Question

我有一个类似于下面的表结构：

declare @t1 table (code varchar)
declare @t2 table (id int, code varchar, closed bit)

@ t2保持打开/关闭@ t1中的代码：

insert into @t1 values ('x')

insert into @t2 values (1, 'x', 1)
insert into @t2 values (2, 'x', 1)
insert into @t2 values (3, 'x', 0)

我正在尝试获取代码的打开/关闭计数：

select 
count (t2_open.id) as opencount
, count (t2_closed.id) as closecount
from @t1 t1
inner join @t2 t2_open on t2_open.code = t1.code AND t2_open.closed = 0
inner join @t2 t2_closed on t2_closed.code = t1.code AND t2_closed.closed = 1

返回2＆amp; 2，我看不出我在这里做错了什么，为什么不回归1＆amp; 2？

不应该从我的查询中选择*返回null并且当记录不匹配时返回正确的计数？我期待select *返回3条记录，但它返回2条记录，其中两行中的id为3。

Answer 1

您计算的总记录数不是单个值

SELECT t1.code,
       Count (CASE WHEN t2.closed = 1 THEN 1 END) AS opencount,
       Count (CASE WHEN t2.closed = 0 THEN 1 END) AS closecount
FROM   @t1 t1
       LEFT JOIN @t2 t2
              ON t2.code = t1.code
GROUP  BY t1.code

更新：

让我们分解联接

SELECT t1.code,t2_open.closed
FROM   @t1 t1
       INNER JOIN @t2 t2_open
               ON t2_open.code = t1.code
                  AND t2_open.closed = 0

在第一次加入时，您将拉出关闭= 0的记录，因此结果将类似于

+------+--------+
| code | closed |
+------+--------+
| x    |      0 |
+------+--------+

现在，在上述结果的基础上，您将使用代码加入@ t2并关闭= 1

SELECT t1.code,t2_open.closed,t2_closed.closed
FROM   @t1 t1
       INNER JOIN @t2 t2_open
               ON t2_open.code = t1.code
                  AND t2_open.closed = 0
       INNER JOIN @t2 t2_closed
               ON t2_closed.code = t1.code
                  AND t2_closed.closed = 1

我们有两个关闭= 1的记录，对于相同的代码，当我们加入前一个结果时，关闭也将重复两次，因为两个关闭= 1

+------+--------+--------+
| code | closed | closed |
+------+--------+--------+
| x    |      0 |      1 |
| x    |      0 |      1 |
+------+--------+--------+

正如你所看到的，我们在加入后有两个关闭= 0，所以在计算时你会看到两个接近计数

具有相同表计数的多个连接

1 个答案: