Question

需要一些查询帮助。我在数据库中有2个表，联系人和任务。

mysql> describe contacts;
+------------+--------------+------+-----+---------+----------------+
| Field      | Type         | Null | Key | Default | Extra          |
+------------+--------------+------+-----+---------+----------------+
| contact_id | int(11)      | NO   | PRI | NULL    | auto_increment |
| last_name  | varchar(100) | YES  |     | NULL    |                |
| first_name | varchar(100) | YES  |     | NULL    |                |
| email      | varchar(50)  | YES  |     | NULL    |                |
| phone      | varchar(20)  | YES  |     | NULL    |                |
| school_id  | varchar(12)  | NO   |     | NULL    |                |
| access     | char(1)      | YES  |     | NULL    |                |
+------------+--------------+------+-----+---------+----------------+

mysql> describe tasks;
+-------------------+--------------+------+-----+---------+----------------+
| Field             | Type         | Null | Key | Default | Extra          |
+-------------------+--------------+------+-----+---------+----------------+
| task_id           | int(11)      | NO   | PRI | NULL    | auto_increment |
| task_name         | varchar(125) | YES  |     | NULL    |                |
| task_date         | date         | YES  |     | NULL    |                |
| task_description  | text         | YES  |     | NULL    |                |
| contact_id1       | int(11)      | YES  |     | NULL    |                |
| contact_id2       | int(11)      | YES  |     | NULL    |                |
| contact_id3       | int(11)      | YES  |     | NULL    |                |
| contact_id4       | int(11)      | YES  |     | NULL    |                |
| contact_id5       | int(11)      | YES  |     | NULL    |                |
| contact_id6       | int(11)      | YES  |     | NULL    |                |
| completed         | char(1)      | YES  |     | NULL    |                |
+-------------------+--------------+------+-----+---------+----------------+

我想从tasks表中获取信息，但我想将contact_id1-contact_id6替换为匹配的contact_id中的一些字段。我一直在搞乱嵌套的select语句，这些语句可以工作但非常混乱。它似乎必须有一个更清洁的方法来做到这一点。我以为我有一些这个，

SELECT tasks.task_id, tasks.task_name, tasks.reminder_time, tasks.reminder_interval, CONCAT (contact_1.first_name, " ", contact_1.last_name) as contact_1_name, CONCAT(contact_2.first_name, contact_2.last_name) as contact_2_name, CONCAT(contact_3.first_name, contact_3.last_name) as contact_3_name
FROM tasks
JOIN contacts contact_1 ON tasks.contact_id1 = contact_1.contact_id
JOIN contacts contact_2 ON tasks.contact_id2 = contact_2.contact_id
JOIN contacts contact_3 ON tasks.contact_id3 = contact_3.contact_id

但我的问题是，它没有显示任何其中contact_id1-contact_id6的任何值的值为0的任务，如果没有为其设置contact_id，则该值是默认值。

您可以提供任何帮助都很棒。

Answer 1

您可以创建一个可以规范化任务表的视图，然后很容易进行连接。您可以创建永久视图，也可以仅选择。

视图需要以某种方式成为UNION

SELECT task_id, task_name, task_description, contact_id1 as contact_id, completed from tasks
UNION ALL
SELECT task_id, task_name, task_description, contact_id2              , completed from tasks
UNION ALL
SELECT task_id, task_name, task_description, contact_id3              , completed from tasks
UNION ALL
SELECT task_id, task_name, task_description, contact_id4              , completed from tasks
UNION ALL
SELECT task_id, task_name, task_description, contact_id5              , completed from tasks
UNION ALL
SELECT task_id, task_name, task_description, contact_id6              , completed from tasks

那么它看起来像这样

WITH tasks_easy as (    
    SELECT task_id, task_name, task_description, contact_id1 as contact_id, completed from tasks
    UNION
    SELECT task_id, task_name, task_description, contact_id2              , completed from tasks
    UNION
    SELECT task_id, task_name, task_description, contact_id3              , completed from tasks
    UNION
    SELECT task_id, task_name, task_description, contact_id4              , completed from tasks
    UNION
    SELECT task_id, task_name, task_description, contact_id5              , completed from tasks
    UNION
    SELECT task_id, task_name, task_description, contact_id6              , completed from tasks
 )
 select *
 from   tasks_easy t
 join   contacts c   on (t.contact_id = c.contact_id)

没有对此进行测试，但应该像这样工作。

编辑：其实正在思考它。 UNION ALL，至少对于临时视图来说是没有必要的。如果您的桌子很大，可能会为您带来一些时间的改善。

将2个表连接成单个查询结果

1 个答案: