？

家具（0）不要在查询中显示。

感谢您的帮助。

选择parent.title，parent.level，coalesce（count（product.title），0）as sum_products从类别作为节点，类别作为父级，产品在哪里 parent.lft和parent.rght以及node.category_id =之间的node.lft product.category_id group by parent.title order by node.lft

http://i.stack.imgur.com/adJd2.png

谢谢jpw的回答

此代码完成

select 
  parent.title, 
  parent.level, 
  count(product.title) as sum_products  
from category as node
join category as parent 
  on node.lft between parent.lft and parent.rght   
left join product 
  on node.category_id = product.category_id  
group by parent.title, parent.level
order by node.lft

|       TITLE | LEVEL | SUM_PRODUCTS |
|-------------|-------|--------------|
| Electronics |     0 |           15 |
| Televisions |     1 |            8 |
|         LCD |     2 |            3 |
|      PLASMA |     2 |            5 |
|     Players |     1 |            7 |
| Mp3 players |     2 |            2 |
|  CD players |     2 |            2 |
| DVD players |     2 |            3 |
|   Furniture |     0 |            0 |

Answer 1

如果问题是排除了空类别家具，我认为解决方案是使用显式左连接而不是隐式内连接到产品表，如下所示：

select 
  parent.title, 
  parent.level, 
  count(product.title) as sum_products  
from category as node
join category as parent 
  on node.lft between parent.lft and parent.rght   
left join product 
  on node.category_id = product.category_id  
group by parent.title, parent.level
order by node.lft

Sample SQL Fiddle

此查询将提供以下输出：

|       TITLE | LEVEL | SUM_PRODUCTS |
|-------------|-------|--------------|
| Electronics |     0 |           15 |
| Televisions |     1 |            8 |
|         LCD |     2 |            3 |
|      PLASMA |     2 |            5 |
|     Players |     1 |            7 |
| Mp3 players |     2 |            2 |
|  CD players |     2 |            2 |
| DVD players |     2 |            3 |
|   Furniture |     0 |            0 |

如何计算此类产品？

1 个答案: