目录
- 1. 基本知识
- 2. Demo
- 2.1 简单聚合
- 2.2 带排序聚合
- 2.2.1 子查询进行排序
- 2.2.2 创建临时表
- 2.3 带条件聚合
- 2.4 多列聚合
- 2.5 嵌套 JSON 结构
1. 基本知识
JSON_ARRAYAGG
为 SQL 聚合函数,用于将一组值聚合为一个 JSON 数组
- 多行结果组合成一个 JSON 数组形式的场景中非常有用
JSON_ARRAYAGG
可以与其他 JSON 处理函数(如JSON_OBJECTAGG
)结合使用,以构建复杂的 JSON 结构
其语法结构如下:
JSON_ARRAYAGG(expression [ORDER BY ...])
expression
:要聚合的列或表达式ORDER BY
:可选,指定聚合值的排序顺序
2. Demo
为更好的加深印象,以Demo的方式进行展示
示例如下:
CREATE TABLE employees (id INT PRIMARY KEY,name VARCHAR(100),department VARCHAR(100),salary DECIMAL(10, 2)
);INSERT INTO employees (id, name, department, salary) VALUES
(1, 'Alice', 'Engineering', 60000),
(2, 'Bob', 'Engineering', 70000),
(3, 'Charlie', 'HR', 50000),
(4, 'David', 'Engineering', 80000),
(5, 'Eve', 'HR', 55000);
由于我的Navicat版本较低,无法输出JSON_ARRAYAGG的相关结果
- 要么使用其他工具,要么升级Navicat软件(后续以命令行的结果进行展示)
2.1 简单聚合
SELECT JSON_ARRAYAGG(name ORDER BY salary DESC) AS employees_names
FROM employees;
截图如下:
2.2 带排序聚合
内部嵌套排序,有些数据库是不支持的,即使8的版本号
SELECT JSON_ARRAYAGG(name ORDER BY salary DESC) AS employees_names
会输出如下:
[Err] 1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ORDER BY salary DESC) AS employees_names
FROM employees' at line 1
对于上述情况,使用如下方式进行处理
2.2.1 子查询进行排序
SELECT JSON_ARRAYAGG(name) AS employees_names
FROM (SELECT nameFROM employeesORDER BY salary DESC
) AS sorted_employees;
截图如下:
2.2.2 创建临时表
-- 创建临时表
CREATE TEMPORARY TABLE sorted_employees AS
SELECT name
FROM employees
ORDER BY salary DESC;-- 对临时表进行聚合
SELECT JSON_ARRAYAGG(name) AS employees_names
FROM sorted_employees;-- 删除临时表
DROP TEMPORARY TABLE sorted_employees;
截图如下:
2.3 带条件聚合
SELECT JSON_ARRAYAGG(name) AS engineering_employees
FROM employees
WHERE department = 'Engineering';
截图如下:
2.4 多列聚合
将员工的 name 和 salary 作为对象聚合为一个 JSON 数组
SELECT JSON_ARRAYAGG(JSON_OBJECT('name', name, 'salary', salary)) AS employees_info
FROM employees;#输出如下:
{"employees_info": [{"name": "Alice", "salary": 60000},{"name": "Bob", "salary": 70000},{"name": "Charlie", "salary": 50000},{"name": "David", "salary": 80000},{"name": "Eve", "salary": 55000}
]}
截图如下:
2.5 嵌套 JSON 结构
按部门聚合员工信息:
SELECT department, JSON_ARRAYAGG(JSON_OBJECT('name', name, 'salary', salary)) AS employees
FROM employees
GROUP BY department;# 输出如下
[{"department": "Engineering","employees": [{"name": "Alice", "salary": 60000},{"name": "Bob", "salary": 70000},{"name": "David", "salary": 80000}]},{"department": "HR","employees": [{"name": "Charlie", "salary": 50000},{"name": "Eve", "salary": 55000}]}
]
截图如下: