MySQL5.7中的JSON基本操作指南(2)-木庄网络博客

创建虚拟列:

mysql> ALTER TABLE `test_user` ADD `nick_name` VARCHAR(50) GENERATED ALWAYS AS (info->>'$.nick_name') VIRTUAL;

注意用操作符->>

使用时和普通类型的列查询是一样:

mysql> select name,nick_name from test_user where nick_name='小萌'; 
+----------+-----------+ 
| name  | nick_name | 
+----------+-----------+ 
| xiaoming | 小萌  | 
+----------+-----------+ 
1 row in set (0.05 sec)

更新

使用JSON_INSERT()插入新值，但不会覆盖已经存在的值

mysql> UPDATE test_user SET info = JSON_INSERT(info, '$.sex', 1, '$.nick_name', '小花') where id=2;

看下结果

mysql> select * from test_user where id=2; 
+----+---------+--------------------------------------------+-----------+ 
| id | name | info          | nick_name | 
+----+---------+--------------------------------------------+-----------+ 
| 2 | xiaohua | {"age": 17, "sex": 0, "nick_name": "小花"} | 小花  | 
+----+---------+--------------------------------------------+-----------+ 
1 row in set (0.06 sec)

使用JSON_SET()插入新值，并覆盖已经存在的值

mysql> UPDATE test_user SET info = JSON_INSERT(info, '$.sex', 0, '$.nick_name', '小张') where id=3;

看下结果

mysql> select * from test_user where id=3; 
+----+-----------+---------------------------------------------------------------+-----------+ 
| id | name  | info               | nick_name | 
+----+-----------+---------------------------------------------------------------+-----------+ 
| 3 | xiaozhang | {"age": 19, "sex": 1, "tag": [3, 5, 90], "nick_name": "小张"} | 小张  | 
+----+-----------+---------------------------------------------------------------+-----------+ 
1 row in set (0.06 sec)

使用JSON_REPLACE()只替换存在的值

mysql> UPDATE test_user SET info = JSON_REPLACE(info, '$.sex', 1, '$.tag', '[1,2,3]') where id=2;

看下结果

mysql> select * from test_user where id=2; 
+----+---------+--------------------------------------------+-----------+ 
| id | name | info          | nick_name | 
+----+---------+--------------------------------------------+-----------+ 
| 2 | xiaohua | {"age": 17, "sex": 1, "nick_name": "小花"} | 小花  | 
+----+---------+--------------------------------------------+-----------+ 
1 row in set (0.06 sec)

可以看到tag没有更新进去

删除

使用JSON_REMOVE()删除JSON元素

mysql> UPDATE test_user SET info = JSON_REMOVE(info, '$.sex', '$.tag') where id=1;

看下结果

mysql> select * from test_user where id=1; 
+----+----------+----------------------------------+-----------+ 
| id | name  | info        | nick_name | 
+----+----------+----------------------------------+-----------+ 
| 1 | xiaoming | {"age": 18, "nick_name": "小萌"} | 小萌  | 
+----+----------+----------------------------------+-----------+ 
1 row in set (0.05 sec)

总结

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