Question

SELECT GUEST.FIRSTNAME,GUEST.LASTNAME, CHARGETYPE.DESCRIPTION
FROM GUEST
INNER JOIN CHARGES
ON GUEST.GUESTID = CHARGES.GUESTID
INNER JOIN CHARGETYPE
ON  CHARGETYPE.CHARGEID = CHARGES.CHARGEID
WHERE CHARGES.CHARGEID IN ('07','03' );

-DEAN JOHNSON晚餐服务

-JANE ADAMS晚餐服务

-PETER ADAMS晚餐服务

-JOHN SMITH晚餐服务

-DEAN JOHNSON ATHLETIC CHARGE

-JOHN SMITH ATHLETIC CHARGE

输出应该只显示DEAN和JOHN有运动和晚餐服务费。有没有简单的方法来解决这个问题？

Answer 1

在count(distinct col)子句中使用having：

select 
    guest.firstname
  , guest.lastname
from guest
  inner join charges 
    on guest.guestid = charges.guestid
  inner join chargetype 
    on chargetype.chargeid = charges.chargeid
where charges.chargeid in ('07', '03')
group by
    guest.firstname
  , guest.lastname
having count(distinct charges.chargeid) = 2;

Answer 2

使用复制连接非常简单

SELECT
  GUEST.FIRSTNAME,
  GUEST.LASTNAME,
  CHARGETYPE.DESCRIPTION
FROM GUEST
INNER JOIN CHARGES
  ON GUEST.GUESTID = CHARGES.GUESTID
INNER JOIN CHARGETYPE c1
  ON c1.CHARGEID = CHARGES.CHARGEID
  and c1.CHARGEID = '07'
INNER JOIN CHARGETYPE c2
  ON c2.CHARGEID = CHARGES.CHARGEID
  and c2.CHARGEID = '03'

Answer 3

我会为此添加一个案例，如果需要，您可以将其用作子查询，以过滤到那些有hadDinner = 1和isAnAthlete = 1的那些：

SELECT GUEST.FIRSTNAME
,GUEST.LASTNAME
, sum(case when CHARGETYPE.DESCRIPTION = 'DINNER SERVICE' then 1 else 0 end) hadDinner
, sum(case when CHARGETYPE.DESCRIPTION = 'ATHLETIC CHARGE' then 1 else 0 end) isAnAthlete
FROM GUEST
INNER JOIN CHARGES
ON GUEST.GUESTID = CHARGES.GUESTID
INNER JOIN CHARGETYPE
ON  CHARGETYPE.CHARGEID = CHARGES.CHARGEID
WHERE CHARGES.CHARGEID IN ('07','03' )
Group by GUEST.FIRSTNAME
,GUEST.LASTNAME;

编写一个SQL查询来显示同时收费的客人的姓名？

3 个答案: