MYSQL获取观看过所有足球比赛的人

Question

fan(*IDfan*,name,surname)
match(*IDmatch*, type_of_competition, teamA, teamB, date, scoreA, scoreB)
booking(*IDfan,IDmatch*,price)...
 *name* : name is the primary key ` 

例如。

粉丝:

IDfan     Name   Surname
---------------------
1     | A      |A
2     | B      |B
3     | C      |C

匹配:

IDmatch   type        teamA    teamB         date...
----------------------------------------------------
1     | champions |Juve     |Barcellona |01/10/17
2     | league    |Milan    |Real       |02/11/17
3     | TIM cup   |Inter    |Napoli     |03/12/17

预订:

IDfan     IDmatch   price
---------------------
1     | 1      |5
1     | 2      |6
1     | 3      |7
2     |1       |8
2     |1       |8
2     |1       |8
3     |2       |5
3     |3       |10

早上好，考虑到球迷可以为一场比赛购买更多门票，我不知道如何选择观看所有比赛的球迷。例如，应用于上述数据库的脚本的结果必须是:

IDfan 姓名

<小时/>

1     | A      |A

Answer 1

已编辑

在 booking 表中，相同值的行出现 3 次，这很奇怪。

IDfan     IDmatch   price
---------------------
2     |1       |8
2     |1       |8
2     |1       |8

假设三次出现没有错， 这将是正确的答案(未经测试)。

SELECT fan.*
FROM (SELECT DISTINCT IDfan, IDmatch
    FROM booking) AS distinct_booking
JOIN fan ON distinct_booking.IDfan = fan.IDfan
GROUP BY fan.IDfan, fan.Name, fan.Surname
HAVING COUNT(*) = (SELECT COUNT(*) FROM match)
;

另一方面，假设booking表具有IDfan和IDmatch的唯一约束，则三次出现是错误的， 尝试这个。经过测试并有效

SELECT fan.*
FROM booking
JOIN fan ON booking.IDfan = fan.IDfan
GROUP BY fan.IDfan, fan.Name, fan.Surname
HAVING COUNT(*) = (SELECT COUNT(*) FROM match)
;

对于 mysql，这也可以工作(已测试)

SELECT fan.*
FROM booking
JOIN fan ON booking.IDfan = fan.IDfan
GROUP BY fan.IDfan
HAVING COUNT(*) = (SELECT COUNT(*) FROM match)
;

--

对此评论的解释。

Can you explain me why you did this: GROUP BY fan.IDfan, fan.Name, fan.Surname ?

某些数据库有一个规则:SELECT 中的列必须位于 GROUP BY 子句中。

为了获取 IDfan Name Surname，我编写了 GROUP BY fan.IDfan, fan.Name, fan.Surname 但mysql不是这样，如果主键在GROUP BY中就可以了。

所以，我再次阅读了你的问题，我意识到你使用mysql，那么这也适用于mysql。

SELECT fan.*
FROM booking
JOIN fan ON booking.IDfan = fan.IDfan
GROUP BY fan.IDfan
HAVING COUNT(*) = (SELECT COUNT(*) FROM match)
;