如何返回仅与特定属性有关的SQL结果？_Sql

如何返回仅与特定属性有关的SQL结果？

sql

如何返回仅与特定属性有关的SQL结果？,sql,Sql,我有一张类似于以下内容的表格： REL_Aid_To_AttributeValue: ID AttributeValueID 101 1319 101 1320 101 1344 101 1345 102 1319 102 1320 102 1321 102 1336 103 1320 103 1336 SELECT a.ID, a.Title, c.AttributeTypeID, COUNT(b.AttributeValueID)AS Tot

我有一张类似于以下内容的表格：

REL_Aid_To_AttributeValue:
ID    AttributeValueID
101   1319
101   1320
101   1344
101   1345
102   1319
102   1320
102   1321
102   1336
103   1320
103   1336

SELECT a.ID, a.Title, c.AttributeTypeID, 
COUNT(b.AttributeValueID)AS Total
FROM TBL_Aids as a,
REL_Aid_To_AttributeValue as b,
TBL_AttributeValues as c
WHERE a.ID=b.AidID
AND b.AttributeValueID=c.ID
AND a.Status=2
AND AttributeTypeID=1
AND (b.AttributeValueID=1319
OR b.AttributeValueID=1320)
GROUP BY a.ID, a.Title,
c.AttributeTypeID
HAVING COUNT(b.AttributeValueID) <= 2

我有另一个表，它按类型对这些AttributeValueId进行分组：

TBL_AttributeValues:
AttributeValueID    AttributeTypeID
1319                1
1320                1
1321                1
1336                2
1344                3 
1345                3

我有第三个表，提供了第一个表中每个ID的信息：

TBL_Aids:
ID     Title                Author       etc.
101    Aid About Spiders    John Doe
102    Aid About Mites      Jane Doe 
103    Aid About Beetles    Joe Schmo

我想返回只包含AttributeValueIDs 1319或（包括）1320的ID的结果（以及关于它们的信息，如标题、作者等），在AttributeTypeID=1内。所以我希望返回101，因为它的AttributeType 1只有1319和1320，还有103，因为它的AttributeType 1只有1320。我不想返回102，因为尽管它有1319和1320，但它也有1321，这也是AttributeType 1

我尝试了以下方法：

REL_Aid_To_AttributeValue:
ID    AttributeValueID
101   1319
101   1320
101   1344
101   1345
102   1319
102   1320
102   1321
102   1336
103   1320
103   1336

SELECT a.ID, a.Title, c.AttributeTypeID, 
COUNT(b.AttributeValueID)AS Total
FROM TBL_Aids as a,
REL_Aid_To_AttributeValue as b,
TBL_AttributeValues as c
WHERE a.ID=b.AidID
AND b.AttributeValueID=c.ID
AND a.Status=2
AND AttributeTypeID=1
AND (b.AttributeValueID=1319
OR b.AttributeValueID=1320)
GROUP BY a.ID, a.Title,
c.AttributeTypeID
HAVING COUNT(b.AttributeValueID) <= 2

选择a.ID、a.Title、c.AttributeTypeID、，
将（b.AttributeValueID）计数为总计
从TBL_Aids作为一个，
将属性值与辅助值关联为b，
TBL_属性值为c
其中a.ID=b.AidID
b.AttributeValueID=c.ID
a.状态=2
和AttributeTypeID=1
和（b.AttributeValueID=1319
或b.AttributeValueID=1320）
按a.ID、a.标题分组，
c、 属性肽
拥有COUNT（b.AttributeValueID）尝试：
尝试：
您想限制您的ID。在where子句中，可以包含以下内容：
where id in (select distinct av.id
             from AttributeValueType avt join
                  AttributeValue av
                   on avt.AttributeValueId = av.AttributeValueId
             where avt.AttributeValueId in (1319, 1320) and avt.AttributeTypeID = 1
            )

如果仅当这两个属性存在时才需要，请尝试以下操作：
Where id in (
    Select av.id
    From AttributeValueType avt join
         AttributeValue av
         on avt.AttributeValueId = av.AttributeValueId
    Group by av.id
    Having max(case when avt.AttributeValueId in (1319, 1320) and avt.AttributeTypeID = 1 then 0 else 1)=0
    Having max(case when 

您想限制您的ID。在where子句中，可以包含以下内容：
where id in (select distinct av.id
             from AttributeValueType avt join
                  AttributeValue av
                   on avt.AttributeValueId = av.AttributeValueId
             where avt.AttributeValueId in (1319, 1320) and avt.AttributeTypeID = 1
            )

如果仅当这两个属性存在时才需要，请尝试以下操作：
Where id in (
    Select av.id
    From AttributeValueType avt join
         AttributeValue av
         on avt.AttributeValueId = av.AttributeValueId
    Group by av.id
    Having max(case when avt.AttributeValueId in (1319, 1320) and avt.AttributeTypeID = 1 then 0 else 1)=0
    Having max(case when 

我认为您需要以某种方式添加“不存在”：
SELECT DISTINCT ID
FROM TBL_Aids x
WHERE AttributeValueID IN
(
    SELECT AttributeValueID
    FROM AttributeValueIDs
    WHERE AttributeValueID IN (1319, 1320) AND AttributeTypeID=1
) AND NOT EXISTS
(
    SELECT *
    FROM AttributeValueIDs a
    WHERE a.AttributeValueID NOT IN (1319, 1320) AND a.AttributeTypeID=1
    AND a.AttributeValueIDs = x.AttributeValueIDs
)

我认为您需要以某种方式添加“不存在”：
SELECT DISTINCT ID
FROM TBL_Aids x
WHERE AttributeValueID IN
(
    SELECT AttributeValueID
    FROM AttributeValueIDs
    WHERE AttributeValueID IN (1319, 1320) AND AttributeTypeID=1
) AND NOT EXISTS
(
    SELECT *
    FROM AttributeValueIDs a
    WHERE a.AttributeValueID NOT IN (1319, 1320) AND a.AttributeTypeID=1
    AND a.AttributeValueIDs = x.AttributeValueIDs
)

您需要两次加入到REL_Aid_to_AttributeValue，以确保存在两行
SELECT a.ID, a.Title, c.AttributeTypeID, 2 AS Total
FROM TBL_Aids a
join REL_Aid_To_AttributeValue b on a.ID=b.AidID and b.AttributeValueID = 1319
join REL_Aid_To_AttributeValue b2 on a.ID=b2.AidID and b2.AttributeValueID = 1320
join TBL_AttributeValues c on b.AttributeValueID=c.ID
WHERE a.Status=2
AND AttributeTypeID=1;

SELECT a.ID, a.Title, c.AttributeTypeID, 2 AS Total
FROM TBL_Aids a
join REL_Aid_To_AttributeValue b on a.ID=b.AidID and b.AttributeValueID = 1319
join REL_Aid_To_AttributeValue b2 on a.ID=b2.AidID and b2.AttributeValueID = 1320
join TBL_AttributeValues c on b.AttributeValueID=c.ID
WHERE a.Status=2
AND AttributeTypeID=1;

另外，如果你知道你有什么值，你就知道有多少，所以这是一个文本2
，你需要加入两次REL\u Aid\u to\u AttributeValue以确保两行存在
SELECT a.ID, a.Title, c.AttributeTypeID, 2 AS Total
FROM TBL_Aids a
join REL_Aid_To_AttributeValue b on a.ID=b.AidID and b.AttributeValueID = 1319
join REL_Aid_To_AttributeValue b2 on a.ID=b2.AidID and b2.AttributeValueID = 1320
join TBL_AttributeValues c on b.AttributeValueID=c.ID
WHERE a.Status=2
AND AttributeTypeID=1;

SELECT a.ID, a.Title, c.AttributeTypeID, 2 AS Total
FROM TBL_Aids a
join REL_Aid_To_AttributeValue b on a.ID=b.AidID and b.AttributeValueID = 1319
join REL_Aid_To_AttributeValue b2 on a.ID=b2.AidID and b2.AttributeValueID = 1320
join TBL_AttributeValues c on b.AttributeValueID=c.ID
WHERE a.Status=2
AND AttributeTypeID=1;

SELECT  id
FROM    (
        SELECT  DISTINCT id
        FROM    REL_Aid_To_AttributeValue a
        WHERE   attributeValueId IN (1319, 1320)
        ) q
WHERE   NOT EXISTS
        (
        SELECT  id
        FROM    (
                SELECT  AttributeValueID
                FROM    TBL_AttributeValues
                WHERE   AttributeTypeID = 1
                        AND AttributeValueID NOT IN (1319, 1320)
                ) t
        JOIN    REL_Aid_To_AttributeValue a
        ON      a.id = q.id
                AND a.AttributeValueID = t.AttributeValueID
        )

