是否有一個查詢來根據字段中以逗號分隔的值的計數來獲取排序結果集的結果？

Question

嗨，我最近開始研究 Symfony 4。我有 4 個按照 EAV 規則設計的表。 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 | raj@gmail.com        | 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 | ank@gmail.com        | 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 | bas@gmail.com        | 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 | vij@gmail.com        | 2020-05-19 21:47:32 | 2020-05-19 21:47:32 |
| 82 |          54 |           34 | abd@gmail.com        | 2020-05-19 21:49:07 | 2020-05-19 21:49:07 |
| 83 |          55 |           34 | jam@gmail.com        | 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 | raja@gmail.com       | 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 |
+----+-------------+--------------+----------------------+---------------------+---------------------+

項目和用戶是具有不同屬性的兩個實體。

在 eav_entity_instance 中，您可以看到用戶和員工具有不同的屬性 ID。

我的問題是我需要生成查詢，以便以沒有項目用戶參與的方式返回用戶的 instance_id。

在 eav_attribute 表中有一個名為 player 的屬性，其 id 為 46。

this 的值存儲在 eav_value_text 表中，其中 user 的 instance_id 用逗號分隔。

任何人都可以幫我查詢，以便每個 instance_id 在玩家逗號分隔值中匹配，並計算每個用戶參與的項目數量，並按計數的排序順序返回結果。

例如 instance_id 38 是用戶的，它在 2 個項目中重復，他們的 instance_id 是 41 和 42。所以類似地，實例 id 為 39 的用戶只能在一個值為 42 的值中找到。

因此，由於實例 id 為 38 的用戶有 2 個項目，而實例 id 的用戶有 1 個項目，如果我們在 ASC 中排序，結果應該如下所示，如果我們在 DESC 中排序，結果應該如下所示。

+---+
|id +
+---+
|38 |
|39 |
+---+

Answer 1

在 SQL 服務器中，您可以使用 function STRING_SPLIT 但快速谷歌搜索顯示它在 MySql 中不存在。 但是我的搜索結果確實將我指向了這個站點您可以通過以下方式使用該信息來解決您的問題（ 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;