mysql - 选择的结果是数组，我想将它用于 "where in"另一个选择

Question

我有两个表(t1 和 t2):

t1(第二列是数组)

name  | code        

ee    | 123, 124, 125

ef    | 121, 123    
______________________

t2

code_id | code_desc

121     | xxxxx        

123     | yyyyyyy      

124     | xxxxxxxx   

<小时/>

如果我执行此查询，一切正常:

SELECT * FROM t2 where code_id in (121,122)

但是如果我执行此查询，我会得到 NULL 单元格/结果

SELECT * FROM t2 where code_id in (SELECT code FROM t1 where name = ee)

如何从一个查询中获取两个表中的所有信息？

这是代码，我找不到好的sql在线工具

CREATE TABLE t1 (name VARCHAR(200), codes VARCHAR(200));
CREATE TABLE t2 (codes_id VARCHAR(200), codes_desc VARCHAR(200));

INSERT INTO t1 (name, codes) VALUES ('ee', '123,124,125');
INSERT INTO t1 (name, codes) VALUES ('ef', '121,124');
INSERT INTO t1 (name, codes) VALUES ('eh', '123,124,125');
INSERT INTO t2 (codes_id, codes_desc) VALUES ('121', 'yyyyyyyyy');
INSERT INTO t2 (codes_id, codes_desc) VALUES ('122', 'xxxxxxxxx');
INSERT INTO t2 (codes_id, codes_desc) VALUES ('123', 'zzzzzzzzzzz');



SELECT * FROM t2 where code_id in (121,122)
SELECT * FROM t2 where code_id in (SELECT codes FROM t1 where name = 'ee')

Answer 1

您可以使用find_in_set函数:

select *
from t2
where exists (
        select 1
        from t1
        where name = 'ee'
        and find_in_set(t2.code_id, t1.code) > 0
        )

不过，我建议您标准化您的表结构。因为即使上面的查询有效，它也是非 sargable .