MYSQL JSON字段操作

create

CREATE TABLE `t_test`  (
  `salary_data` json NULL COMMENT 'JSON类型字段'
);

insert

INSERT INTO t_test(salary_data) VALUES('{"param_1": "111","param_2": "222","param_3": "333","param_4": "444"}');
INSERT INTO t_test(salary_data) VALUES('{"param_1": "xxx","param_2": "yyy","param_3": "333","param_4": 444}');

select -> json_keys

SELECT json_keys(salary_data) from t_test;

select -> one param

SELECT salary_data->'$.param_4' FROM t_test;

select -> json_object

select salary_data from t_test; 

select -> where

SELECT salary_data->'$.param_4' FROM t_test where salary_data->'$.param_4'='444';
SELECT salary_data->'$.param_4' FROM t_test where salary_data->'$.param_4'=444;
SELECT salary_data->'$.param_4' FROM t_test where JSON_CONTAINS(salary_data, '"444"', '$.param_4');
SELECT salary_data->'$.param_4' FROM t_test where JSON_CONTAINS(salary_data, '444', '$.param_4');

update -> json_insert

UPDATE t_test SET salary_data = json_insert(salary_data, '$.param_5', 555, '$.param_6', 666) WHERE salary_data->'$.param_1'= 'xxx';

update -> json_set

UPDATE t_test SET salary_data = json_set(salary_data, '$.param_2', 222, '$.param_3', 333) WHERE salary_data->'$.param_1'= '111';

update -> json_remove

UPDATE t_test SET salary_data = json_remove(salary_data, '$.param_5', '$.param_6') WHERE salary_data->'$.param_1'= 'xxx';
posted @ 2019-09-18 14:50  BlueBerry006  Views(3837)  Comments(0Edit  收藏  举报