查找另一个集合中的所有集合/实体 [英] Find all sets/entities that are in another set
问题描述
答案在摘要中找到
The answer is found in the abstract here but I'm looking for the concrete SQL solution.
Given the following tables:
------------ -----------
| F_Roles | | T_Roles |
------+----- -----+-----
| FId | RId| |TId | RId|
------+------ -----+-----
| f1 | 2 | | t1 | 1 |
| f1 | 3 | | t1 | 2 |
| f2 | 2 | | t1 | 3 |
| f2 | 4 | | t1 | 4 |
| f2 | 9 | | t1 | 5 |
| f3 | 6 | | t1 | 6 |
| f3 | 7 | | t1 | 7 |
------------ ----------
(F_Roles) is a join table between F (not shown) and Roles (also not shown) (T_Roles) is a join table between T (not shown) and Roles (not shown)
I need to return:
all (DISTINCT) FId's where the set of RId's for a given FId is a subset of (or 'IN') Roles. (I know I'm mixing Set Theory with database terms but only in the interest of better conveying the idea, I hope). So, f1 and f3 should be returned in this case, because the set of RId's for f1, {2,3}, and for f3, {6,7}, are subsets of T_Roles.
the list of RId's in T_Roles NOT found in any of the functions returned above. (T_Roles - (f1 Union f3)), or {1,4,5} in this example.
Let's define the following sample data:
DECLARE @F_Roles TABLE
(
[FID] CHAR(2)
,[RID] TINYINT
);
DECLARE @Roles TABLE
(
[RID] TINYINT
);
INSERT INTO @F_Roles ([FID], [RID])
VALUES ('f1', 2)
,('f1', 3)
,('f2', 2)
,('f2', 4)
,('f2', 9)
,('f3', 6)
,('f3', 7);
INSERT INTO @Roles ([RID])
VALUES (1), (2), (3), (4), (5), (6), (7);
No, the first query can be solved using the T-SQL statement below:
SELECT F.[FID]
FROM @F_Roles F
LEFT JOIN @Roles R
ON F.[RID] = R.[RID]
GROUP BY F.[FID]
HAVING SUM(CASE WHEN R.[RID] IS NULL THEN 0 ELSE 1 END) = COUNT(F.[RID]);
The idea is pretty simple. We are using LEFT
join in order to check which RID
from the @F_Roles
table has corresponding RID
in the @Roles
table. If it has not, the value returned by the query for the corresponding row is NULL
. So, we just need to count the RIDs
for each FID
and to check if this count is equal to the count of values returned by the second table (NULL
values are ignored).
The latter query is simple, too. Having the FID
from the first, we just can use EXCEPT
in order to found RIDs
which are not matched:
SELECT [RID]
FROM @Roles
EXCEPT
SELECT [RID]
FROM @F_Roles
WHERE [FID] IN
(
SELECT F.[FID]
FROM @F_Roles F
LEFT JOIN @Roles R
ON F.[RID] = R.[RID]
GROUP BY F.[FID]
HAVING SUM(CASE WHEN R.[RID] IS NULL THEN 0 ELSE 1 END) = COUNT(F.[RID])
);
Here is the result of the execution of the queries:
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