我有一个json类型的MySQL列。在那里,存储了像JSON对象这样的字典。现在,我想从此JSON对象提取值并创建一个JSON数组。
我该如何实现?
with json_objs(json_col) as (
select CAST('{"key1": "value1", "key2": "value2"}' AS JSON)
UNION ALL
select CAST('{"key3": "value3", "key4": "value4"}' AS JSON)
)
select SOME_EXPR_I_CAN_T_FIGURE_OUT from json_objs
+----------------------+
| resulting_column |
+----------------------+
| ["value1", "value2"] |
| ["value3", "value4"] |
+----------------------+
CREATE TABLE `json_objs` (
`json_col` json DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8
答案 0 :(得分:2)
您可以像在表达式->
中那样使用column -> path
运算符,如下所示:
create table table1 (
json_dict JSON
);
insert into table1 values('{"ak":"av","bk":"bv"}');
insert into table1 values('{"ak2":"av2","bk2":"bv2"}');
select * from table1;
+------------------------------+
| json_dict |
+------------------------------+
| {"ak": "av", "bk": "bv"} |
| {"ak2": "av2", "bk2": "bv2"} |
+------------------------------+
2 rows in set (0.00 sec)
select json_dict->"$.*" from table1;
+------------------+
| json_dict->"$.*" |
+------------------+
| ["av", "bv"] |
| ["av2", "bv2"] |
+------------------+
2 rows in set (0.00 sec)
https://dev.mysql.com/doc/refman/8.0/en/json-search-functions.html#operator_json-column-path
答案 1 :(得分:1)
除了->
运算符之外,您还可以使用Json_Extract()
函数:
模式(MySQL v8.0)
create table table1 (
json_dict JSON
);
insert into table1 values('{"ak":"av","bk":"bv"}');
insert into table1 values('{"ak2":"av2","bk2":"bv2"}');
查询#1
select JSON_EXTRACT(json_dict, '$.*') from table1;
| JSON_EXTRACT(json_dict, '$.*') |
| ------------------------------ |
| ["av", "bv"] |
| ["av2", "bv2"] |