Question

我的数据库中有这些表：

游客 - 这是第一张表

Tourist_ID - primary key
name...etc...

EXTRA_CHARGES

Extra_Charge_ID - primary key 
Extra_Charge_Description
Amount

Tourist_Extra_Charges

Tourist_Extra_Charge_ID
Extra_Charge_ID - foreign key
Tourist_ID - foreign key

所以这是示例

我和Tourist_ID有一个游客 - 86。这位身份为86的游客使用Extra_Charge_ID - 7和Extra_charge_ID - 11收取额外费用;

我尝试进行查询，以便我可以将游客的名字和EXTRA_CHARGES表格中的所有费用不属于这位游客。

这是我尝试的查询 - 但它不会返回任何内容。

SELECT
    Tourist.Name
    , EXTRA_CHARGES.Extra_Charge_Description
    , EXTRA_CHARGES.Amount 
FROM 
    Tourist 
    INNER JOIN TOURIST_EXTRA_CHARGES 
        ON Tourist.Tourist_ID = TOURIST_EXTRA_CHARGES.Tourist_ID 
    INNER JOIN EXTRA_CHARGES 
        ON TOURIST_EXTRA_CHARGES.Extra_Charge_ID = EXTRA_CHARGES.Extra_Charge_ID 
WHERE
    Tourist.Tourist_ID= 86 
    and EXTRA_CHARGES.Extra_Charge_ID NOT IN
    (   SELECT Extra_Charge_ID 
        FROM TOURIST_EXTRA_CHARGES te
        WHERE te.Tourist_ID = 86
    )

我当然只能通过此查询获得收费

SELECT * FROM EXTRA_CHARGES e
WHERE e.Extra_Charge_ID NOT IN
 (SELECT Extra_Charge_ID from TOURIST_EXTRA_CHARGES te

  WHERE te.Tourist_ID = 86
 )

但我无法找到获得这个游客名字的方法

Answer 1

您可以在select语句中包含旅游名称（对于Tourist_ID = 86）作为子查询：

SELECT (SELECT Tourist.Name FROM Tourist WHERE Tourist_ID=86) TouristName, e.*
FROM   EXTRA_CHARGES e
WHERE  e.Extra_Charge_ID NOT IN
       (SELECT Extra_Charge_ID 
        FROM   TOURIST_EXTRA_CHARGES te
        WHERE  te.Tourist_ID = 86
       )

Answer 2

你可以尝试一下它的效率更高：

SELECT t.Name 
FROM Tourist t JOIN 
 (SELECT * FROM EXTRA_CHARGES e 
 JOIN TOURIST_EXTRA_CHARGE tec ON e.Extra_Charge_ID = tec.Extra_Charge_ID AND tec.TOURIST_EXTRA_CHARGES != 86)
WHERE t.Tourist_ID = 86

顺便说一句。您不需要Tourist_Extra_Charges中的Tourist_Extra_Charge_ID列

Answer 3

您可以使用两个选项，两者都非常相似，但one may perform better than the other取决于您的DBMS。

两者的原则是相同的，获得旅游和额外费用的交叉加入，因此您需要为所有游客收取所有额外费用，然后使用NOT EXISTS或LEFT JOIN/IS NULL来消除所有额外费用游客有：

SELECT  Tourist.Name,
        EXTRA_CHARGES.Extra_Charge_Description,
        EXTRA_CHARGES.Amount 
FROM    Tourist
        CROSS JOIN EXTRA_CHARGES
WHERE   Tourist.Tourist_ID= 86 
AND     NOT EXISTS
        (   SELECT  1
            FROM    TOURIST_EXTRA_CHARGES
            WHERE   TOURIST_EXTRA_CHARGES.Tourist_ID = Tourist.Tourist_ID
            AND     TOURIST_EXTRA_CHARGES.Extra_Charge_ID = EXTRA_CHARGES.Extra_Charge_ID
        );


SELECT  Tourist.Name,
        EXTRA_CHARGES.Extra_Charge_Description,
        EXTRA_CHARGES.Amount 
FROM    Tourist
        CROSS JOIN EXTRA_CHARGES
        LEFT JOIN TOURIST_EXTRA_CHARGES
            ON TOURIST_EXTRA_CHARGES.Tourist_ID = Tourist.Tourist_ID
            AND TOURIST_EXTRA_CHARGES.Extra_Charge_ID = EXTRA_CHARGES.Extra_Charge_ID
WHERE   Tourist.Tourist_ID = 86 
AND     TOURIST_EXTRA_CHARGES.Tourist_Extra_Charge_ID IS NULL;

修改

由于您应用的两个条件在逻辑上是不同的，因此您需要使用两个查询来获取它。第一个是和以前一样，游客没有额外收费，第二个是收取所有额外费用的游客

SELECT Tourist.Name, EXTRA_CHARGES.Extra_Charge_Description, EXTRA_CHARGES.Amount FROM Tourist CROSS JOIN EXTRA_CHARGES LEFT JOIN TOURIST_EXTRA_CHARGES ON TOURIST_EXTRA_CHARGES.Tourist_ID = Tourist.Tourist_ID AND TOURIST_EXTRA_CHARGES.Extra_Charge_ID = EXTRA_CHARGES.Extra_Charge_ID WHERE Tourist.Tourist_ID = 1 AND TOURIST_EXTRA_CHARGES.Tourist_Extra_Charge_ID IS NULL UNION ALL SELECT Tourist.Name, NULL, NULL FROM Tourist CROSS JOIN EXTRA_CHARGES LEFT JOIN TOURIST_EXTRA_CHARGES ON TOURIST_EXTRA_CHARGES.Tourist_ID = Tourist.Tourist_ID AND TOURIST_EXTRA_CHARGES.Extra_Charge_ID = EXTRA_CHARGES.Extra_Charge_ID WHERE Tourist.Tourist_ID = 1 GROUP BY Tourist.Name HAVING COUNT(*) = COUNT(TOURIST_EXTRA_CHARGES.Tourist_Extra_Charge_ID);

<强> Example on SQL Fiddle

Answer 4

只需使用INNER JOIN ... ON .. <> ...

即可

SELECT
    Tourist.Name
    , EXTRA_CHARGES.Extra_Charge_Description
    , EXTRA_CHARGES.Amout 
FROM 
    Tourist 
    INNER JOIN TOURIST_EXTRA_CHARGES 
        ON Tourist.Tourist_ID <> TOURIST_EXTRA_CHARGES.Tourist_ID 
    INNER JOIN EXTRA_CHARGES 
        ON TOURIST_EXTRA_CHARGES.Extra_Charge_ID = EXTRA_CHARGES.Extra_Charge_ID 
WHERE
    Tourist.Tourist_ID= 86

见这里： http://sqlfiddle.com/#!2/28665/2

内部联接的sql子查询问题

4 个答案: