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在 MySQL 中过滤一对多关系

[英]Filtering on a one-to-many relationship in MySQL

I have a question regarding filtering in MySQL.我有一个关于在 MySQL 中过滤的问题。 I have two tables one with members and one with membership data.我有两张表,一张是会员,一张是会员数据。 I like to be able to join the two tables and be able to filter on:我喜欢能够加入两个表并能够过滤:

  • members that still have an active membership (date of October 6th 2020 as reference date)仍为活跃会员的会员(以 2020 年 10 月 6 日为参考日期)
  • memebers that do not have an active membership anymore不再是活跃会员的会员

Can anyone give me the correct SQL code the filter on this?任何人都可以给我正确的 SQL 代码过滤器吗? I am not realy good in this more advanced SQL statements.我不太擅长这个更高级的 SQL 语句。 Many thanks in advance!提前谢谢了!

Members会员

+----+------------+
| ID |   Name     |
+----+------------+
| 01 | Gerico     |
| 02 | Stefan     |
+----+------------+

Membership会员资格

+----+------------+-------------+------------+
| ID | MemberID   | From        | To         |
+----+------------+-------------+------------+
| 01 |         01 | 01/01/1990  | 01/01/2000 |
| 02 |         01 | 01/01/2005  | 31/12/2154 |
| 03 |         02 | 01/01/1990  | 01/01/2000 |
| 04 |         02 | 01/01/1992  | 31/12/1999 |
+----+------------+-------------+------------+

I am assuming an active member is one who has a membership whose date range includes the current date.我假设一个活跃会员是一个会员,其日期范围包括当前日期。 Then to select active members:然后选择活跃成员:

SELECT DISTINCT Members.* FROM
Members JOIN Membership ON Members.ID = Membership.MemberID
WHERE CURDATE() BETWEEN Membership.`From` AND Membership.`To`

(A SELECT DISTINCT is done in case a member has multiple, overlapping memberships.) (如果成员具有多个重叠的成员资格,则执行SELECT DISTINCT 。)

Or you can use:或者你可以使用:

SELECT * FROM Members
WHERE EXISTS (
    SELECT ID FROM Membership WHERE CURDATE() BETWEEN `From` AND `To` AND MemberID = Members.ID
)

To select non-active members:选择非活跃成员:

SELECT * FROM Members
WHERE NOT EXISTS (
    SELECT ID FROM Membership WHERE CURDATE() BETWEEN `From` AND `To` AND MemberID = Members.ID
)

See db-fiddle见 db-fiddle

The comment made by @Strawberry refers to the responsibility of the OP to create the above db-fiddle or equivalent for people to be able to test against us. @Strawberry 发表的评论指的是 OP 有责任创建上述 db-fiddle 或等效项,以便人们能够对我们进行测试。

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