如何将选择语句结果转换为JSONARRAY并更新另一个表

Question

我想将选择结果转换为JSON并将其写入另一个表：

update patrol_patrol a, position_user b
set a.route = json_array(select coordinate from b )
where a.id = 1;

并得到错误：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'select coordinate from b ) where a.id = 1' at line 2 

 select route from patrol_patrol;                                                                                                                                                    
+--------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| route                                                                                                                                                              |
+--------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| ["112.58006496213066,22.311484443420195"] |
+--------------------------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set

 select coordinate from position_user;                                                                                                                                                    
+---------------------------------------+
| coordinate                            |
+---------------------------------------+
| 112.701036,22.738611                  |
| 112.701036,22.738632                  |
| 112.701036,22.738632                  |
| 112.701036,22.738652                  

position_user.coordinate应该是[“ 112.701036,22.738611”，“ 112.701036,22.738632”，“ 112.701036,22.738652”，....]

Answer 1

由于您仅更新patrol_patrol表，因此只应将其包括在update语句的第一部分中。为了获得所需的信息，我建议使用JSON_ARRAYAGG函数，该函数会将您的结果组合到一个数组中，然后可以将其用于将结果分配给a.route：

UPDATE patrol_patrol a
SET a.route = (SELECT JSON_ARRAYAGG(coordinate) FROM position_user)
WHERE a.id = 1;

可以找到here的dbfiddle来证明这种方法。