另外，如果你知道你有什么值，你就知道有多少，所以这是一个文本2
，你需要加入两次REL\u Aid\u to\u AttributeValue以确保两行存在
SELECT a.ID, a.Title, c.AttributeTypeID, 2 AS Total
FROM TBL_Aids a
join REL_Aid_To_AttributeValue b on a.ID=b.AidID and b.AttributeValueID = 1319
join REL_Aid_To_AttributeValue b2 on a.ID=b2.AidID and b2.AttributeValueID = 1320
join TBL_AttributeValues c on b.AttributeValueID=c.ID
WHERE a.Status=2
AND AttributeTypeID=1;

SELECT a.ID, a.Title, c.AttributeTypeID, 2 AS Total
FROM TBL_Aids a
join REL_Aid_To_AttributeValue b on a.ID=b.AidID and b.AttributeValueID = 1319
join REL_Aid_To_AttributeValue b2 on a.ID=b2.AidID and b2.AttributeValueID = 1320
join TBL_AttributeValues c on b.AttributeValueID=c.ID
WHERE a.Status=2
AND AttributeTypeID=1;

SELECT  id
FROM    (
        SELECT  DISTINCT id
        FROM    REL_Aid_To_AttributeValue a
        WHERE   attributeValueId IN (1319, 1320)
        ) q
WHERE   NOT EXISTS
        (
        SELECT  id
        FROM    (
                SELECT  AttributeValueID
                FROM    TBL_AttributeValues
                WHERE   AttributeTypeID = 1
                        AND AttributeValueID NOT IN (1319, 1320)
                ) t
        JOIN    REL_Aid_To_AttributeValue a
        ON      a.id = q.id
                AND a.AttributeValueID = t.AttributeValueID
        )

注意，我已将您的查询转换为更现代的“join”语法。我建议你学一下
还要注意的是，如果您知道自己有哪些值，那么您就知道有多少个值，因此这是一个文本2
，您需要两次连接到REL\u Aid\u to\u AttributeValue以确保存在两行
SELECT a.ID, a.Title, c.AttributeTypeID, 2 AS Total
FROM TBL_Aids a
join REL_Aid_To_AttributeValue b on a.ID=b.AidID and b.AttributeValueID = 1319
join REL_Aid_To_AttributeValue b2 on a.ID=b2.AidID and b2.AttributeValueID = 1320
join TBL_AttributeValues c on b.AttributeValueID=c.ID
WHERE a.Status=2
AND AttributeTypeID=1;

SELECT a.ID, a.Title, c.AttributeTypeID, 2 AS Total
FROM TBL_Aids a
join REL_Aid_To_AttributeValue b on a.ID=b.AidID and b.AttributeValueID = 1319
join REL_Aid_To_AttributeValue b2 on a.ID=b2.AidID and b2.AttributeValueID = 1320
join TBL_AttributeValues c on b.AttributeValueID=c.ID
WHERE a.Status=2
AND AttributeTypeID=1;

SELECT  id
FROM    (
        SELECT  DISTINCT id
        FROM    REL_Aid_To_AttributeValue a
        WHERE   attributeValueId IN (1319, 1320)
        ) q
WHERE   NOT EXISTS
        (
        SELECT  id
        FROM    (
                SELECT  AttributeValueID
                FROM    TBL_AttributeValues
                WHERE   AttributeTypeID = 1
                        AND AttributeValueID NOT IN (1319, 1320)
                ) t
        JOIN    REL_Aid_To_AttributeValue a
        ON      a.id = q.id
                AND a.AttributeValueID = t.AttributeValueID
        )

注意，我已将您的查询转换为更现代的“join”语法。我建议你学一下
还要注意的是，如果您知道自己有哪些值，那么您就知道有多少个值，因此它是一个文本2
，这要感谢Quassnoi：
SELECT  id
FROM    (
        SELECT  DISTINCT id
        FROM    REL_Aid_To_AttributeValue a
        WHERE   attributeValueId IN (1319, 1320)
        ) q
WHERE   NOT EXISTS
        (
        SELECT  id
        FROM    (
                SELECT  AttributeValueID
                FROM    TBL_AttributeValues
                WHERE   AttributeTypeID = 1
                        AND AttributeValueID NOT IN (1319, 1320)
                ) t
        JOIN    REL_Aid_To_AttributeValue a
        ON      a.id = q.id
                AND a.AttributeValueID = t.AttributeValueID
        )

