是否有一个查询以基于以逗号分隔存储在字段中的值的计数将结果作为排序结果集?

问题描述 投票:0回答:1

嗨,我最近开始研究Symfony4。我有4个表是按照EAV规则设计的。 eav_entity,eav_attribute,eav_entity_instance,eav_value_text具有以下字段。


mysql> select * from eav_entity;
+----+---------+---------+---------------------+---------------------+
| id | code    | label   | created_at          | updated_at          |
+----+---------+---------+---------------------+---------------------+
|  5 | user    | User    | 2020-05-07 21:53:48 | 2020-05-07 21:53:48 |
|  6 | project | Project | 2020-05-07 21:53:48 | 2020-05-07 21:53:48 |
+----+---------+---------+---------------------+---------------------+


mysql> select * from eav_attribute;
+----+-----------+-----------------+-----------+-----------+-------------+---------------------+---------------------+
| id | entity_id | code            | type      | is_unique | is_required | created_at          | updated_at          |
+----+-----------+-----------------+-----------+-----------+-------------+---------------------+---------------------+
| 33 |         5 | full_name       | text      |         0 |           1 | 2020-05-07 21:59:20 | 2020-05-07 21:59:20 |
| 34 |         5 | email           | text      |         1 |           1 | 2020-05-07 22:00:32 | 2020-05-07 22:00:32 |
| 35 |         5 | phone_number    | varchar   |         1 |           1 | 2020-05-07 22:00:51 | 2020-05-07 22:00:51 |
| 36 |         5 | password        | varchar   |         1 |           1 | 2020-05-07 22:01:02 | 2020-05-07 22:01:02 |
| 43 |         6 | project_name    | varchar   |         0 |           1 | 2020-05-11 00:49:41 | 2020-05-11 00:49:41 |
| 44 |         6 | project_id      | varchar   |         1 |           1 | 2020-05-11 00:49:41 | 2020-05-11 00:49:41 |
| 45 |         6 | project_manager | varchar   |         0 |           1 | 2020-05-11 00:49:41 | 2020-05-11 00:49:41 |
| 46 |         6 | players         | text      |         0 |           1 | 2020-05-11 00:49:41 | 2020-05-11 00:49:41 |
| 47 |         5 | is_admin        | boolean   |         0 |           0 | 2020-05-12 10:21:33 | 2020-05-12 10:21:33 |
| 49 |         5 | is_enabled      | boolean   |         0 |           0 | 2020-05-16 17:34:03 | 2020-05-16 17:34:03 |
| 51 |         5 | last_activity   | timestamp |         0 |           0 | 2020-05-16 22:20:45 | 2020-05-16 22:20:45 |
| 58 |         5 | experience      | varchar   |         0 |           0 | 2020-05-18 18:26:30 | 2020-05-18 18:26:30 |
| 59 |         5 | branch          | varchar   |         0 |           0 | 2020-05-18 22:18:53 | 2020-05-18 22:18:53 |
+----+-----------+-----------------+-----------+-----------+-------------+---------------------+---------------------+

mysql> select * from eav_entity_instance;
+----+-----------+---------------------+---------------------+
| id | entity_id | created_at          | updated_at          |
+----+-----------+---------------------+---------------------+
| 38 |         5 | 2020-05-16 22:21:50 | 2020-05-16 22:21:50 |
| 39 |         5 | 2020-05-17 21:52:03 | 2020-05-17 21:52:03 |
| 40 |         5 | 2020-05-17 21:53:10 | 2020-05-17 21:53:10 |
| 41 |         6 | 2020-05-17 21:57:24 | 2020-05-17 21:57:24 |
| 42 |         6 | 2020-05-17 22:20:38 | 2020-05-17 22:20:38 |
| 53 |         5 | 2020-05-19 21:47:32 | 2020-05-19 21:47:32 |
| 54 |         5 | 2020-05-19 21:49:07 | 2020-05-19 21:49:07 |
| 55 |         5 | 2020-05-19 21:49:09 | 2020-05-19 21:49:09 |
| 56 |         5 | 2020-05-20 20:57:05 | 2020-05-20 20:57:05 |
+----+-----------+---------------------+---------------------+

