我最近开始使用MySQL 5.7.10,我很喜欢原生的JSON数据类型。
但是在更新JSON类型值时我遇到了一个问题。
问题:
下面是表格格式,这里我想在data
表的JSON t1
列中再添加1个键。现在我必须获取值修改它并更新表。所以它涉及额外的SELECT
声明。
我可以像这样插入
INSERT INTO t1 values ('{"key2":"value2"}', 1);
mysql> select * from t1;
+--------------------+------+
| data | id |
+--------------------+------+
| {"key1": "value1"} | 1 |
| {"key2": "value2"} | 2 |
| {"key2": "value2"} | 1 |
+--------------------+------+
3 rows in set (0.00 sec)
mysql>Show create table t1;
+-------+-------------------------------------------------------------
-------------------------------------------------------+
| Table | Create Table |
+-------+--------------------------------------------------------------------------------------------------------------------+
| t1 | CREATE TABLE `t1` (
`data` json DEFAULT NULL,
`id` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1 |
+-------+--------------------------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)
有解决方法吗?
答案 0 :(得分:52)
感谢@wchiquito指出我正确的方向。我解决了这个问题。我就是这样做的。
mysql> select * from t1;
+----------------------------------------+------+
| data | id |
+----------------------------------------+------+
| {"key1": "value1", "key2": "VALUE2"} | 1 |
| {"key2": "VALUE2"} | 2 |
| {"key2": "VALUE2"} | 1 |
| {"a": "x", "b": "y", "key2": "VALUE2"} | 1 |
+----------------------------------------+------+
4 rows in set (0.00 sec)
mysql> update t1 set data = JSON_SET(data, "$.key2", "I am ID2") where id = 2;
Query OK, 1 row affected (0.04 sec)
Rows matched: 1 Changed: 1 Warnings: 0
mysql> select * from t1;
+----------------------------------------+------+
| data | id |
+----------------------------------------+------+
| {"key1": "value1", "key2": "VALUE2"} | 1 |
| {"key2": "I am ID2"} | 2 |
| {"key2": "VALUE2"} | 1 |
| {"a": "x", "b": "y", "key2": "VALUE2"} | 1 |
+----------------------------------------+------+
4 rows in set (0.00 sec)
mysql> update t1 set data = JSON_SET(data, "$.key3", "I am ID3") where id = 2;
Query OK, 1 row affected (0.07 sec)
Rows matched: 1 Changed: 1 Warnings: 0
mysql> select * from t1;
+------------------------------------------+------+
| data | id |
+------------------------------------------+------+
| {"key1": "value1", "key2": "VALUE2"} | 1 |
| {"key2": "I am ID2", "key3": "I am ID3"} | 2 |
| {"key2": "VALUE2"} | 1 |
| {"a": "x", "b": "y", "key2": "VALUE2"} | 1 |
+------------------------------------------+------+
4 rows in set (0.00 sec)
答案 1 :(得分:4)
现在,对于MySQL 5.7.22+,像这样在单个查询中更新json的整个片段(多个键值,甚至是嵌套的)非常容易和直接:
update t1 set data =
JSON_MERGE_PATCH(`data`, '{"key2": "I am ID2", "key3": "I am ID3"}') where id = 2;
希望它可以帮助访问此页面并寻找“更好的” JSON_SET
的人:)
有关JSON_MERGE_PATCH
的更多信息,请参见:
https://dev.mysql.com/doc/refman/5.7/en/json-modification-functions.html#function_json-merge-patch