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  • 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';
    
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  • 原文地址:https://www.cnblogs.com/blueberry006/p/11542157.html
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