需求描述:

在看mysql中关于json的内容,通过json_replace函数可以实现对json值的替换,

在此记录下.

操作过程:

1.查看带有json数据类型的表

mysql> select * from tab_json;

+----+---------------------------------------------------------------------------------------+

| id | data |

+----+---------------------------------------------------------------------------------------+

| 1 | {"age": "33", "tel": 13249872314, "passcode": "654567"} |

| 2 | {"age": "33", "tel": 189776542, "name": "David", "olds": "12", "address": "Hangzhou"} |

+----+---------------------------------------------------------------------------------------+

2 rows in set (0.00 sec)

2.使用json_replace函数对json值进行操作

mysql> select json_replace(data,'$.age',54,'$.tel',15046464563) from tab_json where id = 1; #使用json_replace进行查询处理,对已经存在的key值进行替换

+-------------------------------------------------------+

| json_replace(data,'$.age',54,'$.tel',15046464563) |

+-------------------------------------------------------+

| {"age": 54, "tel": 15046464563, "passcode": "654567"} |

+-------------------------------------------------------+

1 row in set (0.00 sec)

mysql> select json_replace(data,'$.age',54,'$.tel',15046464563,'$.sex',"male") from tab_json where id = 1; #对于不存在key,是没有增加新的key-value值的

+------------------------------------------------------------------+

| json_replace(data,'$.age',54,'$.tel',15046464563,'$.sex',"male") |

+------------------------------------------------------------------+

| {"age": 54, "tel": 15046464563, "passcode": "654567"} |

+------------------------------------------------------------------+

1 row in set (0.00 sec)

3.通过update语句对json中的值进行替换操作

mysql> update tab_json set data = json_replace(data,'$.age',54,'$.tel',15046464563) where id = 1; #对id=1的行进行更新操作,更新之后,age和tel的值发生了变化

Query OK, 1 row affected (0.10 sec)

Rows matched: 1 Changed: 1 Warnings: 0

mysql> select * from tab_json;

+----+---------------------------------------------------------------------------------------+

| id | data |

+----+---------------------------------------------------------------------------------------+

| 1 | {"age": 54, "tel": 15046464563, "passcode": "654567"} |

| 2 | {"age": "33", "tel": 189776542, "name": "David", "olds": "12", "address": "Hangzhou"} |

+----+---------------------------------------------------------------------------------------+

2 rows in set (0.00 sec)

mysql> update tab_json set data = json_replace(data,'$.age',54,'$.tel',15046464563,'$.sex',"male") where id = 1; 对id=1的行进行更新操作,更新之后,age和tel的值发生了变化,但是并没有增加新的key

Query OK, 0 rows affected (0.00 sec)

Rows matched: 1 Changed: 0 Warnings: 0

mysql> select * from tab_json;

+----+---------------------------------------------------------------------------------------+

| id | data |

+----+---------------------------------------------------------------------------------------+

| 1 | {"age": 54, "tel": 15046464563, "passcode": "654567"} |

| 2 | {"age": "33", "tel": 189776542, "name": "David", "olds": "12", "address": "Hangzhou"} |

+----+---------------------------------------------------------------------------------------+

2 rows in set (0.00 sec)

备注:所以json_replace的主要作用是替换,如果存在key就替换对应的值,如果不存在key也不会增加,与json_insert的使用有区别.

文档创建时间:2018年6月6日09:48:10