Mysql 根据嵌套查询的结果替换sql查询中的值

我有两个表：请求表和位置表

请求的样本数据

request id | requestor | locations
     1     |   ankur   |   2,5
     2     |   akshay  |   1  
     3     |   avneet  |   3,4
     4     |   priya   |   4     

位置的示例数据

loc_id     |  loc_name |
     1     |  gondor   
     2     |  rohan
     3     |  mordor   
     4     |  bree   
     5     |  shire    

我想查找特定位置的请求id。如果我使用location_id执行此操作，则会得到正确的结果

select request_id from request where locations like "%,3%" or locations like "%3,%";

此查询为我提供了针对位置id=3的请求

我如何才能为loc_名称实现这一点？将查询“like”部分中的数字替换为

select loc_id from locations where loc_name = "mordor"

---

这方面的任何帮助都会非常有用。谢谢。

如果位置是varchar字段，则解决方案

用一个相似的字符串连接表（在r.locations处用逗号开头和结尾）

代码未经测试：

SELECT
    r.request_id
FROM location l
INNER JOIN request r
    ON CONCAT(',', r.locations, ',') LIKE CONCAT('%,',l.loc_id,',%')
WHERE l.loc_name = 'mordor'

---

如果位置是varchar字段，则解决方案

用一个相似的字符串连接表（在r.locations处用逗号开头和结尾）

代码未经测试：

SELECT
    r.request_id
FROM location l
INNER JOIN request r
    ON CONCAT(',', r.locations, ',') LIKE CONCAT('%,',l.loc_id,',%')
WHERE l.loc_name = 'mordor'

---

您可以使用

FIND\u IN\u SET（）

这里是演示

但是，您最好通过引入一个看起来像的多对多表来规范化数据

CREATE TABLE request_location
(
  request_id INT NOT NULL,
  loc_id INT NOT NULL,
  PRIMARY KEY (request_id, loc_id),
  FOREIGN KEY (request_id) REFERENCES request (request_id),
  FOREIGN KEY (loc_id) REFERENCES locations (loc_id)
);

SELECT *
  FROM request_location rl JOIN request r 
    ON rl.request_id = r.request_id JOIN locations l
    ON rl.loc_id = l.loc_id
 WHERE l.loc_name = 'mordor'

从长远来看，这将带来巨大的回报，使您能够正常地维护和查询数据

那么您的查询可能如下所示

CREATE TABLE request_location
(
  request_id INT NOT NULL,
  loc_id INT NOT NULL,
  PRIMARY KEY (request_id, loc_id),
  FOREIGN KEY (request_id) REFERENCES request (request_id),
  FOREIGN KEY (loc_id) REFERENCES locations (loc_id)
);

SELECT *
  FROM request_location rl JOIN request r 
    ON rl.request_id = r.request_id JOIN locations l
    ON rl.loc_id = l.loc_id
 WHERE l.loc_name = 'mordor'

甚至

SELECT rl.request_id
  FROM request_location rl JOIN locations l
    ON rl.loc_id = l.loc_id
 WHERE l.loc_name = 'mordor';

如果您只需要返回

请求\u id

---

下面是演示

您可以使用

在集合中查找（）

这里是演示

但是，您最好通过引入一个看起来像的多对多表来规范化数据

CREATE TABLE request_location
(
  request_id INT NOT NULL,
  loc_id INT NOT NULL,
  PRIMARY KEY (request_id, loc_id),
  FOREIGN KEY (request_id) REFERENCES request (request_id),
  FOREIGN KEY (loc_id) REFERENCES locations (loc_id)
);

SELECT *
  FROM request_location rl JOIN request r 
    ON rl.request_id = r.request_id JOIN locations l
    ON rl.loc_id = l.loc_id
 WHERE l.loc_name = 'mordor'

从长远来看，这将带来巨大的回报，使您能够正常地维护和查询数据

那么您的查询可能如下所示

CREATE TABLE request_location
(
  request_id INT NOT NULL,
  loc_id INT NOT NULL,
  PRIMARY KEY (request_id, loc_id),
  FOREIGN KEY (request_id) REFERENCES request (request_id),
  FOREIGN KEY (loc_id) REFERENCES locations (loc_id)
);

SELECT *
  FROM request_location rl JOIN request r 
    ON rl.request_id = r.request_id JOIN locations l
    ON rl.loc_id = l.loc_id
 WHERE l.loc_name = 'mordor'

甚至

SELECT rl.request_id
  FROM request_location rl JOIN locations l
    ON rl.loc_id = l.loc_id
 WHERE l.loc_name = 'mordor';

如果您只需要返回
请求\u id

这里是演示
试试这个

SELECT  *
FROM    request 
        CROSS JOIN
        (
        SELECT  1 AS loc
        UNION ALL
        SELECT  2 AS loc
        UNION ALL
        SELECT  3 AS loc
        UNION ALL
        SELECT  4 AS loc
        UNION ALL
        SELECT  5 AS loc
        ) q
JOIN    Location
ON      loc_id  = CAST(NULLIF(SUBSTRING_INDEX(locations, ',', -loc), SUBSTRING_INDEX(locations, ',', 1 - loc)) AS UNSIGNED)

试试这个

SELECT  *
FROM    request 
        CROSS JOIN
        (
        SELECT  1 AS loc
        UNION ALL
        SELECT  2 AS loc
        UNION ALL
        SELECT  3 AS loc
        UNION ALL
        SELECT  4 AS loc
        UNION ALL
        SELECT  5 AS loc
        ) q
JOIN    Location
ON      loc_id  = CAST(NULLIF(SUBSTRING_INDEX(locations, ',', -loc), SUBSTRING_INDEX(locations, ',', 1 - loc)) AS UNSIGNED)