Question

我一直在尝试使用我在SQL server

中构建的以下示例来完全理解联接

DECLARE @tablePlace TABLE (ID INT, someplace varchar(10))
DECLARE @tableType TABLE (ID INT, sometype varchar(10))
DECLARE @tableOrders TABLE (ID INT, type int , place int)

INSERT INTO @tablePlace (ID, someplace) values (1,'Place A')
INSERT INTO @tablePlace (ID, someplace) values (2,'Place B')

INSERT INTO @tableType (ID, sometype) VALUES (1,'Type 1')
INSERT INTO @tableType (ID, sometype) VALUES (2,'Type 2')
INSERT INTO @tableType (ID, sometype) VALUES (3,'Type 3')
INSERT INTO @tableType (ID, sometype) VALUES (4,'Type 4')

INSERT INTO @tableOrders (ID, place, type) values (1 , 1 , 1)  -- PLACE A TYPE 1
INSERT INTO @tableOrders (ID, place, type) values (2 , 1 , 2)  -- PLACE A TYPE 2
INSERT INTO @tableOrders (ID, place, type) values (3 , 2 , 2)  -- PLACE B TYPE 2

现在我要做的是链接三个表以获得以下结果

╔═════════╦════════╦═══════╗
║  PLACE  ║  TYPE  ║ COUNT ║
╠═════════╬════════╬═══════╣
║ PLACE A ║ TYPE 1 ║     1 ║
║ PLACE A ║ TYPE 2 ║     1 ║
║ PLACE A ║ TYPE 3 ║     0 ║
║ PLACE A ║ TYPE 4 ║     0 ║
║ PLACE B ║ TYPE 1 ║     0 ║
║ PLACE B ║ TYPE 2 ║     1 ║
║ PLACE B ║ TYPE 3 ║     0 ║
║ PLACE B ║ TYPE 4 ║     0 ║
╚═════════╩════════╩═══════╝

所以我要做的就是链接两个地方，并根据从@tableorders表中检索到的记录显示每种类型的计数。

到目前为止我的查询：

SELECT place.someplace,
       type.sometype,
       Count(*) AS count
FROM   @tableOrders orders
       INNER JOIN @tableplace place
               ON orders.place = place.id
       INNER JOIN @tabletype type
               ON place.id = type.id
GROUP  BY someplace,
          sometype
ORDER  BY someplace,
          sometype

有人可以解释我应该遵循的逻辑来实现我想要的结果吗？提前谢谢！

Answer 1

您希望交叉连接生成行。然后left join（或相关子查询）获取值：

select p.someplace, t.sometype, count(o.id) as count
from @tableplace p cross join
     @tabletype t left join
     @tableOrders o
     on o.type = t.id and o.place = p.id 
group by p.someplace, t.sometype
order by p.someplace, t.sometype;

Answer 2

试试这个：

SELECT place.someplace, [type].sometype , T.cnt
FROM @tablePlace place
CROSS join @tabletype [type] 
OUTER APPLY 
(   SELECT COUNT(1) cnt
    FROM @tableOrders tableOrders 
    WHERE tableOrders.place=place.ID AND tableOrders.[type] = [type].ID
)T
ORDER BY someplace,sometype

输出：

Place A Type 1  1
Place A Type 2  1
Place A Type 3  0
Place A Type 4  0
Place B Type 1  0
Place B Type 2  1
Place B Type 3  0
Place B Type 4  0

Answer 3

使用CROSS JOIN生成MxN表（将所有地点与类型相关联），然后使用LEFT JOIN生成订单表以获取每对(someplace, sometype)的计数：

select tp.someplace, tt.sometype, count(to.id) as count
from @tablePlace tp
cross join @tableType tt
left join @tableOrders to on 
  to.place = tp.id and to.type = tt.id
group by tp.someplace, tt.sometype
order by tp.someplace, tt.sometype

Answer 4

试试这个：

SELECT A.someplace,B.sometype,COUNT(C.ID) count_val FROM @tablePlace A CROSS JOIN 
@tableType B LEFT JOIN @tableOrders C ON C.place = A.ID AND C.type = B.ID 
GROUP BY someplace,sometype

希望它有所帮助。：）

通过示例了解联接

4 个答案: