Sql server 在存储过程中左联接的正确方法

我有一些关于左撇子的问题

我有这些桌子：

dbo.ARoom_型

id | roomtype_name |  isactive
---+---------------+-----------
1  | Presidential  |    1
2  |   De Luxe     |    1

弱势群体

id | amenities_name | isactive
---+----------------+----------
1  |   Internet     |   1
2  |   Cable TV     |   1
3  |   Bath Tub     |   1
4  |   Air-con      |   1
5  |   Telephone    |   1

dbo.A详细信息：

id | roomtype_id | amenities_id
---+-------------+--------------
1  |      1      |      1
2  |      1      |      2
3  |      1      |      4
4  |      2      |      1

我试过这个问题

SELECT
    B.id as 'details_id',
    C.roomtype_name,
    A.amenities_name
FROM 
    AAmenities AS A
LEFT JOIN 
    AAmenities_details AS B ON B.amenities_id = A.id
LEFT JOIN 
    ARoom_type as C ON C.id = B.roomtype_id

  @id INT
  AS
  BEGIN TRY

  SELECT
      B.id AS 'details_id',
      C.roomtype_name,
      A.amenities_name
  FROM
      AAmenities AS A
  LEFT JOIN 
      AAmenities_details AS B ON B.amenities_id = A.id
  LEFT JOIN 
      ARoom_type AS C ON C.id = B.roomtype_id
  WHERE 
      C.id = @id

这就是我得到的

结果:

details_id | roomtype_name  | amenities_name 
-----------+----------------+----------------
    1      |   Presidential |   Internet
    2      |   De Luxe      |   Internet
    3      |   Presidential |   Cable TV
    4      |   NULL         |   Bath Tub
    5      |   Presidential |   Air-con
    6      |   NULL         |   Telephone

我可以得到所有的便利设施和所有的房间类型名称，这是我想要的，但我希望这是由房间类型，所以如果总统只有总统，我尝试了这个问题

SELECT
    B.id as 'details_id',
    C.roomtype_name,
    A.amenities_name
FROM 
    AAmenities AS A
LEFT JOIN 
    AAmenities_details AS B ON B.amenities_id = A.id
LEFT JOIN 
    ARoom_type as C ON C.id = B.roomtype_id

  @id INT
  AS
  BEGIN TRY

  SELECT
      B.id AS 'details_id',
      C.roomtype_name,
      A.amenities_name
  FROM
      AAmenities AS A
  LEFT JOIN 
      AAmenities_details AS B ON B.amenities_id = A.id
  LEFT JOIN 
      ARoom_type AS C ON C.id = B.roomtype_id
  WHERE 
      C.id = @id

我只能得到这些结果

  id  |  roomtype_name  | amenities_name
  ----+-----------------+----------------
   1  |    Presidential |   Internet
   2  |    Presidential |   Cable TV
   4  |    Presidential |   Air-con

这就是结果，当我想要的是这样：

  details_id | roomtype_name  | amenities_name 
  -----------+----------------+-----------------
      1      |   Presidential |   Internet
      3      |   Presidential |   Cable TV
      4      |   NULL         |   Bath Tub
      5      |   Presidential |   Air-con
      6      |   NULL         |   Telephone

---

当我想要id并且仍然获得所有便利设施的名称时，哪种查询是正确的？即使我的roomtype\u名称的值为空，如何左键连接答案？非常感谢您

您需要

aameanities\u详细信息的所有数据

，然后再试一次

SELECT
   B.id as 'details_id',
   C.roomtype_name,
   A.amenities_name
 FROM 
 AAmenities_details as B
 LEFT JOIN AAmenities  as A
   on B.amenities_id = A.id
 LEFT JOIN ARoom_type as C
   on C.id = B.roomtype_id
 WHERE B.roomtype_id = @id

但我认为每个

AAmenities\u详细信息

必须具有

AAmenities

因此您可以在

AAmenities\u details

和

AAmenities

SELECT
   B.id as 'details_id',
   C.roomtype_name,
   A.amenities_name
 FROM 
 AAmenities_details as B
 INNER JOIN AAmenities  as A
   on B.amenities_id = A.id
 LEFT JOIN ARoom_type as C
   on C.id = B.roomtype_id
 WHERE B.roomtype_id = @id

---

您需要

aamenilities\u details

的所有数据，然后重试

SELECT
   B.id as 'details_id',
   C.roomtype_name,
   A.amenities_name
 FROM 
 AAmenities_details as B
 LEFT JOIN AAmenities  as A
   on B.amenities_id = A.id
 LEFT JOIN ARoom_type as C
   on C.id = B.roomtype_id
 WHERE B.roomtype_id = @id

但我认为每个

AAmenities\u详细信息

必须具有

AAmenities

因此您可以在

AAmenities\u details

和

AAmenities

SELECT
   B.id as 'details_id',
   C.roomtype_name,
   A.amenities_name
 FROM 
 AAmenities_details as B
 INNER JOIN AAmenities  as A
   on B.amenities_id = A.id
 LEFT JOIN ARoom_type as C
   on C.id = B.roomtype_id
 WHERE B.roomtype_id = @id

---

mysql还是sql server？正确标记您的rdbms。这是

mysql

还是

sql server

？哦，对不起，是sql server mysql还是sql server？正确标记您的rdbms。这是

mysql

还是

sql server

？哦，很抱歉，它的sql server我只想看到所有设施的名称，即使roomtype名称为空，所以如果我输入id 1，结果将是：'details_id | roomtype | name |设施| name 1 | Presidential | Internet |3 |总统|有线电视4 |空|浴缸5 |总统|空调6 |空|电话|我编辑。你不应该使用

c.id=@id

，因为它可能是空的。更改为

B.roomtypeid=@id

不能相同。当

detail\u id

=4时，其

roomtype

=2，您的结果中不存在该名称。我只想看到所有设施的名称，即使roomtype名称为空，因此如果我输入id 1，结果将是：“详细信息”\u id | roomtype |名称|设施|名称1 |总统|互联网3 |总统|有线电视4 |空|浴缸5 |总统|空调6 |空|电话|我编辑。你不应该使用

c.id=@id

，因为它可能是空的。更改为

B.roomtypeid=@id

不能相同。当

detail\u id

=4时，则其

roomtype

=2，它不能存在于您的结果中。