在PostgreSQL的使用过程中发现了一个很有意思的功能，就是对于需要类似于树状结构的结果可以使用递归查询实现。比如说我们常用的公司部门这种数据结构，一般我们设计表结构的时候都是类似下面的SQL，其中parent_id为NULL时表示顶级节点，否则表示上级节点ID。
CREATE TABLE DEPARTMENT (
ID INTEGER PRIMARY KEY,
NAME VARCHAR(32),
PARENT_ID INTEGER REFERENCES DEPARTMENT(ID)
);
下面我们造几条测试数据
INSERT INTO DEPARTMENT(ID, NAME, PARENT_ID) VALUES(1, 'DEPARTMENT_1', NULL);
INSERT INTO DEPARTMENT(ID, NAME, PARENT_ID) VALUES(11, 'DEPARTMENT_11', 1);
INSERT INTO DEPARTMENT(ID, NAME, PARENT_ID) VALUES(12, 'DEPARTMENT_12', 1);
INSERT INTO DEPARTMENT(ID, NAME, PARENT_ID) VALUES(111, 'DEPARTMENT_111', 11);
INSERT INTO DEPARTMENT(ID, NAME, PARENT_ID) VALUES(121, 'DEPARTMENT_121', 12);
INSERT INTO DEPARTMENT(ID, NAME, PARENT_ID) VALUES(122, 'DEPARTMENT_122', 12);
其中
- DEPARTMENT_1是顶级节点，它有两个子节点DEPARTMENT_11和DEPARTMENT_12。
- DEPARTMENT_11节点又有一个子节点DEPARTMENT_111。
- DEPARTMENT_12节点有两个子节点DEPARTMENT_121和DEPARTMENT_122。
下面是递归查询生成树状结构查询语句
WITH RECURSIVE T (ID, NAME, PARENT_ID, PATH, DEPTH) AS (
SELECT ID, NAME, PARENT_ID, ARRAY[ID] AS PATH, 1 AS DEPTH
FROM DEPARTMENT
WHERE PARENT_ID IS NULL
UNION ALL
SELECT D.ID, D.NAME, D.PARENT_ID, T.PATH || D.ID, T.DEPTH + 1 AS DEPTH
FROM DEPARTMENT D
JOIN T ON D.PARENT_ID = T.ID
)
SELECT ID, NAME, PARENT_ID, PATH, DEPTH FROM T
ORDER BY PATH;
ID NAME PARENT_ID PATH DEPTH
1 DEPARTMENT_1 1 1
11 DEPARTMENT_11 1 1,11 2
111 DEPARTMENT_111 11 1,11,111 3
12 DEPARTMENT_12 1 1,12 2
121 DEPARTMENT_121 12 1,12,121 3
122 DEPARTMENT_122 12 1,12,122 3
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