将两个JOIN查询组合在一起

Question

我有3个表Appliance，Tx_property，Rx_property

#Appliance Table structure:
  -id
  -tx_property_id
  -rx_property_id
  -...

#Tx_property Table structure:
  -id
  -...

#Rx_property Table structure:
  -id
  -...

在rx_property.id = Appliance.id和tx_property.id = Appliance.id上设置约束

是否可以仅运行一个查询以从Appliance表中检索具有相应条目的所有记录（基于rx_property_id和tx_property_id）？

总而言之，我想将以下两个查询合并为一个。

select * from appliance INNER JOIN rx_property ON rx_property.id= appliance.rx_property_id;
select * from appliance INNER JOIN tx_property.id ON tx_property.id= appliance.tx_property_id;

有任何想法吗？ 谢谢！

编辑：

因此，如果我有以下示例记录：

appliance records:
id = 70
tx_property_id = 11
rx_property_id = null

id = 71
tx_property_id = 12
rx_property_id = null

id = 72
tx_property_id = null
rx_property_id = 11

id = 73
tx_property_id = null
rx_property_id = 12

tx_property records:
id = 11
name = 'tx_aa'

id = 12
name = 'tx_bb'

rx_property records:
id = 11
name = 'rx_aa'

id = 12
name = 'rx_bb'

我希望其中4个被检索为：

appliance.id = 70
appliance.tx_property_id = 11
appliance.rx_property_id = null
tx_property.id = 11
tx_property.name = 'tx_aa'

appliance.id = 71
appliance.tx_property_id = 12
appliance.rx_property_id = null
tx_property.id = 12
tx_property.name = 'tx_bb'

appliance.id = 72
appliance.tx_property_id = null
appliance.rx_property_id = 11
tx_property.id = 11
tx_property.name = 'rx_aa'

appliance.id = 73
appliance.tx_property_id = null
appliance.rx_property_id = 12
tx_property.id = 11
tx_property.name = 'rx_aa'

Answer 1

就这么简单：

select * from appliance
INNER JOIN rx_property ON rx_property.id= appliance.rx_property_id
INNER JOIN tx_property ON tx_property.id= appliance.tx_property_id

您可以根据需要将多个连接放入单个SELECT语句中。

Answer 2

您可以在单个查询中将两个表连接到appliance ：

select * 
from appliance a
join rx_property r on r.id = a.rx_property_id
join tx_property t on t.id = a.tx_property_id

上面的查询仅返回引用这两个表的这些行。 如果要获取所有行，请使用left join ：

select * 
from appliance a
left join rx_property r on r.id = a.rx_property_id
left join tx_property t on t.id = a.tx_property_id;

 id | rx_property_id | tx_property_id | id | name  | id | name  
----+----------------+----------------+----+-------+----+-------
 70 |             11 |                | 11 | rx_aa |    | 
 71 |             12 |                | 12 | rx_bb |    | 
 72 |                |             11 |    |       | 11 | tx_aa
 73 |                |             12 |    |       | 12 | tx_bb
(4 rows)    

您可以在选择列表中使用coalesce()选择不为空的列，例如：

select 
    a.id, rx_property_id, tx_property_id,
    coalesce(r.name, t.name) as property_name
from appliance a
left join rx_property r on r.id = a.rx_property_id
left join tx_property t on t.id = a.tx_property_id

 id | rx_property_id | tx_property_id | property_name 
----+----------------+----------------+---------------
 70 |             11 |                | rx_aa
 71 |             12 |                | rx_bb
 72 |                |             11 | tx_aa
 73 |                |             12 | tx_bb
(4 rows)    

还要注意，我为表使用了别名，通常使查询更具可读性。

在文档中阅读有关联接类型的更多信息。