4

我有 2 张桌子:

客户表:

siteid nvarchar(2) PRIMARY KEY,
custid int PRIMARY KEY,
fname varchar(30),
lname varchar(30)

儿童表:

childid1 nvarchar(2) PRIMARY KEY,
childid2 int PRIMARY KEY,
siteid nvarchar(2),
custid int,
lname varchar(30),
lname varchar(30),
FOREIGN KEY(siteid)REFERENCES Cust(siteid),
FOREIGN KEY(custid)REFERENCES Cust(custid)

价值观:

客户:

siteid | custid | fname | lname
A1     | 1111   | John  | S
A2     | 1111   | Steve | H
B1     | 2222   | Paul  | N
C3     | 3333   | Mary  | J

孩子们:

childid1 | childid2 | siteid | custid | fname | lname
A6       | 1010     | A1     | 1111   | Lisa  | S
A8       | 1011     | A1     | 1111   | Linda | S
A9       | 1012     | A1     | 1111   | Jose  | S
D9       | 1013     | A2     | 1111   | Jake  | H
D1       | 1014     | B1     | 2222   | Judy  | N
D1       | 1015     | B1     | 2222   | Judy  | N

我正在寻找没有孩子的卡斯特,这是我的查询:

SELECT * FROM Cust WHERE Cust.siteid NOT IN(
SELECT Children.siteid FROM Children
) AND Cust.custid NOT IN(
SELECT Children.custid FROM Children
)

但结果是空的。由于表具有复合主键,这里的正确查询是什么?

4

2 回答 2

4

表没有 2 个主键。有些有复合主键,这里似乎就是这种情况。如果Cust(siteid, custid)具有PRIMARY KEY

CREATE TABLE Cust
( siteid nvarchar(2),
  custid int,
  fname varchar(30),
  lname varchar(30),
  PRIMARY KEY (siteid, custid)              -- one primary key
) ;

那么你的外键定义是错误的。您应该有一个(复合)外键,引用(复合)主键:

CREATE TABLE Children
( childid1 nvarchar(2),
  childid2 int,
  siteid nvarchar(2),
  custid int,
  fname varchar(30),
  lname varchar(30),
  PRIMARY KEY (childid1, childid2),
  FOREIGN KEY (siteid, custid)               -- one foreign key 
    REFERENCES Cust(siteid, custid)
) ;

那么编写查询的一种方法是(更正:这是 ANSI SQL,适用于其他 DBMS,但不适用于 SQL-Server):

SELECT * 
FROM Cust 
WHERE (siteid, custid) NOT IN
      ( SELECT siteid, custid  
        FROM Children
      ) ;

或者更好,NOT EXISTS因为它避免了可能的NULL值的陷阱,这将使NOT IN版本显示非预期的结果(第二,因为 NOT IN 在 SQL-Server 中不起作用:)

SELECT * 
FROM Cust AS c
WHERE NOT EXISTS
      ( SELECT * 
        FROM Children AS ch
        WHERE ch.siteid = c.siteid
          AND ch.custid = c.custid
      ) ;

SQL-Fiddle中测试

于 2013-05-27T19:14:28.163 回答
2

表达查询的方式不止一种。在选择一个之前,请查看用于指定集群和非集群主键的选项、执行计划以及 ypercube 关于表的注释。(我相信这NOT IN不是可耻的。)

select cust.siteid, cust.custid
from cust
left join children
       on cust.siteid = children.siteid
      and cust.custid = children.custid
where children.siteid is null;
于 2013-05-27T19:23:26.333 回答