    SELECT ID, Title, Author
    FROM TBL_Aids X
    JOIN (
    SELECT AidID
    FROM    (
            SELECT  DISTINCT AidID
            FROM    REL_Aid_To_AttributeValue a
            WHERE   AttributeValueID IN (1319, 1320)
            ) q
    WHERE   NOT EXISTS
            (
            SELECT  AidID
            FROM    (
                    SELECT  ID
                    FROM    TBL_AttributeValues
                    WHERE   AttributeTypeID = 1
                            AND ID NOT IN (1319, 1320)
                    ) t
            JOIN    REL_Aid_To_AttributeValue a
            ON      a.AidID = q.AidID
                    AND a.AttributeValueID = t.ID
            ))
    Y ON X.ID=Y.AidID
    WHERE Status=2

多亏了卡西诺：
    SELECT ID, Title, Author
    FROM TBL_Aids X
    JOIN (
    SELECT AidID
    FROM    (
            SELECT  DISTINCT AidID
            FROM    REL_Aid_To_AttributeValue a
            WHERE   AttributeValueID IN (1319, 1320)
            ) q
    WHERE   NOT EXISTS
            (
            SELECT  AidID
            FROM    (
                    SELECT  ID
                    FROM    TBL_AttributeValues
                    WHERE   AttributeTypeID = 1
                            AND ID NOT IN (1319, 1320)
                    ) t
            JOIN    REL_Aid_To_AttributeValue a
            ON      a.AidID = q.AidID
                    AND a.AttributeValueID = t.ID
            ))
    Y ON X.ID=Y.AidID
    WHERE Status=2

这不太正确，但对我来说似乎有道理。它只返回大约100个结果，但我估计结果集接近700个。而且更重要的是，结果是不准确的。它们既有属性值1319和1320，也有其他属性值。是的，我很少使用join，因为我没有看到它比我使用的方式有什么好处。如果它在这种情况下起作用，我会被说服的！谢谢你的建议。这不太正确，但对我来说似乎有意义。它只返回大约100个结果，但我估计结果集接近700个。而且更重要的是，结果是不准确的。它们既有属性值1319和1320，也有其他属性值。是的，我很少使用join，因为我没有看到它比我使用的方式有什么好处。如果它在这种情况下起作用，我会被说服的！谢谢你的建议。“不存在”似乎是个好主意，但我仍然不知道这将如何工作。AttributeValueID应该从REL_Aid_To_AttributeValue中选择，而AttributeTypeID只在TBL_AttributeValues中。这里不需要其他连接来关联这两个吗？这会起作用，因为它会拒绝AttributeTypeID属于给定类型但AttributeValueID不属于给定列表的行。对不起，我应该更好地解释它。我对上面的问题进行了编辑，以包括我需要从中获取信息的另一个表。AttributeValueID不在TBL_Aids中，AttributeTypeID也不在您所谓的AttributeValueID中（我想您指的是REL_Aid_To_AttributeValue？），没问题。我试图编辑答案以使用正确的表名，但结果对我来说太复杂了。我没弄错，“不存在”似乎是个好主意，但我还是不明白这是怎么回事。AttributeValueID应该从REL_Aid_To_AttributeValue中选择，而AttributeTypeID只在TBL_AttributeValues中。这里不需要其他连接来关联这两个吗？这会起作用，因为它会拒绝AttributeTypeID属于给定类型但AttributeValueID不属于给定列表的行。对不起，我应该更好地解释它。我对上面的问题进行了编辑，以包括我需要从中获取信息的另一个表。AttributeValueID不在TBL_Aids中，AttributeTypeID也不在您所谓的AttributeValueID中（我想您指的是REL_Aid_To_AttributeValue？），没问题。我试图编辑答案以使用正确的表名，但结果对我来说太复杂了。Quassnoi没有弄错。自行运行此子查询将返回除1319和1320（仍然是AttributeValueTypeID=1）之外的其他AttributeValueId的结果。那会发生吗？糟糕透了。这似乎是个好主意，但结果是0。谢谢你再试一次！单独运行此子查询将返回除1319和1320（仍然为AttributeValueTypeID=1）之外的其他AttributeValueId的结果。那会发生吗？糟糕透了。这似乎是个好主意，但结果是0。谢谢你再试一次！对不起，那部分我一定解释得不太清楚。我应该解释第三张表和标签