PostgreSQL遞歸查詢實現樹狀結構查詢

在Postgresql的使用過程中發現了一個非常有意思的功能，就是對於須要相似於樹狀結構的結果能夠使用遞歸查詢實現。比方說我們經常使用的公司部門這樣的數據結構。一般我們設計表結構的時候都是相似以下的SQL，當中parent_id為NULL時表示頂級節點，否則表示上級節點ID。

CREATE TABLE DEPARTMENT ( ID INTEGER PRIMARY KEY, NAME VARCHAR(32), PARENT_ID INTEGER REFERENCES DEPARTMENT(ID) );

以下我們造幾條測試數據

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);

當中
- DEPARTMENT_1是頂級節點。它有兩個子節點​DEPARTMENT_11和​DEPARTMENT_12。
- DEPARTMENT_11節點又有一個子節點​DEPARTMENT_111。
​- DEPARTMENT_12節點有兩個子節點​DEPARTMENT_121和​DEPARTMENT_122。

​

以下是遞歸查詢生成樹狀結構查詢語句

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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