mysql> select * from eav_value_text;
+----+-------------+--------------+----------------------+---------------------+---------------------+
| id | instance_id | attribute_id | value                | created_at          | updated_at          |
+----+-------------+--------------+----------------------+---------------------+---------------------+
| 63 |          38 |           33 | Raj                  | 2020-05-16 22:21:50 | 2020-05-16 22:21:50 |
| 64 |          38 |           34 | [email protected]        | 2020-05-16 22:21:50 | 2020-05-16 22:21:50 |
| 65 |          39 |           33 | Ank                  | 2020-05-17 21:52:03 | 2020-05-17 21:52:03 |
| 66 |          39 |           34 | [email protected]        | 2020-05-17 21:52:03 | 2020-05-17 21:52:03 |
| 67 |          40 |           33 | Bas                  | 2020-05-17 21:53:10 | 2020-05-17 21:53:10 |
| 68 |          40 |           34 | [email protected]        | 2020-05-17 21:53:10 | 2020-05-17 21:53:10 |
| 69 |          41 |           46 | 38                   | 2020-05-17 21:57:24 | 2020-05-17 21:57:24 |
| 70 |          42 |           46 | 38,39                | 2020-05-17 22:20:38 | 2020-05-17 22:20:38 |
| 81 |          53 |           34 | [email protected]        | 2020-05-19 21:47:32 | 2020-05-19 21:47:32 |
| 82 |          54 |           34 | [email protected]        | 2020-05-19 21:49:07 | 2020-05-19 21:49:07 |
| 83 |          55 |           34 | [email protected]        | 2020-05-19 21:49:09 | 2020-05-19 21:49:09 |
| 84 |          53 |           33 | Vij                  | 2020-05-19 21:59:35 | 2020-05-19 21:59:35 |
| 85 |          54 |           33 | Abd                  | 2020-05-19 22:04:59 | 2020-05-19 22:04:59 |
| 86 |          56 |           34 | [email protected]       | 2020-05-20 20:57:05 | 2020-05-20 20:57:05 |
| 87 |          55 |           33 | Jam                  | 2020-05-21 16:07:30 | 2020-05-21 16:07:30 |
+----+-------------+--------------+----------------------+---------------------+---------------------+

项目和用户是两个具有不同属性的实体。

在eav_entity_instance中,您可以看到用户和雇员具有不同的attribute_id。

我的问题是我需要生成查询,以便以没有用户参与的项目的方式返回用户的instance_id。

在eav_attribute表中,有一个名为player的属性,其ID为46。

此值存储在eav_value_text表中,其中用户的instance_id以逗号分隔。

任何人都可以帮助我进行查询,以使每个instance_id与玩家之间用逗号分隔的值匹配,并计算每个用户参与的项目数,并按计数的排序顺序返回结果。

例如,instance_id 38是user的用户,并且在2个项目中重复使用instance_id为41和42。因此,类似地,实例ID为39的用户只能在一个值为42的情况下找到。

所以,由于实例ID为38的用户有2个项目,实例ID为38的用户有1个项目,如果在ASC中对它进行排序,而在DESC中对它进行排序,则结果应如下所示。

+---+
|id +
+---+
|38 |
|39 |
+---+
php mysql sql symfony4
1个回答
0
投票

在SQL Server中,您可以使用功能STRING_SPLIT,但是谷歌快速搜索显示它在MySql中不存在。但是我的搜索结果确实指向this site,您可以通过以下方式使用信息来解决您的问题(SQL fiddle):

CREATE TABLE ProjectMembers 
    ( instance_id int
    , value varchar(500)
    );
INSERT INTO ProjectMembers(instance_id, value)
VALUES    (41, '38')
        , (42, '38,39');

/* Be sure that the amount of numbers in this table is at least as long as the 
   maximum amount of project members in a project */
CREATE TABLE Numbers
  (Number int);
INSERT INTO Numbers(Number)
VALUES (1), (2), (3), (4), (5), (6), (7), (8), (9);

SELECT Member, COUNT(*) AS Nr FROM ( 
  SELECT 
    instance_id, 
    substring_index(
      substring_index(value, ',', Number), 
      ','
      , -1
    ) AS Member
  FROM ProjectMembers
  JOIN Numbers
    ON char_length(value) 
      - char_length(replace(value, ',', '')) 
      >= Number - 1
) t
GROUP BY 1
ORDER BY Nr DESC;
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