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跪求一句SQL语句坐等。解决办法

发布时间: 2012-05-28 17:59:33 作者: rapoo

跪求一句SQL语句,坐等。
4个表,

box是盒子,
note是消息,
box_user是哪些用户在哪些盒子里聊天。
box_note是哪些消息在哪些盒子里(消息可在多盒共享)。

盒子就和微信的一个聊天框一样,消息就是里面的消息。

SQL code
mysql> describe box;      +---------------------------+------------+------+-----+---------+----------------+| Field                     | Type       | Null | Key | Default | Extra          |+---------------------------+------------+------+-----+---------+----------------+| box_id                    | bigint(20) | NO   | PRI | NULL    | auto_increment || type                      | tinyint(4) | NO   |     | NULL    |                || status_type               | char(1)    | NO   |     | NULL    |                || create_time               | datetime   | NO   |     | NULL    |                || delete_time_from_one_part | datetime   | NO   |     | NULL    |                |+---------------------------+------------+------+-----+---------+----------------+5 rows in set (0.00 sec)mysql> describe box_user;+---------+------------+------+-----+---------+-------+| Field   | Type       | Null | Key | Default | Extra |+---------+------------+------+-----+---------+-------+| user_id | bigint(20) | NO   | PRI | 0       |       || box_id  | bigint(20) | NO   | PRI | 0       |       |+---------+------------+------+-----+---------+-------+2 rows in set (0.01 sec)mysql> describe note;+-------------+-------------+------+-----+---------+----------------+| Field       | Type        | Null | Key | Default | Extra          |+-------------+-------------+------+-----+---------+----------------+| note_id     | bigint(20)  | NO   | PRI | NULL    | auto_increment || user_id     | bigint(20)  | NO   | MUL | NULL    |                || type        | tinyint(4)  | NO   |     | NULL    |                || content     | text        | NO   |     | NULL    |                || mood        | tinyint(4)  | NO   |     | NULL    |                || locate      | varchar(30) | NO   |     | none    |                || privacy     | char(1)     | NO   |     | 1       |                || create_time | datetime    | NO   | MUL | NULL    |                || delay       | int(11)     | NO   |     | 0       |                || festival    | char(30)    | NO   |     | NULL    |                || delete_time | datetime    | NO   |     | NULL    |                |+-------------+-------------+------+-----+---------+----------------+11 rows in set (0.00 sec)mysql> describe box_note;+---------+------------+------+-----+---------+-------+| Field   | Type       | Null | Key | Default | Extra |+---------+------------+------+-----+---------+-------+| note_id | bigint(20) | NO   | PRI | 0       |       || box_id  | bigint(20) | NO   | PRI | 0       |       |+---------+------------+------+-----+---------+-------+2 rows in set (0.00 sec)


想法很简单,试了半天都失败:想根据user_id获取该user的所有盒子以及每个盒子最后一条消息的时间。

我准备了这两个基本SQL语句,
SQL code
//获取所有user_id用户的盒子select box.box_id,type,status_type from box,box_user where box.box_id=box_user.box_id and box_user.user_id=1;//获取所有内部具有消息的user_id用户的盒子以及最后更新时间select box_id,max(create_time) as time from note,box_note where note.note_id=box_note.note_id and note.user_id=1 group by box_id;

它们分别执行结果:
SQL code
mysql> select box.box_id,type,status_type from box,box_user where box.box_id=box_user.box_id and box_user.user_id=1;+--------+------+-------------+| box_id | type | status_type |+--------+------+-------------+|      1 |    0 | 0           ||      6 |    1 | 0           ||      7 |    3 | 0           ||      8 |    3 | 0           |+--------+------+-------------+4 rows in set (0.00 sec)mysql> select box_id,max(create_time) as time from note,box_note where note.note_id=box_note.note_id and note.user_id=1 group by box_id;+--------+---------------------+| box_id | time                |+--------+---------------------+|      1 | 2012-05-21 00:00:00 ||      6 | 2012-05-30 00:00:00 |+--------+---------------------+2 rows in set (0.00 sec) 



给这个例子的目的就是说,我要用left join把两个结果集串起来,以第一个结果集为基准,没有对应的最后时间就显示个null或者0000-00-00 00:00:00之类的都可以了。

我彻头彻尾的失败了,left join子查询一直报各种错,坐等解答。


[解决办法]
SELECT 1 box_id, 0
TYPE , 0 status_type
UNION ALL SELECT 6 , 1, 0
UNION ALL SELECT 7 , 3, 0
UNION ALL SELECT 8 , 3, 0

这是构造你的第一个SQL语句执行结果

我测试,懒得建表,插入数据【虽然你所贴很详细,但没有直接的SQL语句】

后面那个()里的union all类似的,构造出第二个语句结果
[解决办法]
你应该贴出你的
create table.............语句,
inset into table..................语句。

还有期望结果及简单描述。这样别人可以搭建与你相同的环境进行测试。

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