为什么使用另一张表的筛选结果，放到where id in （）里面，结果不对？

Question

为什么使用另一张表的筛选结果，放到where id in （）里面，结果不对？

比如现在固定筛选，5514，5515这两个id是可以的。

mysql> select id from terminal where id in (5514,5515);
+------+
| id   |
+------+
| 5514 |
| 5515 |
+------+
2 rows in set (0.08 sec)

另一张表的某个json里面有个字段，记录的这个id表，也没问题，能查出来5514，5515

mysql> select pgrpdevlist ->> '$.devicelist' from playproject where id = 13;
+--------------------------------+
| pgrpdevlist ->> '$.devicelist' |
+--------------------------------+
| 5514,5515                      |
+--------------------------------+
1 row in set (0.06 sec)

但是两个结合起来就有问题了，只能读出来一个。

mysql> select id from terminal where id in (select pgrpdevlist ->> '$.devicelist' from playproject where id = 13);
+------+
| id   |
+------+
| 5514 |
+------+
1 row in set (0.09 sec)

Answer 1

mysql> select pgrpdevlist ->> '$.devicelist' from playproject where id = 13;
这个结果是 5514,5515 数据类型是varchar吧、
所以mysql> select id from terminal where id in (select pgrpdevlist ->> '$.devicelist' from playproject where id = 13);
= select id from terminal where id in (‘5514,5515’);