请教一个 关联查询 的 条件问题,再线等,急 - V2EX
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1010011010
V2EX    MySQL

请教一个 关联查询 的 条件问题,再线等,急

  •  
  •   1010011010 2017-03-06 18:12:06 +08:00 3348 次点击
    这是一个创建于 3145 天前的主题,其中的信息可能已经有所发展或是发生改变。

    这里有三个表:

    a 表 +----+ | id | +----+ | 1 | | 2 | | 3 | +----+ b 表 +----+------+------+ | id | a_id | c_id | +----+------+------+ | 1 | 1 | 1 | | 2 | 1 | 2 | | 3 | 2 | 2 | | 4 | 2 | 3 | | 5 | 3 | 1 | | 6 | 3 | 3 | +----+------+------+ c 表 +----+ | id | +----+ | 1 | | 2 | | 3 | +----+ 

    b 表 关联了 a 表 和 c 表

    查询全表时:

    SELECT a.id, group_concat(c.id) FROM a LEFT JOIN b ON b.a_id = a.id LEFT JOIN c ON c.id = b.c_id GROUP BY a.id; +----+--------------------+ | id | group_concat(c.id) | +----+--------------------+ | 1 | 1,2 | | 2 | 2,3 | | 3 | 1,3 | +----+--------------------+ 

    然后有条件: where c.id = 2

    我想要的:

    +----+--------------------+ | id | group_concat(c.id) | +----+--------------------+ | 1 | 1,2 | | 2 | 2,3 | +----+--------------------+ 

    然而结果是:

    +----+--------------------+ | id | group_concat(c.id) | +----+--------------------+ | 1 | 2 | | 2 | 2 | +----+--------------------+ 

    如何写 查询条件?

    源氏大雕

    第 1 条附言    2017-03-07 18:12:51 +08:00

    结果:

    先倒过来筛选出 符合条件的 a表 再查询

    SELECT a1.id, group_concat(c.id) FROM ( SELECT a.id FROM a LEFT JOIN b ON b.a_id = a.id LEFT JOIN c ON c.id = b.c_id WHERE c.id = 2 ) AS a1 #筛选出的a表 LEFT JOIN b ON b.a_id = a1.id LEFT JOIN c ON c.id = b.c_id GROUP BY a1.id 

    还有其他答案吗

    6 条回复    2017-03-06 20:50:49 +08:00
    noNOno
        1
    noNOno  
       2017-03-06 18:22:04 +08:00
    b.c_id=2
    1010011010
        2
    1010011010  
    OP
       2017-03-06 18:22:39 +08:00
    @noNOno #1
    一样的。
    noNOno
        3
    noNOno  
       2017-03-06 18:28:32 +08:00
    @1010011010 不连 a 表应该就对了,我在脑补...
    zeraba
        4
    zeraba  
       2017-03-06 18:55:42 +08:00 via Android
    SELECT b.a_id, group_concat(t1.c_id)
    from b
    left join (select a_id from b where b_id = 2) t1
    on t1.a_id = b.a_id
    left join c on c.id = b.c_id
    没测试 大概是需要子查询 拿到 c.id = 2 之后的 b.a_id 然后再合并
    1010011010
        5
    1010011010  
    OP
       2017-03-06 19:57:49 +08:00
    @zeraba #4

    由此扩展得

    SELECT a1.id, group_concat(c.id)

    FROM
    (
    SELECT a.id

    FROM a

    LEFT JOIN b ON b.a_id = a.id

    WHERE b.c_id = 2
    )
    AS a1

    LEFT JOIN b ON b.a_id = a1.id

    LEFT JOIN c ON c.id = b.c_id

    GROUP BY a1.id

    先倒过来筛选出 符合条件的 a 表 再查询

    结果是对的

    有没有更简单的

    (つ*))
    zeraba
        6
    zeraba  
       2017-03-06 20:50:49 +08:00
    @1010011010
    '''
    SELECT
    b.a_id,
    group_concat(b.c_id)
    FROM
    (SELECT a_id FROM b WHERE c_id = 2) t1
    LEFT JOIN b ON t1.a_id = b.a_id
    LEFT JOIN c ON c.id = b.c_id
    GROUP BY
    b.a_id;

    理论上更优?本来要打这个的
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