加入NULL

时间:2016-12-14 01:55:34

标签: mysql

关于连接和null有很多问题,但我找不到与此特定模式匹配的问题。我有3个非常简单的表。

+---------+
| service |
+---------+
| id      |
| name    |
+---------+

+-----------+
| propnames |
+-----------+
| id        |
| name      |
| sort      |
+-----------+

+----------+
| props    |
+----------+
| sid      |
| pid      |
| value    |
+----------+

我希望能够添加属性(到propnames)并查询我的服务,加入我的属性(props)并知道尚未设置哪些属性。

如果这是服务

(id), (name)
1, "AAA"
2, "BBB"

这是propnames

(id), (name), (sort)
1, "property_a", 1
2, "property_b", 2
3, "property_c", 3

这是道具

(service.id), (propname.id), (value)
1, 1, "Service AAA has property_a value"
1, 2, "Service AAA has property_b value"
2, 1, "Service BBB has property_a value"

然后我的查询最终会产生这个:

(service.id), (service.name), (property.id), (property.name), (props.value)
1, "AAA", 1, "property_a", "Service AAA has property_a value"
1, "AAA", 2, "property_b", "Service AAA has property_b value"
1, "AAA", 3, "property_c", NULL
2, "BBB", 1, "property_a", "Service BBB has property_a value"
2, "BBB", 2, "property_b", NULL
2, "BBB", 3, "property_c", NULL

理想情况下,它将按service.name ASC -then- property.sort

排序

目前不完整的查询是:

SELECT s.id, s.name, p.id, p.name, props.value 
FROM service.s 
LEFT JOIN propnames p ON s.id = p.sid 
LEFT JOIN props ON props.pid = p.id 
ORDER BY s.name ASC, p.sort ASC

1 个答案:

答案 0 :(得分:1)

使用cross join生成行。然后使用left join查找匹配项:

select s.id, s.name, p.id, p.name, props.value 
from services s cross join
     propnames p left join
     props
     on props.sid = s.id and props.pid = p.id;