在MySQL中解析JSON或将表中字段值合并为JSON问题

CREATE TABLE my_table (
    id INT PRIMARY KEY,
    json_data JSON
);

INSERT INTO my_table (id, json_data) VALUES
(1, '{"name": "John", "age": 25, "city": "New York"}'),
(2, '{"name": "Alice", "age": 30, "city": "Los Angeles"}');

-- 提取name字段的值
SELECT id, JSON_UNQUOTE(JSON_EXTRACT(json_data, '$.name')) AS name
FROM my_table;

-- 提取age字段的值
SELECT id, JSON_EXTRACT(json_data, '$.age') AS age
FROM my_table;

-- 提取city字段的值
SELECT id, JSON_UNQUOTE(JSON_EXTRACT(json_data, '$.city')) AS city
FROM my_table;

-- 提取name字段的值
SELECT id, json_data->'$.name' AS name
FROM my_table;

-- 提取age字段的值
SELECT id, json_data->'$.age' AS age
FROM my_table;

-- 提取city字段的值
SELECT id, json_data->'$.city' AS city
FROM my_table;

CREATE TABLE person (
    id INT PRIMARY KEY,
    name VARCHAR(50),
    age INT,
    city VARCHAR(50)
);

INSERT INTO person (id, name, age, city) VALUES
(1, 'John', 25, 'New York'),
(2, 'Alice', 30, 'Los Angeles');

-- 将id、name、age、city字段合并为JSON字符串
SELECT id, JSON_OBJECT('name', name, 'age', age, 'city', city) AS json_data
FROM person;

+----+---------------------------------------------+
| id | json_data                                   |
+----+---------------------------------------------+
|  1 | {"name": "John", "age": 25, "city": "New York"} |
|  2 | {"name": "Alice", "age": 30, "city": "Los Angeles"} |
+----+---------------------------------------------+