mysql> select * from tab_json;
+----+-----------------------------------------------------------------------------------+
| id | data |
+----+-----------------------------------------------------------------------------------+
| 1 | {"Tel": "132223232444", "name": "david", "address": "Beijing"} |
| 2 | {"Tel": "13390989765", "name": "Mike", "address": "Guangzhou"} |
| 3 | {"names": "Smith"} |
| 4 | {"names": "Smith", "address": "Beijing"} |
| 5 | {"names": "Smith", "address": "Beijing", "birthday": "2018-09-09"} |
| 6 | {"Max": "true", "names": "Smith", "address": "Beijing", "birthday": "2018-09-09"} |
| 7 | {"max": "true", "names": "Smith", "address": "Beijing", "birthday": "2018-09-09"} |
| 8 | {"oax": "true", "names": "Smith", "address": "Beijing", "birthday": "2018-09-09"} |
+----+-----------------------------------------------------------------------------------+
8 rows in set (0.00 sec)
2.将names的值与address的值进行合并
mysql> select json_extract(data,'$.names'),json_extract(data,'$.address') from tab_json;
+------------------------------+--------------------------------+
| json_extract(data,'$.names') | json_extract(data,'$.address') |
+------------------------------+--------------------------------+
| NULL | "Beijing" |
| NULL | "Guangzhou" |
| "Smith" | NULL |
| "Smith" | "Beijing" |
| "Smith" | "Beijing" |
| "Smith" | "Beijing" |
| "Smith" | "Beijing" |
| "Smith" | "Beijing" |
+------------------------------+--------------------------------+
8 rows in set (0.00 sec)
mysql> select json_merge(json_extract(data,'$.names'),json_extract(data,'$.address')) from tab_json;
+-------------------------------------------------------------------------+
| json_merge(json_extract(data,'$.names'),json_extract(data,'$.address')) |
+-------------------------------------------------------------------------+
| NULL |
| NULL |
| NULL |
| ["Smith", "Beijing"] |
| ["Smith", "Beijing"] |
| ["Smith", "Beijing"] |
| ["Smith", "Beijing"] |
| ["Smith", "Beijing"] |
+-------------------------------------------------------------------------+
8 rows in set (0.00 sec)
3.如果多个对象含有相同的key,那么也会进行合并为具体的values
mysql> SELECT JSON_MERGE('{"a": 1, "b": 2}', '{"c": 3, "a": 4}');
+----------------------------------------------------+
| JSON_MERGE('{"a": 1, "b": 2}', '{"c": 3, "a": 4}') |
+----------------------------------------------------+
| {"a": [1, 4], "b": 2, "c": 3} |
+----------------------------------------------------+
1 row in set (0.00 sec)
备注:将两个对象的值合并成一个,a这个key的值也增加到了2个.
文档创建:2018年6月6日17:49:18