Mysql实现递归树查询的使用示例

WITH RECURSIVE hierarchy AS (
	-- 基础情况
	...
	UNION ALL 
	-- 递归情况
	...
)
select xxx from hierarchy

DROP TABLE IF EXISTS employees;
CREATE TABLE employees (
    id INT PRIMARY KEY AUTO_INCREMENT,
    name VARCHAR(100),
    manager_id INT,
    FOREIGN KEY (manager_id) REFERENCES employees(id)
);
-- 新增根节点数据
INSERT INTO employees (name) VALUES ('CEO');
-- 创建执行过程
DELIMITER $$
CREATE PROCEDURE InsertLargeAmountOfData()
BEGIN
    DECLARE id INT DEFAULT 2;
    DECLARE parentId INT DEFAULT 1;
    DECLARE counter INT DEFAULT 0;
    -- 第2层
    WHILE counter < 10 DO
        INSERT INTO employees (name, manager_id) VALUES (CONCAT('L2-', id), 1);
        SET id = id + 1;
        SET counter = counter + 1;
    END WHILE;
    SET counter = 0;
    -- 第3层
    WHILE counter < 100 DO
        SET parentId = 1 + FLOOR(counter / 10) + 1;
        INSERT INTO employees (name, manager_id) VALUES (CONCAT('L3-', id), parentId);
        SET id = id + 1;
        SET counter = counter + 1;
    END WHILE;
    SET counter = 0;
    -- 第4层
    WHILE counter < 1000 DO
        SET parentId = 11 + FLOOR(counter / 100);
        INSERT INTO employees (name, manager_id) VALUES (CONCAT('L4-', id), parentId);
        SET id = id + 1;
        SET counter = counter + 1;
    END WHILE;
    SET counter = 0;
    -- 第5层
    WHILE counter < 10000 DO
        SET parentId = 111 + FLOOR(counter / 1000);
        INSERT INTO employees (name, manager_id) VALUES (CONCAT('L5-', id), parentId);
        SET id = id + 1;
        SET counter = counter + 1;
    END WHILE;
    SET counter = 0;
    -- 第6层
    WHILE counter < 88889 DO
        SET parentId = 1111 + FLOOR(counter / 10000);
        INSERT INTO employees (name, manager_id) VALUES (CONCAT('L6-', id), parentId);
        SET id = id + 1;
        SET counter = counter + 1;
    END WHILE;
END$$
DELIMITER ;
-- 执行执行过程
CALL InsertLargeAmountOfData();

WITH RECURSIVE hierarchy AS (
	-- 基础情况
	select id,name,manager_id from employees where manager_id is null
	UNION ALL 
	-- 递归情况
	select e.* from employees as e  inner join hierarchy as h on e.manager_id = h.id
)
select id,name,manager_id from hierarchy

WITH RECURSIVE hierarchy AS (
    -- 基础情况：从CEO开始
    SELECT id, name, manager_id, CAST(name AS CHAR(255)) AS hierarchy_path
    FROM employees
    WHERE manager_id IS NULL 
    UNION ALL
    -- 递归情况：为每个下级员工添加上级
    SELECT e.id, e.name, e.manager_id, CONCAT(h.hierarchy_path, ' > ', e.name)
    FROM employees e
    JOIN hierarchy h ON e.manager_id = h.id
)
SELECT id, name, manager_id, hierarchy_path 
FROM hierarchy
-- 查询某条数据条件
WHERE name = 'L3-111'
ORDER BY id;