Question

给出以下SQL Server架构：

CREATE TABLE #TableA (Id int);
CREATE TABLE #TableB (Id int, TableA_Id int, Status_Id int)
CREATE TABLE #Statuses (Id int)

SELECT
    *
FROM
    #TableA AS A
    INNER JOIN #TableB AS B
        ON A.Id = B.TableA_Id
    INNER JOIN #Statuses AS S
        ON B.Status_Id = S.Id

如何获取TableA中dbo.Statuses中每行至少有一个条目的#TableA中的所有行？

例如，在以下数据集中，只应返回#TableA [1] [2] [3] [4] #Statuses [1] [2] [3] #TableB [1][1][1] [2][1][3] [3][2][1] [4][2][2] [5][2][3] [6][3][1]的第2行：

{{1}}

Answer 1

SELECT B.TableA_Id
FROM #Statuses S
LEFT JOIN #TableB B
    ON B.Status_Id = S.Id
GROUP BY B.TableA_Id
HAVING COUNT(DISTINCT S.Id) = COUNT(DISTINCT B.Status_Id)

Answer 2

你可以用简单的＆＃34; where条款。假设#TableB中的状态列仅具有有效状态值：

select a.*
from #TableA a
where (select count(*) from statuses) = 
      (select count(distinct status) from #TableB b where b.TableA_Id = a.id);

你可以放松这个假设。这是一种方式：

select a.*
from #TableA a join
      (select count(*) as numstatuses from #statuses) s
where (select count(distinct b.status)
       from #TableB b join
            #statuses s
            on b.id = s.id
       where b.TableA_Id = a.id
      ) = numstatuses

Answer 3

SQL Fiddle：

SELECT TableA_Id
FROM TableB
GROUP BY TableA_Id
HAVING COUNT(*) >= (SELECT COUNT(*) FROM Statuses);

从SQL表中选择行，其中连接表中存在所有参数集

3 个答案: